jQuery Ajax Request - Lose Method - ajax

I have a page index.php that uses a modal to upload files. After those have uploaded I use the following to update my database and load in the new images to a list.
$('#sortableImages').load('../includes/sortImages.php?edit=' + edit);
Executes:
<script type="text/javascript">
$(document).ready(function(){
$(function() {
$("#sortableImages ul").sortable({
opacity: 0.6, cursor: 'move', update: function() {
var order = $(this).sortable("serialize") + '&action=updateRecordsListings';
$.post("../albumUploader/queries/sort.php", order);
}
});
});
});
</script>
echo "<ul class='revisionList'>";
while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$sortImageName = $row['OrgImageName'];
$sortPath = "../data/gallery/" . $getGalleryID . "/images/album/" . $sortImageName;
echo "<li class='sortPhotos' id='recordsArray_{$row['id']}' >";
echo '<img src="'. $sortPath .'"/>';
echo "</li>";
}
echo "</ul>";
The images populate in a div #sortableImages on the index page. However it seems that I lose my method of sortable() from the js file that was originally loaded in the index.php or after the ajax request it's not reading the js. What am I missing here?
Thanks a million.

When you load script from a remote page using ajax, it is important to realize that the ready event has already occured in page you are loading into.
This means that code wrapped in $(function(){}) will fire as soon as it is received. If that code preceeds the html it refers to, it will not find that html, as it doesn't exist yet.
If you move the same code below the html it refers to, it will fire after the html exists and therefore will find it.
EDIT: My answer presumes that all the code shown after "Executes:" in OP is contained in remote page

You have no handler for the result of the sort.php. Invoking this will only load the data into cache.
You need a complete handler function to refresh the data, not to mention add it to the dom. You should clarify your question and make it obvious that those are two different pages.
$.post("../albumUploader/queries/sort.php", order).complete = func...

Related

How can I return an id value from a div already populated through ajax

I am having some difficulty passing a correct id function back to AJAX.
I'm creating a product bulletin generator that lets items to be added by their SKU code (which works fine). My problem is that when a bulletin is clicked on, a preview of that bulletin is loaded into a div and shows all products associated with that bulletin.
From inside those results, I am trying to add the ability to delete a product from the bulletin. The problem is that the value being passed back to AJAX belongs to the first product only. It won't send the value belonging to the particular item if it is any other item than the first one.
This is the code (belonging to main.php) that gets loaded via AJAX into a div and is looped with each product associated with a selected bulletin
echo "<form name='myDelForm'>
$news_gen_id<br>
<input type='hidden' id='delccode' value='".$news_gen_id."'>
<input type='hidden' id='deledit' value='".$edit."'>
<input type='button' onclick='ajaxDelCcode()' value='Delete' /><br></form>
</td>";
The AJAX code (on index.php, where the div that calls in main.php is also located) is this
function ajaxDelCcode(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
} catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try{
ajaxRequest = new
ActiveXObject("Microsoft.XMLHTTP");
} catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
// Create a function that will receive data sent from the server
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
var ajaxDisplay = document.getElementById("ajaxMain2");
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
var deledit = document.getElementById("deledit").value;
var delccode = document.getElementById("delccode").value;
var queryString = "?delccode=" + delccode + "&deledit=" + deledit;
ajaxRequest.open("GET", "main.php" + queryString, true);
ajaxRequest.send(null);
}
//-->
</script>
Currently, using those two pieces of code, I can successfully delete only the first product. The delccode variables do not seem to change when the products are looped (although when I echo the variables during the loop, it is definitely changing to the appropriate value...it's just not passing it correctly back to AJAX.)
I tried taking the AJAX code, putting it inside the main.php product loop, and change the function name during each loop (so ajaxDelCcode$news_gen_id() for example) and also to the button itself so that it is calling the AJAX specific to it. And it works if you are visiting main.php directly...but not from index.php after main.php has been called into the div.
I can't figure out how to pass the correct looped value from main.php within the div, back to the AJAX code on index.php
Can anyone help me with this?
Thanks,
Dustin
Instead of storing the id in the input, just pass it as an argument to the function:
function ajaxDelCcode(delccode) { ...
<input type='button' onclick='ajaxDelCcode(\"".$news_gen_id."\")' value='Delete' />
Also, I'd swap the quotes if I were you. Or better yet, instead of using echo, break the PHP code and just write HTML:
<? ... ?><input type="button" onclick="ajaxDelCcode('<?= $news_gen_id ?>')" value="Delete" /><? ... ?>
What does the code you use to delete look like? Is it in the same php file as the form you posted above? If so, is the form getting submitted to itself accidentally? Like perhaps when a user presses enter while on an input type=text control? I understand that you want to do this by ajax but I am suspecting that the form is your problem.
Seconding the jQuery comment.
Here try this
1) add jquery to your document.
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
2) give your inputs name attributes
<input type='hidden' name='delcode' id='delccode' value='".$news_gen_id."'>
<input type='hidden' name='deledit' id='deledit' value='".$edit."'>
3) Use a function something like this instead of all that code above
function ajaxDelCcode() {
$.ajax({
url: "main.php",
type: "GET",
dataType: "text",
data: $("#myDelForm").serialize(),
success: function(rText) {
$("#ajaxMain2").text(rText);
}
});
}

