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Back from ApacheCon North America 2016

May 18, 2016 in Apache Cordova, JavaScript, mobile

I just get back from ApacheCon North America that was be held from 11 May to 13 May @Vancouver, Canada. The conference organization was really great and there were a lot of attendees in the conference sessions.

HazemSaleh_ApacheCon2016

I had the chance to present “Advanced Apache Cordova” in 12 May, the session was interactive, contained three live demos and I was really amazed by the energy, enthusiasm, and responsiveness of the attendees during the session.

My session is uploaded below.

Finally, I would like to thank all the organizers of ApacheCon conference for making the conference looks so great.

iOS 9, Cordova, and jQuery mobile apps Integration tips

November 16, 2015 in Apache Cordova, iOS, JavaScript, jQuery Mobile

Unfortunately after having iOS 9 update, you may find your Cordova jQuery mobile app has the following problems:

  1. All of your app’s popups are suddenly closed after the first opening time!
  2. All your app’s back button are not working anymore!

The root cause of these problems is a bug in iOS9 UIWebView‘s window.location.hash as shown in the link below:
https://openradar.appspot.com/22186109

Fixes

In order to fix these issues, you can do the following workarounds.

Fixing the sudden closing pop-up

In order to fix this issue, make sure to set the data-history attribute of jQuery mobile popup to "false". An example of this is shown in the link below:
https://github.com/hazems/cordova-mega-app/blob/master/www/index.html

Fixing the broken back button

In order to fix this issue, make sure to disable the hash listening behaviour for jQuery mobile when the device is ready. You need to be careful when applying this fix, make sure that this fix will be applied for only iOS version 9 as shown in the code snippet below.

if (device.platform === "iOS"  && parseInt(device.version) === 9) {
  $.mobile.hashListeningEnabled=false;
} /* when the device is ready */

Checking a working Cordova jQuery mobile app in iOS 9

Finally, you can check a complete working Cordova jQuery mobile app in iOS 9. This app is one of the examples of the “JavaScript Mobile Application Development” book which can be reached at:
http://www.amazon.com/JavaScript-Native-Mobile-Apps-Development/dp/1783554177/
https://www.packtpub.com/web-development/javascript-native-mobile-apps-development

The example app uses Apache Cordova 5.3.3 and jQM 1.4.5. Enjoy!

[JavaScript Quiz #16] Smallest number of flights

August 15, 2015 in JavaScript, Quiz

Consider you have a set of places (CAI, FRA, MAD, NYC, SFO) and there are available flights between some of these destinations as shown below.

CAI -> FRA
FRA -> MAD
MAD -> NYC
CAI -> MAD
NYC -> SFO
FRA -> CAI
MAD -> FRA

Design and develop a JavaScript function that can get the shortest possible path from a destination X to destination Y if there is any.
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In order to develop a JavaScript function which can get the shortest path between two destinations, first of all, we need to think about the proper data structure which can represent these places and the connections between them. If we think about it, we will find that these placed can be represented as nodes and the connections between these places can be represented as edges which is typically a Graph. We can represent this Graph as a 2D N X N array as shown below.

    CAI FRA MAD NYC SFO
CAI  0   1   1   0   0
FRA  1   0   1   0   0
MAD  0   1   0   1   0
NYC  0   0   0   0   1
SFO  0   0   0   0   0

The previous matrix can be translated to the following JavaScript code.

var mapping = {
    "CAI": 0,
    "FRA": 1,
    "MAD": 2,
    "NYC": 3,
    "SFO": 4
};

var routeGraph = 
[
    [0, 1, 1, 0, 0],
    [1, 0, 1, 0, 0],
    [0, 1, 0, 1, 0],
    [0, 0, 0, 0, 1],
    [0, 0, 0, 0, 0]    
];

After make the proper representation, we need to think about how to get the shortest path between any two nodes in this unweighted graph. Since this graph is unweighted, we can use directly BFS (Breath First Search) in order to get the shortest path. For more information about BFS, check: https://en.wikipedia.org/wiki/Breadth-first_search.

In order to implement BFS, there are some notes for our implementation below:

  1. We utilized the Queue data structure for storing the nodes in a first-in, first-out (FIFO) style, we explore the oldest unexplored nodes first. Thus our explorations radiate out slowly from the starting node, defining a breadth-first search.
  2. In order to keep track of the shortest path, we have a JavaScript map nodeParent object which stores the first visited parent of the current visited node.
  3. Finally, when the search finds the destination then we construct the final path from the nodeParent object and stores it in the finalPath array.

