NOTE: This question is a split from the post: jQuery not working on elements created by jQuery
I am dynamically adding list items to a list in jQuery through an ajax call that is called every second.
Below is the code for the ajax call.
$.ajax({
url: 'php/update_group_list.php',
data: '',
dataType: 'json',
success: function(data) {
var id = data.instructor_id;
group_cnt = data.group_cnt,
group_name = data.group_name,
group_code = data.group_code;
for (i = current_row; i < group_cnt; i++)
{
//setInterval(function() { $('#group-list-div').load('php/group_list.php'); }, 5000);
$('#group-list').append("<li><a href='#' data-role='button' class='view-group-btns' id='"+group_code[i]+"' value='"+id+"' text='"+group_name[i]+"'>"+group_name[i]+"</a></li>");
$('#delete-group-list').append("<fieldset data-role='controlgroup data-iconpos='right'>" +
"<input id='"+group_code[i]+i+"' value='"+group_code[i]+"' type='checkbox' name='groups[]'>" +
"<label for='"+group_code[i]+i+"'>"+group_name[i]+"</label>" +
"</fieldset>");
}
current_row = i;
$('#group-list').listview('refresh');
$('#delete-group-list').trigger('create');
}
});
when I try to send the form data for the checkboxes (referencing line $('#delete-group-list').blah...blah in the ajax call code above) the post returns the error unexpected token <
What am I doing wrong? I think the two problems are related as I am creating the list items that are used dynamically.
Here is extra code relating to the problem
HTML:
<form id='delete-group-form' action='php/delete_groups.php' method='post'>
<h3 style='text-align: center;'>Check the Box Beside the Groups you Would Like to Delete </h3>
<div style='margin-top: 20px;'></div>
<div id='delete-group-list'>
</div>
<div style='margin-top: 20px;'></div>
<input type='submit' id='delete-groups-btn' data-theme='b' value='Delete Groups(s)'>
</form>
JS Code
$('#delete-group-form').submit(function(e)
{
e.preventDefault();
alert($('#delete-group-form').serialize());
if ($('#delete-group-form').serialize() == "")
{
alert('No groups selected to be deleted.')
return false;
}
else
if ($('#delete-groups-form').serialize() == null)
{
alert('No groups selected to be deleted.')
return false;
}
else
{
$.post('php/delete_groups.php',$('#delete-groups-form').serialize()).done(function(data)
{
obj = jQuery.parseJSON(data);
var group_codes = obj.group_list;
alert(group_codes);
alert("The selected groups have been deleted");
window.setTimeout(2000);
return false;
});
}
return false;
});
delete_groups.php
<?php
$group_codes = $_POST['groups'];
$items = array('group_list'=>$group_codes); //creating an array of data to be sent back to js file
echo json_encode($items); //sending data back through json encoding
?>
I think the root of the SECOND problem is the line $group_codes = $_POST['groups']; specfically the $_POST['groups'] because when I replace it with $group_codes = 'test'; (just for debugging purposes) , the code works as expected.
ok, previously i thought the mistake with the content type header of php file, still its there (you need to specify the header type to parse with jQuery JSON). But found one more mistake which i suppose the cause of the problem.
<form id='delete-group-form' action='php/delete_groups.php' method='post'>
and the form you wish to serialize is
$('#delete-groups-form').serialize()
You can see the difference in IDs. This will return an empty object, where in php file it expects the index called groups which never exists. Therefore it'll return an undefined index error as an html content type which starts with <html>, with that JSON parser return an error with unexpected token <
Hope this helps
When you call $_POST['groups']; (in your delete_groups.php) it looks for the content sent from the form element with an attribute name="groups", but your delete-group-form doesn't have any such element.
The "error unexpected token <" message may be from a separate issue. Validating your HTML might reveal the issue.
Also, I'm not entirely clear on why you're using two separate functions (the $.post in your JS and the $.ajax). Unless there's something else going on that I can't see from the code you've posted, you could simplify it considerably by just using one call to ajax. Put that one call inside the $('#delete-group-form').submit(function(e) in place of the $.post, and pass your groups variable using the ajax call's data parameter. Modify delete_groups.php to return the necessary information (currently returned from update_group_list.php) to update the display on your page.