AJAX Div Refresh with PHP

I am trying to refresh some elements on my page every so often. I know theres a million topics on here about that and I have tried to get mine working, but here is what I need to refresh..
This is the code that gets generated when the page loads:
<div id="galleria">
<?php
$a = array();
$dir = '../public/wp-content/uploads/2012/01';
if ($handle = opendir($dir)) {
while (false !== ($file = readdir($handle))) {
if (preg_match("/\.png$/", $file)) $a[] = $file;
elseif (preg_match("/\.jpg$/", $file)) $a[] = $file;
elseif (preg_match("/\.jpeg$/", $file)) $a[] = $file;
}
closedir($handle);
}
$totalImgs = count($a);
$imgUsed = array();
for ($j = 0; $j < 100; $j++)
{
do
{
$randIndex = mt_rand(0, $totalImgs);
}
while ($imgUsed[$randIndex] === TRUE);
$imgUsed[$randIndex] = TRUE;
echo "<img src='" . $dir . '/' . $a[$randIndex] . "' />";
}
?>
</div>
I would like to automatically refresh this every 10 seconds but not reload the page. I have read up on ajax and it seems this is possible but I cannot seem to get it to work.
All this is doing is showing the galleria div, and loading the 100 images inside the div. Then the galleria script takes over and displays it nicely. Will AJAX work better or JQuery?
Thank you for your help!
"Will AJAX work better or jQuery?" -- AJAX is a technique, jQuery is a library. As it turns out, jQuery has an excellent API for AJAX.
Let's call this bit of PHP "galleria.php". On original page load, it is inserted into the parent PHP page using good ol' <?php include('galleria.php')?>. Now the end user is seeing the full initialized page.
To update it, you have a number of AJAX options available, but the easiest is to include jQuery on your page and then you can use .load() in a script:
var updateGallery = setInterval(function() {
$('#someDiv').load('galleria.php');
}, 10000);
There's room for tweaking... maybe galleria.php doesn't include the <div id="galleria">, which is set on the page. In which case you would load right into #galleria instead of #someDiv and save yourself an unnecessary container. Maybe you cache the $('#someDiv') object by declaring it in a different scope so that it can be re-used. But this is the general gist.
Use setInterval function with ajax call.
http://jquery-howto.blogspot.com/2009/04/ajax-update-content-every-x-seconds.html
As I wrote here you can fill a div with a jQuery ajax call.
<html>
<head>
<script type="text/javascript">
function refresh_gallery(){
$.ajax({
type: "POST",
url: "generate_gallery.php", // your PHP generating ONLY the inner DIV code
data: "showimages=100",
success: function(html){
$("#output").html(html);
}
});
}
$(function() {
refresh_gallery(); //first initialize
setTimeout(refresh_gallery(),10000); // refresh every 10 secs
});
</script>
</head>
<body>
<div id="output"></div>
</body>
</html>

How can I make an AJAX link work once it is moved out of the iFrame?

I am using an iFrame with a form that return some content with an AJAX link.
I am then moving the returned content out of the iFrame into the main page.
However, then the ajax link does not work and the error "Element is null" is created once the link is clicked.
How can I move content from the iFrame and still have the AJAX link working?
Here's the code returned by the iFrame:
<span id="top">
<a id="link8" onclick=" event.returnValue = false; return false;" href="/item_pictures/delete/7">
<img src="/img/delete.bmp"/>
</a>
<script type="text/javascript">
parent.Event.observe('link8', 'click', function(event) {
new Ajax.Updater('top','/item_pictures/delete/3', {
asynchronous:true, evalScripts:true,
onCreate:function(request, xhr) {
document.getElementById("top").innerHTML = "<img src=\"/img/spinner_small.gif\">";
},
requestHeaders:['X-Update', 'top']
})
}, false);
</script>
</span>
I see two problems with your code.
First the solution (I think :-))
When your iframe loads, the javascript in it runs. Javascript attaches an observer to parent document's link8.
Inside the observer you define another function (onCreate). This function will run in iframe context, making document object refer to iframe and not to main document. When you remove link8 from iframe to move it to main document, document.getElementById("top") will become null - hence error.
Perhaps change it to:
parent.document.getElementById("top").innerHTML = "<img src=\"/img/spinner_small.gif\">";
Second problem (that is not really a problem in this particular case) is, if you move whole span (including the script) to main document, the javascript will run again in main document's context. In your case, you should see an error or warning after you move the content, stating that parent is null (or similar).
To remove the second problem, return your iframe data in two divs or similar. Then copy only div with html to main document.
What I did was move the AJAX call out to an external js file and called the function once the link was clicked. It works now.