The code below shows our BFS implementation in order to get the shortest path between the start and destination nodes.

var ShortestPathFinder = function() {
}

ShortestPathFinder.find = function (start, destination) {
    var visited = {}; // A map containing all of the visited nodes ...
    var nodeParent = {}; // An map containing the node parents ...
    var queue = [start], i, found = false, finalPath = [], neighbours = [];    
    
    nodeParent[start] = null;
    
    var current = queue.shift();
    
    while ((neighbours = ShortestPathFinder.getNeighbours(current)) != null) {
    
        for (i = 0; i < neighbours.length; ++i) {
            
            // If we find the destination then exit these loops ...
            if (neighbours[i] == destination) {
                found = true;
                nodeParent[neighbours[i]] = current;
                break;
            }
        
            if (! visited[neighbours[i]]) {
                
                // Mark this node as "visited" ...
                visited[neighbours[i]] = true;
            
                // Add element to the queue ...
                queue.push(neighbours[i]);
                nodeParent[neighbours[i]] = current;
            }
        }
        
        if (queue.length == 0) {
            break;
        }
        
        current = queue.shift();
    }
    
    if (! found) {
        console.error("No path is found between " + start + " and " + destination);
        
        return null;
    }
    
    // Construct the final path from the node parent map ...
    current = destination;
    
    finalPath.push(destination);
    
    while (nodeParent[current] != start) {
        finalPath.push(nodeParent[current]);
        
        current = nodeParent[current];
    }
    
    finalPath.push(start);    
    
    return finalPath.reverse();
}

In order to get the neighbours of the current node, a simple function is implemented to get the neighbours from the routeGraph matrix as follows.

ShortestPathFinder.getNeighbours = function(current) {
    var currentIndex = mapping[current], i, result = [];
    
    for (i = 0; i < routeGraph[currentIndex].length; ++i) {
        if (routeGraph[currentIndex][i] == 1) {
            result.push(mapping.getKey(i));
        }
    }
    
    return result;
}

Finally, we can test our developed JavaScript function by having the following test cases.

console.log(ShortestPathFinder.find("CAI", "NYC"));
console.log(ShortestPathFinder.find("CAI", "SFO"));
console.log(ShortestPathFinder.find("FRA", "NYC"));
console.log(ShortestPathFinder.find("SFO", "CAI"));
console.log(ShortestPathFinder.find("MAD", "CAI"));

Below is the output of the console.

["CAI", "MAD", "NYC"]
["CAI", "MAD", "NYC", "SFO"]
["FRA", "MAD", "NYC"]
No path is found between SFO and CAI
null
["MAD", "FRA", "CAI"]

Attached below, the complete solution code for your reference, if you have comments or a better solution, feel free to comment below.

/*
CAI -> FRA
FRA -> MAD
MAD -> NYC
CAI -> MAD
NYC -> SFO
FRA -> CAI
MAD -> FRA

    CAI FRA MAD NYC SFO
CAI  0   1   1   0   0
FRA  1   0   1   0   0
MAD  0   1   0   1   0
NYC  0   0   0   0   1
SFO  0   0   0   0   0
*/

var mapping = {
    "CAI": 0,
    "FRA": 1,
    "MAD": 2,
    "NYC": 3,
    "SFO": 4
};

var routeGraph = 
[
    [0, 1, 1, 0, 0],
    [1, 0, 1, 0, 0],
    [0, 1, 0, 1, 0],
    [0, 0, 0, 0, 1],
    [0, 0, 0, 0, 0]    
];

Object.prototype.getKey = function(value){
  for(var key in this){
    if(this[key] == value){
      return key;
    }
  }
  return null;
};


var ShortestPathFinder = function() {
}

ShortestPathFinder.find = function (start, destination) {
    var visited = {}; // A map containing all of the visited nodes ...
    var nodeParent = {}; // An map containing the node parents ...
    var queue = [start], i, found = false, finalPath = [], neighbours = [];    
    
    nodeParent[start] = null;
    
    var current = queue.shift();
    
    while ((neighbours = ShortestPathFinder.getNeighbours(current)) != null) {
    
        for (i = 0; i < neighbours.length; ++i) {
            
            // If we find the destination then exit these loops ...
            if (neighbours[i] == destination) {
                found = true;
                nodeParent[neighbours[i]] = current;
                break;
            }
        
            if (! visited[neighbours[i]]) {
                
                // Mark this node as "visited" ...
                visited[neighbours[i]] = true;
            