Related
When my page opens I call a PHP file and output a form in the form of an echo.
A simplified example below:
echo "<table id='results'>";
echo "<form method = 'post'><input id='".$row['batchname']."'><button class='btn1'>Query this record</button></form>";
</table>
There will be many versions of the above form as table rows are pulled from the database.
I am usin AJAX to handle the output:
$(document).ready(function(){
$(".btn1").click(function(e){
e.preventDefault();
var bname = ("#<?php echo $row['batchname'];?>").val();
$post(
"somephp.php",
{name : bname},
function(response, status){
$("#results").replaceWith(response);
}
);
});
});
When I input an non PHP ID into the jQuery the AJAX work but I always post the first returned row for every form produced as the ids output in the PHP are the same. Can I echo PHP variable into jQuery like this? Is there a better way of getting dynamic ID's into jQuery.
Rather than hacking about like this, do it properly.
$(".btn1").click(function(e) {
e.preventDefault();
var form = $(this).parents("form");
var value = form.find("input")[0].value; // or something else here
});
I have a database table which I am trying to retrieve data from using JQUERY AJAX. When my first page loads it does a php call to a table and populates a select form element. - This works
I then want to select one of the options submit the form and have the row returned via Ajax.
Previously I had the script working with just PHP files but am having trouble getting it to work. When submitting the form my URL is changing:
http://localhost/FINTAN/testertester.php?name=Specifics.
I am not getting anything back. In addition when looking at my console I get a jquery not defined
factory (jquery). I can find the line in question in my jquery ui.js. Not sure if this is the issue or my code has caused the issue. I have cleard the firefox cache and due to the fact I have not had a successful AJAX call via jquery method am guessing it my code.
To get the code below I have mixed and matched a book and an online tutorial and many other sources and this is not my first attempt. Ideally I would like to output table row. However just getting a request working and knowing its not a conflict or compatability issue would makeme feel better and not hindered before I start
<script src="jquery/jquery-ui-1.11.2/jquery-ui.js"></script>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#btn").click(function(){
var vname = $("#name").val;
}
}
$.post("addithandle1.php",
{
name:vname};
function(response,status){
alert("recieved data-------*\n\nResponse : " + response
+"\n\nStatus : " + status);
}
}
</script>
</head>
<body>
<?php
include "config.php";
if (mysqli_connect_errno($con))
{
}
else
{
$result = mysqli_query($con, "SELECT * FROM script ");
echo " <Form method='post'> <label>Script :</label> <select id='name' name='name' >";
}
while($row = mysqli_fetch_array($result))
{
echo "<option value = '".$row['scriptname']."'>".$row['scriptname']."</option>";
}
echo "</select>";
echo "<button id='btn' class='btn-search'>Load Script </button></form>";
?>
</body></html>
This is my PHP file that I am trying to retrieve from
<?php
include 'config.php';
$batchtype2 = $_POST['name'];
$batchtype2 = mysqli_real_escape_string($con,$batchtype2);
$sql = "SELECT * FROM script WHERE scriptname = '".$batchtype2."' ";
$result = mysqli_query($con,$sql);
$count=mysqli_num_rows($result);
if($count==0 ){
echo "</br></br></br></br></br></br></br><p> No Matching results found</p>";
}
else{
while($row = mysqli_fetch_array($result)) {
echo '<tr><td>'.$row['scriptname'].'</td></tr>';
echo '<tr><td>'.$row['scripthours'].'</td></tr>';
echo '<tr><td>'.$row['scripttotal'].'</td></tr>';
}
}
mysqli_close($con);
?>
Thanks in advance for any help
By making the following corrections (you have some syntax issues as well as usage issues which should be revealed in your browser's console when you load this page) in your JavaScript/jQuery this will work like you expect -
Make sure to change this line -
var vname = $("#name").val;
to this -
var vname = $("#name").val(); // note the parentheses
in your function -
$(document).ready(function(){
$("#btn").click(function(e){
e.preventDefault(); // prevent the default action of the click
var vname = $("#name").val();
$.post("addithandle1.php", {name:vname}, function(response, status) { // POST instead of GET
// never use alert() for troubleshooting
// output for AJAX must be in the callback for the AJAX function
console.log("recieved data-------*\n\nResponse : " + response +"\n\nStatus : " + status);
$('#table').html(response); // put response in div
});
});
});
Now $_POST['name'] should get populated properly.