jquery ajax post callback - manipulation stops after the "third" call

EDIT: The problem is not related to Boxy, I've run into the same issue when I've used JQuery 's load method.
EDIT 2: When I take out link.remove() from inside the ajax callback and place it before ajax load, the problem is no more. Are there restrictions for manipulating elements inside an ajax callback function.
I am using JQuery with Boxy plugin.
When the 'Flag' link on the page is clicked, a Boxy modal pops-up and loads a form via ajax. When the user submits the form, the link (<a> tag) is removed and a new one is created from the ajax response. This mechanism works for, well, 3 times! After the 3rd, the callback function just does not remove/replace/append (tested several variations of manipulation) the element.
The only hint I have is that after the 3rd call, the parent of the link becomes non-selectable. However I can't make anything of this.
Sorry if this is a very trivial issue, I have no experience in client-side programming.
The relevant html is below:
<div class="flag-link">
<img class="flag-img" style="width: 16px; visibility: hidden;" src="/static/images/flag.png" alt=""/>
<a class="unflagged" href="/i/flag/showform/9/1/?next=/users/1/ozgurisil">Flag</a>
</div>
Here is the relevant js code:
$(document).ready(function() {
$('div.flag-link a.unflagged').live('click', function(e){
doFlag(e);
return false;
});
...
});
function doFlag(e) {
var link = $(e.target);
var url = link.attr('href');
Boxy.load(url, {title:'Inappropriate Content', unloadOnHide:true, cache:false, behaviours: function(r) {
$("#flag-form").live("submit", function(){
var post_url = $("#flag-form").attr('action');
boxy = Boxy.get(this);
boxy.hideAndUnload();
$.post(post_url, $("#flag-form").serialize(), function(data){
par = link.parent();
par.append(data);
alert (par.attr('class')); //BECOMES UNDEFINED AT THE 3RD CALL!!
par.children('img.flag-img').css('visibility', 'visible');
link.remove();
});
return false;
});
}});
}
Old and late reply, but.. I found this while googling for my answer, so.. :)
I think this is a problem with the "notmodified" error being thrown, because you return the same Ajax data.
It seems that this is happening even if the "ifModified" option is set to false (which is also the default).
Returning the same Ajax data three times will cause issues for me (jQuery 1.4). Making the data unique (just adding time/random number in the response) removes the problem.
I don't know if this is a browser (Firefox), jQuery or server (Apache) issue though..
I have had the same problem, I could not run javascript after I call boxy. So I put all my javascript code in afterShow:function one of boxy attributes. I can run almost except submit my form. My be my way can give you something.

Ajax append load

it must be jquery
I have file text.html with 6 div in (a,b,c,d,e,f)
In another file i have a div, i like it to populate the content of a+b+c+d+e+f into that single div
I have try .load = but b remplace a
i have try append, but i need a temp var
so now i am stuck
That code get the content from the file textes.html ... div #a and put the content into div #right, but the second libe REMPLACE the content of right a with right b
I like to append the content a + b NOT a over b
$(document).ready(function(){
var temp = load('textes.html #nicolas');
$('#right').append(temp);
var temp = load('textes.html #antoine');
$('#right').append(temp);
.
.
.
.
return false;
});
that code is the idea behind what should work, but i cannot make a ajax .load() to load content into a variable to append the content to the div...
<script type="text/javascript">
$(document).ready(function(){
$.ajax({
url: "textes.html",
cache: false,
success: function(html){
$("#right").append(html);
}
});
});
</script>
That code load the WHOLE html file, i like to get only some selected DIV #
$(document).ready(function(){
$.get("textes.html",function(data){
$("#right").append($("#nicolas",data)).end().append($("#antoine",data));
},'html');
});
I had a similar issue just now and I think I figured out a way to do what we want using the .load() function. It's not pretty but never mind ;)
First off, I added a "TempDiv" to my html with a "visibility:hidden" style.
<div id="TempDiv" style="visibility:hidden"></div>
Then you run the jQuery :
$(document).ready(function(){
$('#TempDiv').load('textes.html #nicolas', function(){
$('#right').append($('#TempDiv').html());
});
});
I'm not sure it's the best way !
PS : That is my first stackoverflow post ;)
try,
$.get('url.php', function(data) {
$("#right").append(data);
});
This sounds like jQuery? Please state what framework you are using since I can't really see any mention of it. Anyway, append should work. Just do something like:
mydiv.append(a.text());
mydiv.append(b.text());
mydiv.append(c.text());
mydiv.append(d.text());
mydiv.append(e.text());
mydiv.append(f.text());
They should all be appended into mydiv.
NOTE: if you also want the html, use the .html() function instead of .text().
JQuery ( http://jquery.com/ ) is a good javascript library that you can use to do an AJAX request to get the other file. See this question for more: Use jQuery to replace XMLHttpRequest

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