                // Add element to the queue ...
                queue.push(neighbours[i]);
                nodeParent[neighbours[i]] = current;
            }
        }
        
        if (queue.length == 0) {
            break;
        }
        
        current = queue.shift();
    }
    
    if (! found) {
        console.error("No path is found between " + start + " and " + destination);
        
        return null;
    }
    
    // Construct the final path from the node parent map ...
    current = destination;
    
    finalPath.push(destination);
    
    while (nodeParent[current] != start) {
        finalPath.push(nodeParent[current]);
        
        current = nodeParent[current];
    }
    
    finalPath.push(start);    
    
    return finalPath.reverse();
}

ShortestPathFinder.getNeighbours = function(current) {
    var currentIndex = mapping[current], i, result = [];
    
    for (i = 0; i < routeGraph[currentIndex].length; ++i) {
        if (routeGraph[currentIndex][i] == 1) {
            result.push(mapping.getKey(i));
        }
    }
    
    return result;
}

console.log(ShortestPathFinder.find("CAI", "NYC"));
console.log(ShortestPathFinder.find("CAI", "SFO"));
console.log(ShortestPathFinder.find("FRA", "NYC"));
console.log(ShortestPathFinder.find("SFO", "CAI"));
console.log(ShortestPathFinder.find("MAD", "CAI"));

Creating Cordova jQuery Mobile Apps Rapidly

August 9, 2015 in Apache Cordova, JavaScript, jQuery Mobile

Cordova jQuery npm plugin allows you to add jQuery mobile’s ready-made templates to your existing Apache Cordova app using a neat easy-to-use CLI.

You can install Cordova jQuery npm plugin using the following npm command:
npm install -g cordova-jquery

Once you install it, you can start using it by executing cordova-jquery command from your Apache Cordova app’s root directory. The next six videos shows you how to create six Cordova jQuery mobile apps with different templates (Multipages, Header Navigation Bar, Persistent Navigation Bar, External Panel, Accordion, and ListView).

As shown in the videos below, after executing cordova-jquery command, all what you need to do is to choose the template you would like to apply to your existing Cordova app. All of the examples below work on Apache Cordova version 5.1.1 (The latest Cordova release until the moment).

Cordova jQuery plugin supports the following templates.

  1. Multiple Pages
    This template will apply a simple two page jQuery mobile template to your index.html file.
  2. Header Navbar
    This template will apply a three page jQuery mobile template to your index.html file that includes a header navbar for the header on each page.
  3. Persistent Navbar
    This template will apply a three page jQuery mobile template to your index.html file that includes a persistent navbar for the footer on each page.
  4. External Panel
    This template will apply an external panel to jQuery mobile. When this option is selected you will also be prompted for additional information such as which side of the page you’d like the panel to open on (left, right) and which reveal style would you like (reveal, push, overlay)
  5. Accordion
    This template will apply three sections’ jQuery mobile accordion template to your index.html file. Your existing content will be placed as part of the first section. It is important to note that you have to make sure that your existing content parent element does not use CSS “absolute” position in order to be controlled by the accordion section.
  6. List View
    This template will apply a jQuery mobile list view template to your index.html file. The first item of the list view points to your existing content page, and the second and the third items of the list view point to two placeholder pages.

The main objective of this plugin is to facilitate the usage of the jQuery Mobile in Apache Cordova, so feel free to use it and submit any issues you are facing to:
https://github.com/Open-I-Beam/cordova-jquery-npm/issues

Related Plugin Videos:

My Talk about Automated Jasmine 2.x tests in GeeCON

July 29, 2015 in JavaScript

Finally, my Talk about Automated Jasmine 2.x tests in GeeCON Poland is now published, I hope that you will find it useful.

[JavaScript Quiz #15] All possible compositions of a number

July 4, 2015 in JavaScript

Today’s JavaScript quiz is about finding all the possible compositions of a number. For example if we have a number 4, we would like to find all the possible compositions that sums up to 4 as follows.