To get the table to appear in your requesting page first make sure that your PHP forms the table completely.
Add a div to your requesting page and modify the AJAX call above as shown.
<div id="table"></div>
Now, when you make a request the div on the requesting page will be updated with whatever comes back from the PHP script.
There are a couple of things about your script.
First make sure you write well structured code and that it is nothing in the wrongplace / broken.
You have in the $(document).ready(function(){ only the .click event of the button, but you left the ajax request outside, I imagine you did that so it will also make the ajax request in the first page load
The problem is that now it will only make it in the first page load, but not when you click the button, on clicking button you are only getting the value of name.
I recommend you to try something like this:
<script>
$(document).ready(function() {
// bind button click and load data
$("#btn").click(function(){
loadData();
return false; // prevent browser behaviour of the button that would submit the form
}
// load data for the first time
loadData();
};
function loadData() {
var vname = $("#name").val;
$.post("addithandle1.php", { name:vname }, function(response, status) {
alert("recieved data-------*\n\nResponse : " + response
+"\n\nStatus : " + status);
});
}
</script>
A few notes:
I would recommend always putting jquery code inside $(document).ready since that guarantees that jquery was loaded before running it
By default a form that has a submit button that you click, will get the form submitted by the browser, if you use ajax, you should prevent that behaviour, either on the button click event or on form with onsubmit="return false".
Ok, so I've gotten most of this thing done.. Now comes, for me, the hard part. This is untreaded territory for me.
How do I update my mysql database, with form data, without having the page refresh? I presume you use AJAX and\or Jquery to do this- but I don't quite grasp the examples being given.
Can anybody please tell me how to perform this task within this context?
So this is my form:
<form name="checklist" id="checklist" class="checklist">
<?php // Loop through query results
while($row = mysql_fetch_array($result))
{
$entry = $row['Entry'];
$CID = $row['CID'];
$checked =$row['Checked'];
// echo $CID;
echo "<input type=\"text\" value=\"$entry\" name=\"textfield$CID;\" id=\"textfield$CID;\" onchange=\"showUser(this.value)\" />";
echo "<input type=\"checkbox\" value=\"\" name=\"checkbox$CID;\" id=\"checkbox$CID;\" value=\"$checked\"".(($checked == '1')? ' checked="checked"' : '')." />";
echo "<br>";
}
?>
<div id="dynamicInput"></div>
<input type="submit" id="checklistSubmit" name="checklistSubmit" class="checklist-submit"> <input type="button" id="CompleteAll" name="CompleteAll" value="Check All" onclick="javascript:checkAll('checklist', true);"><input type="button" id="UncheckAll" name="UncheckAll" value="Uncheck All" onclick="javascript:checkAll('checklist', false);">
<input type="button" value="Add another text input" onClick="addInput('dynamicInput');"></form>
It is populated from the database based on the users session_id, however if the user wants to create a new list item (or is a new visitor period) he can click the button "Add another text input" and a new form element will generate.
All updates to the database need to be done through AJAX\JQUERY and not through a post which will refresh the page.
I really need help on this one. Getting my head around this kind of... Updating method kind of hurts!
Thanks.
You will need to catch the click of the button. And make sure you stop propagation.
$('checklistSubmit').click(function(e) {
$(e).stopPropagation();
$.post({
url: 'checklist.php'
data: $('#checklist').serialize(),
dataType: 'html'
success: function(data, status, jqXHR) {
$('div.successmessage').html(data);
//your success callback function
}
error: function() {
//your error callback function
}
});
});
That's just something I worked up off the top of my head. Should give you the basic idea. I'd be happy to elaborate more if need be.
Check out jQuery's documentation of $.post for all the nitty gritty details.
http://api.jquery.com/jQuery.post/
Edit:
I changed it to use jquery's serialize method. Forgot about it originally.
