Is it possible for Javascript to "target" divs through Ajax requests?

Question

Is it possible for Javascript to "target" divs through Ajax requests?

I am facing a challenge with my main program that includes multiple Javascript routines, one of which involves Ajax. The issue I am encountering is related to modifying the attributes of returned html elements within the calling page using another Javascript routine. Despite running several tests, I have been unsuccessful in getting it to work as intended.

Update

The following script is included in the head section:

    <script type="text/javascript">
        [Javascript code goes here]
    </script>

The Ajax call is initiated using the following script:

    <script type="text/javascript">;
    setTimeout(function() {Ajax();}, 10000);
    </script>;
    <div id="ReloadThis">Default text</div>;

Upon making the Ajax call, the PHP page returns HTML code similar to this:

echo "<ul id=\"gallery\" style=\"height:325;\">";
echo "<li><img src=\"Image1.jpg\" alt=\"\" /></li>";
echo "<li><img src=\"Image2.jpeg\" alt=\"\" /></li>";
echo "</ul>";

While the fader code works autonomously when used separately, it fails to function properly in conjunction with Ajax calls. This discrepancy has proven challenging for me to resolve.

javascript ajax

Answer 1

Answer №1

Do you want to explore the possibility of making an Ajax request that returns HTML content? Once the Ajax call is completed, you can then manipulate the HTML and potentially display it on the website.

let ajaxRequest = $.ajax({
    url: '/example/html/',
    data: {
        html :'<div id="dog">Hello, my name is <span class="name">Sam</span></div>'
    },
    type: 'post'
}).always(function(response){
    let element = $(response);
    let name = element.find('.name');
    name.text('Max');
    element.appendTo(document.body);
})

http://jsfiddle.net/abc123xyz/

Answer 2

Do you want to explore the possibility of making an Ajax request that returns HTML content? Once the Ajax call is completed, you can then manipulate the HTML and potentially display it on the website.

let ajaxRequest = $.ajax({
    url: '/example/html/',
    data: {
        html :'<div id="dog">Hello, my name is <span class="name">Sam</span></div>'
    },
    type: 'post'
}).always(function(response){
    let element = $(response);
    let name = element.find('.name');
    name.text('Max');
    element.appendTo(document.body);
})

http://jsfiddle.net/abc123xyz/

Answer 3

Answer №2

Challenge #1

Upon receiving the HTML in the Ajax() function, you swap out your current gallery with it, hoping that the crossfade() function won't encounter any issues. Unfortunately, the new images are not present in the galleryImages variable, causing the crossfade() function to overlook them.

To rectify this, you must execute another fadeInit() to properly organize all these elements.

Include this call within your Ajax function:

document.getElementById('ReloadThis').innerHTML = $http.responseText;
setTimeout(function(){$self();}, 10000);
fadeInit();    /* register the new elements */

Challenge #2

The slideshow is now running, but it's gaining speed! This acceleration occurs because each update keeps the crossfade() function continuously looping through all previous picture sets. To resolve this issue and start fresh whenever necessary, store the timeout in a variable that can be cleared when needed.

/* At the beginning of your JS code */
var crossTimer;

/* Within your crossfade() function */
crossTimer = window.setTimeout("crossfade("+opacity+")", 30);

Simply clear it within your Ajax function

document.getElementById('ReloadThis').innerHTML = $http.responseText;
setTimeout(function(){$self();}, 10000);
if (typeof crossTimer != 'undefined') clearTimeout(crossTimer);
fadeInit();

JS Fiddle Demo

Answer 4

Challenge #1

Upon receiving the HTML in the Ajax() function, you swap out your current gallery with it, hoping that the crossfade() function won't encounter any issues. Unfortunately, the new images are not present in the galleryImages variable, causing the crossfade() function to overlook them.

To rectify this, you must execute another fadeInit() to properly organize all these elements.

Include this call within your Ajax function:

document.getElementById('ReloadThis').innerHTML = $http.responseText;
setTimeout(function(){$self();}, 10000);
fadeInit();    /* register the new elements */

Challenge #2

The slideshow is now running, but it's gaining speed! This acceleration occurs because each update keeps the crossfade() function continuously looping through all previous picture sets. To resolve this issue and start fresh whenever necessary, store the timeout in a variable that can be cleared when needed.

/* At the beginning of your JS code */
var crossTimer;

/* Within your crossfade() function */
crossTimer = window.setTimeout("crossfade("+opacity+")", 30);

Simply clear it within your Ajax function

document.getElementById('ReloadThis').innerHTML = $http.responseText;
setTimeout(function(){$self();}, 10000);
if (typeof crossTimer != 'undefined') clearTimeout(crossTimer);
fadeInit();

Is it possible for Javascript to "target" divs through Ajax requests?

Update

Answer №1

Answer №2

JS Fiddle Demo

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