1111
112
121
13
211
22
31
4

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In order to develop this utility, it is important to understand its nature. For a number n, it has the following possible compositions:
n (for a composition length of 1)
1 n-1 (for a composition length of 2)
1 1 n-2 (for a composition of length 3)

1 1 1 … 1 (for a composition of length n)

This can be handled using recursive function as follows.

function compositions(n, temp, output) {
    var i, newTemp;
    
    if (n == 0) {
        output.push(temp);
    } else {
        for (i = 1; i <= n; ++i) {
            newTemp = temp + i;
            
            compositions(n - i, newTemp, output);
        }
    }
}

As shown, the base is if n is equal to 0 then we add the temp string (which is initialized to "") to the output list, else we subtract i from n and adds i to the temp string. The following function getAllCompositions calls compositions with the initial values.

function getAllCompositions(n) {
    var out = [];
    
    compositions(n, "", out);
    
    return out;
}

Finally, we can test getAllCompositions as follows.

// Test ...
var num = 4;
var out = getAllCompositions(num), i;

console.log("Compositions number for (" + num + ") = " + out.length);
for (i = 0; i < out.length; ++i) {
	console.log(out[i]);
}

The output will be:

Compositions number for (4) = 8
    
1111 
112 
121  
13 
211  
22  
31  
4

If you have a better solution, feel free to put in the comments below. The current solution complexity is n!.

[JavaScript Quiz #14] Number Text Representation

May 11, 2015 in JavaScript

Today’s JavaScript quiz is about creating a Number to String utility. Assume that we have the following numbers as an input to our utility:

23
1999
199999
1000000999

Our JavaScript utility should output the following results:

twenty three
one thousand nine hundreds ninety nine
one hundred ninety nine thousand nine hundred ninety nine
one billion nine hundred ninety nine

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In order to develop this utility, it is important to break-down this problem into small sub-problems and then solve each of them individually. For example, if we wish to print numbers from 0 to 9, this is an easy Job, just create an array that contains the numbers from 0 to 9 and call it digits as follows.

digits = ["zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"];

If we want to print numbers from 10 to 19, we can create another array that represents teens as follows:

teens = ["ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"];

Then if we want to print numbers from 20 up to 99, we can create another array that represents tys as follows.

tys = ["twenty", "thirty", "fourty", "fifty", "sixty", "seventy", "eighty", "ninety"];

But for numbers from 20 to 99, how can we utilize the previous array for printing their values, if we think little about it, we will find that a number like 39 is about 30 + 9 which will produce at the end "thirty nine". But how to divide a number like 39 to 30 + 9 and get its corresponding text representation, simply this can be done by performing three steps:

  1. Dividing 39 (or generally numbers from 20 to 99) by 10 (Integer Division), and then subtract 2 from the result number in order to pick its text representation "thirty" from its corresponding array (tys array).
  2. Having the mod of 39 (or generally numbers from 20 to 99) and 10 which will give us 9, and then we can get simply pick its text representation "nine" from its corresponding array (digits array).
  3. Finally, augment the two text representations to get the final number text representation which is “thirty nine”.

For numbers from 100 up to 999, For example: 205, we can divide it 205 by 100 to know how many hundred units does it has (2 hundreds) and then have the reminder of this number 205 with 100 which is 5 and “five” representation can be got simply using the previous mentioned procedures. Let’s look into the code which can get the text representation of numbers from 0 to 999.

NumberReader.prototype.readThreeDigitNumber = function(number) { /* 0 ... 999 */
    if (number == 0) {
        return "zero";
    }
    
    var output = "", result, reminder;
    
    if (number > 99) {
        result  = Math.floor(number / 100);
        number = number % 100;
        
        output += this.digits[result] + " hundred";
        
        if (number == 0) {
            return output;
        }
    }
    
    if (number < 10) {
        output = NumberReader.appendToOutput(output, this.digits[number]);    
    } else if (number < 20) {
        output = NumberReader.appendToOutput(output, this.teens[number - 10]);                
    } else {
        result = Math.floor(number / 10);
        reminder = number % 10;
        
        output = NumberReader.appendToOutput(output, this.tys[result - 2]);
        
        if (reminder > 0) {
            output = NumberReader.appendToOutput(output, this.digits[reminder]);
        }
    }
    
    return output;
}

We can apply the same concept with thousands by diving them by 1000 to know the thousands units and having mod with 1000 which can be resolved using the previous procedure. Thankfully, we can apply the same concept with millions and billions as shown below.