More Elaboration:
Basically when the submit button is clicked it will call the function specified. You want to do a stop propagation so that the form will not submit by bubbling up the DOM and doing a normal submit.
The $.post is a shorthand version of $.ajax({ type: 'post'});
So all you do is specify the url you want to post to, pass the form data and in php it will come in just like any other request. So then you process the POST data, save your changes in the database or whatever else and send back JSON data as I have it specified. You could also send back HTML or XML. jQuery's documentation shows the possible datatypes.
In your success function will get back data as the first parameter. So whatever you specified as the data type coming back you simply use it how you need to. So let's say you wanted to return some html as a success message. All you would need to do is take the data in the success function and place it where you wanted to in the DOM with .append() or something like that.
Clear as mud?
You need two scripts here: one that runs the AJAX (better to use a framework, jQuery is one of the easiest for me) and a PHP script that gets the Post data and does the database update.
I'm not going to give you a full source (because this is not the place for that), but a guide. In jQuery you can do something like this:
<script type="text/javascript" src="http://code.jquery.com/jquery-1.10.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function() { // DOM is ready
$("form#checklist").submit(function(evt) {
evt.preventDefault(); // Avoid the "submit" to work, we'll do this manually
var data = new Array();
var dynamicInputs = $("input,select", $(this)); // All inputs and selects in the scope of "$(this)" (the form)
dynamicInputs.each(function() {
// Here "$(this)" is every input and select
var object_name = $(this).attr('name');
var object_value = $(this).attr('value');
data[object_name] = object_value; // Add to an associative array
});
// Now data is fully populated, now we can send it to the PHP
// Documentation: http://api.jquery.com/jQuery.post/
$.post("http://localhost/script.php", data, function(response) {
alert('The PHP returned: ' + response);
});
});
});
</script>
Then take the values from $_POST in PHP (or any other webserver scripting engine) and do your thing to update the DB. Change the URL and the data array to your needs.
Remember that data can be like this: { input1 : value1, input2 : value2 } and the PHP will get something like $_POST['input1'] = value1 and $_POST['input2'] = value2.
This is how i post form data using jquery
$.ajax({
url: 'http://example.com',
type: 'GET',
data: $('#checklist').serialize(),
cache: false,
}).done(function (response) {
/* It worked */
}).fail(function () {
/* It didnt worked */
});
Hope this helps, let me know how you get on!
I have a a script that on click do a ajax call connect to the database get imagename and set the image name inside an < -img - > with the right path also it adds a hidden checkbox after it and then echo it.
i then take the ajax message returned and put it as div's HTML. my question is will i be able to preform more action on the inserted content..
The main goal is to be able to click on the image as if it were a checkbox(this part is already sorted for me) however no matter what i try i cant have a .click function works..
Here is the code.
This is the PHP part that echos the images.
if($_POST['updateIgallery'] == 'ajax'){
global $wpdb;
$table_name= $wpdb->prefix . "table_T";
$imagecounter = 1;
$toecho = '';
$currentselected = $wpdb->get_row("query");
preg_match_all('/\/(.+?\..+?)\//',$currentselected ['image_gal'],$preresualts); // images are stored with /image/.
foreach ($preresualts[1] as $imagename){
$toecho .= '
<img rel="no" id="JustantestID" class="JustaTestClass" src="'.site_url().'/wp-content/plugins/wp-ecommerce-extender/images/uploads/'.$imagename.'">
<input name="DoorIMGtoDeleteIDcheck'.$imagecounter.'" style="display:none;" name="DoorIMGtoDelete['.$imagecounter.']" value="/'.$imagename.'/" type="checkbox">
';
$imagecounter++;
}
echo $toecho;
}
This is the ajax part that send and receive and insert the HTML to the div:
$.ajax({
type: "POST",
url: "/wp-content/plugins/wp-ecommerce-extender/DB_Functions.php",
data: { updateIgallery: "ajax", CurrentDoorIDnum: $('#dooridforgallery').val()}
}).success(function(insertID) {
$("#ImgGalleryID").html(insertID);
});
This so far works what i am having trouble with is the following:
$("#JustantestID").click(function() {
//DoorImageGallery($(this).attr('id')); // the function i will use if the alert actually works
alert("kahdaskjdj");
return true;
});
I hope the question and the code is understandable.