NumberReader.prototype.readNumber = function (number) {
    var output = "", result, reminder;
    
    if (number >= 1e9) {
        result = Math.floor(number / 1000000000);
        reminder = number % 1000000000;
        
        output += this.readNumber(result) + " billion " + ((reminder > 0) ? this.readNumber(reminder) : "");
    } else if (number >= 1e6 && number < 1e9) {
        result = Math.floor(number / 1000000);
        reminder = number % 1000000;
        
        output += this.readNumber(result) + " million " + ((reminder > 0) ? this.readNumber(reminder) : "");
    } else if (number >= 1000 && number < 1e6) {
        result = Math.floor(number / 1000);
        reminder = number % 1000;
        
        output += this.readNumber(result) + " thousand " + ((reminder > 0) ? this.readNumber(reminder) : "");
    } else {
        output += this.readThreeDigitNumber(number);
    }
    
    return output;
}

The complete quiz code is shown below.

var NumberReader = function() {
    this.digits = ["zero", "one", "two", "three", "four", "five",
        "six", "seven", "eight", "nine"];

    this.teens = ["ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"];
    
    this.tys = ["twenty", "thirty", "fourty", "fifty", "sixty", "seventy", "eighty", "ninety"];    
}

NumberReader.prototype.readThreeDigitNumber = function(number) { /* 0 ... 999 */
    if (number == 0) {
        return "zero";
    }
    
    var output = "", result, reminder;
    
    if (number > 99) {
        result  = Math.floor(number / 100);
        number = number % 100;
        
        output += this.digits[result] + " hundred";
        
        if (number == 0) {
            return output;
        }
    }
    
    if (number < 10) {
        output = NumberReader.appendToOutput(output, this.digits[number]);    
    } else if (number < 20) {
        output = NumberReader.appendToOutput(output, this.teens[number - 10]);                
    } else {
        result = Math.floor(number / 10);
        reminder = number % 10;
        
        output = NumberReader.appendToOutput(output, this.tys[result - 2]);
        
        if (reminder > 0) {
            output = NumberReader.appendToOutput(output, this.digits[reminder]);
        }
    }
    
    return output;
}

NumberReader.prototype.readNumber = function (number) {
    var output = "", result, reminder;
    
    if (number >= 1e9) {
        result = Math.floor(number / 1000000000);
        reminder = number % 1000000000;
        
        output += this.readNumber(result) + " billion " + ((reminder > 0) ? this.readNumber(reminder) : "");
    } else if (number >= 1e6 && number < 1e9) {
        result = Math.floor(number / 1000000);
        reminder = number % 1000000;
        
        output += this.readNumber(result) + " million " + ((reminder > 0) ? this.readNumber(reminder) : "");
    } else if (number >= 1000 && number < 1e6) {
        result = Math.floor(number / 1000);
        reminder = number % 1000;
        
        output += this.readNumber(result) + " thousand " + ((reminder > 0) ? this.readNumber(reminder) : "");
    } else {
        output += this.readThreeDigitNumber(number);
    }
    
    return output;
}

NumberReader.appendToOutput = function (output, parameter) {
    if (output == "") {
        output += parameter;
    } else {
        output += " " + parameter;
    }
    
    return output;
}

And this is a test code for NumberReader.

// Test our API ...
var numberReader = new NumberReader();

console.log(numberReader.readNumber(1999999999));
console.log(numberReader.readNumber(1000000999));    
console.log(numberReader.readNumber(100000999));        
console.log(numberReader.readNumber(1000999));
console.log(numberReader.readNumber(199999));
console.log(numberReader.readNumber(100000));
console.log(numberReader.readNumber(1999));    
console.log(numberReader.readNumber(9));        
console.log(numberReader.readNumber(999));        
console.log(numberReader.readNumber(193));
console.log(numberReader.readNumber(23));        
console.log(numberReader.readNumber(333));

The output will be:

one million nine hundred ninety nine
one hundred ninety nine thousand nine hundred ninety nine
one hundred thousand 
one thousand nine hundred ninety nine
nine
nine hundred ninety nine
one hundred ninety three
twenty three
three hundred thirty three

[JavaScript Quiz #13] Finding longest common substring of strings

April 25, 2015 in JavaScript

Finding the longest common substring of strings is one of the interesting problems. Assume that we have two JavaScript strings like “ababccd” and “abccw”, Can we write a JavaScript utility function that can find the common substrings of these two strings which is “abcc” in this case.
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One of the possible solutions to this problem is to create a 2D comparison array (of size [First string length][Second string length]) which will hold the comparisons between every character in the first and the second strings. If the character of the first string does not match the second string character then set the array cell to 0.
If we find the first string current character matches the second string current character then set the corresponding array cell to 1 but take care if the upper left diagonal cell of the current cell in the comparison array is greater than 0 then you need set the current cell value to the upper left diagonal cell value + 1 (Upper left diagonal cell refers to the previous character of both strings).

Coding this is very simple and the complexity of this code is O(length of the first string * length of the second string) as follows.

var StringUtils = function() {
}

StringUtils.findLongestCommonSubstring = function(string1, string2) {
	var comparsions = []; //2D array for the char comparsions ...
	var maxSubStrLength = 0;
	var lastMaxSubStrIndex = -1, i, j, char1, char2, startIndex;

	for (i = 0; i < string1.length; ++i) {
		comparsions[i] = new Array();

		for (j = 0; j < string2.length; ++j) {
			char1 = string1.charAt(i);
			char2 = string2.charAt(j);

			if (char1 === char2) {
				if (i > 0 && j > 0) {
					comparsions[i][j] = comparsions[i - 1][j - 1] + 1;
				} else {
					comparsions[i][j] = 1;
				}
			} else {
				comparsions[i][j] = 0;
			}

			if (comparsions[i][j] > maxSubStrLength) {
				maxSubStrLength = comparsions[i][j];
				lastMaxSubStrIndex = i;
			}
		}
	}

	if (maxSubStrLength > 0) {
		startIndex = lastMaxSubStrIndex - maxSubStrLength + 1;

		return string1.substr(startIndex, maxSubStrLength);
	}

	return null;
}

// Test code
console.log(StringUtils.findLongestCommonSubstring("ababccd", "abccx"));
console.log(StringUtils.findLongestCommonSubstring("ababccd", "ccxaba"));
console.log(StringUtils.findLongestCommonSubstring("becooltopeople", "topeoplebecool"));    
console.log(StringUtils.findLongestCommonSubstring("ababccd", "zzzz")); 

As shown in the implementation, we store two main items, the maximum common substring length and its last index and thankfully using JavaScript substr function, we can get the desired common substring string. The output will be as follows.

abcc
aba
topeople
null

Finally, note that if you have two longest substrings then this code will get the first of them.

Back from ApacheCon North America 2015

April 24, 2015 in Apache Cordova, Conference, JavaScript

I just get back from ApacheCon North America that was be held from 13 April to 16 April @Austin, Texas, USA. The conference organization was really great and there were a lot of attendees in the conference sessions.
IMG_8838

I had the chance to present “Apache Cordova In Action” in 13 April, the session had many attendees and I was really amazed by the energy, enthusiasm, and responsiveness of the attendees during the session.
Feedback

The session went great and I was glad to give a free copy of my JavaScript Mobile Application Development book to one of the attendees who answered a JavaScript quiz at the end of my session.
Books

I uploaded my session below:

Finally, I would like to thank all the organizers of ApacheCon conference for making the conference looks so great.
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Speaking in ApacheCon North America 2015

March 21, 2015 in Apache Cordova, Conference, JavaScript

Screen Shot 2015-03-21 at 1.15.10 AM
@Monday, April 13 10:45 AM, I will be speaking in ApacheCon North America conference (that will be held from 13 April to 17 April @Austin, USA) about Apache Cordova under the session title “Apache Cordova In Action”. My session will be a practical one, it will discuss why there is a need for Hybrid mobile development, the current challenges of mobile development, and how using Apache Cordova can help in overcoming many of these technical challenges. It also highlights the best practices of using Apache Cordova with jQuery mobile. then, it demonstrates a real Cordova mobile app that works on three mobile platforms (Android, iOS, and Windows Phone 8.1).

Finally, I hope it will be an interesting session for all the mobile development passionate :):
https://apacheconna2015.sched.org/speaker/hazems

Personally, it is my first time to visit Austin, beside enjoying technical stuff, I would like to visit some tourist places there, any suggestions are welcome :)?

I really wish to see all of you there in ApacheCon North America 2015!

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