Thanks in advanced.
When you replace element's html, all the elements inside it are removed and gone. That means the event handlers attached to them are removed as well.
You could try attaching an event handler to a higher level element that is static and permanent on your page. Without more info I am going to use document:
$(document).on( "click", "#yaniv", function() {
alert("kahdaskjdj");
});
$('img.JustaTestClass').bind('click', function() {
var checkbox = $(this).siblings('input[type=checkbox]');
if (!checkbox.is(':checked')) checkbox.attr('checked', true);
else checkbox.attr('checked', false);
});
Since the elements are dynamically inserted into the DOM with ajax, you have to delegate events to a parent element that actually exists when binding the click handler, which in this case looks to be #ImgGalleryID
$('#ImgGalleryID').on('click', '#yaniv', function() {
DoorImageGallery(this.id);
alert("kahdaskjdj");
});
I am trying to send a php script some content to be stored in a database via ajax. I am using the jQuery framework. I would like to use a link on a page to send the information. I am having trouble writing the function that will send and receive the information, everything that I have tried is asymptotic.
EDIT
The idea is that the user will click the link, and a column called "show_online" (a tiny int) in a table called "listings" will update to either 1 or 0 (**a basic binary toggle!) On success, specific link that was clicked will be updated (if it sent a 1 before, it will be set as 0).
EDIT
There will be 20-30 of these links on a page. I have set each containing div with a unique id ('onlineStatus'). I would rather not have a separate js function for every instance.
Any assistance is much appreciated. The essential code is below.
<script type="text/javascript">
function doAjaxPostOnline( shouldPost, bizID ){
load("ajaxPostOnline.php?b='+bizID+'&p='+shouldPost", jsonData, callbackFunction);
function callbackFunction(responseText, textStatus, XMLHttpRequest)
{
// if you need more functionality than just replacing the contents, do it here
}
}
}
</script>
<!-- the link that submits the info -->:
<div id='onlineStatus<?php echo $b_id ?>'>
<a href='#' onclick="doAjaxPostOnline( 0, <?php echo $b_id ?> ); return false;" >Post Online</a>
</div>
ajaxPostOnline.php
<!-- ajaxPostOnline.php ... the page that the form posts to -->
<?php
$id = mysql_real_escape_string($_GET['b']);
$show = mysql_real_escape_string($_GET['p']);
if( $id && ctype_digit($id) && ($show == 1 || $show == 0) ) {
mysql_query( "UPDATE listing SET show_online = $show
WHERE id = $id LIMIT 1" );
}
if($result) {
if($show == '0'){
$return = "<a class='onlineStatus' href='#' onchange='doAjaxPostOnline( 1, <?php echo $b_id ?> ); return false;' >Post Online</a>";
}
if($show == '1'){
$return = "<a class='onlineStatus' href='#' onchange='doAjaxPostOnline( 0, $b_id ); return false;' >Post Online</a>";
}
print json_encode(array("id" => $id, "return" => $return));
}
?>
The load() function in jQuery is really cool for this sort of thing.
Here's an example. Basically, you have an outer div as a container. You call a script/service which returns html. You have a div in that html with an id that you will refer to later in the ajax call. The replacement div replaces the inner html of the container div. You pass your data as a json object as the second parameter to the load method, and you can pass a reference to a callback function as the third parameter. The callback function will receive every possible piece of information from the response (the full response text for further parsing/processing, the http status code, and the XMLHttpRequest object associated with this ajax call).
$("#id_of_some_outer_div").load("somepage.php #id_of_replacement_div", jsonData, callbackFunction);
function callbackFunction(responseText, textStatus, XMLHttpRequest)
{
// if you need more functionality than just replacing the contents, do it here
}
so, in your case you're talking about replacing links. Put the original link inside of a div on both sides of the operation.
Here's the link to the jQuery api doc for load():
load
EDIT:
In response to your comment about doing multiple replacements in one pass:
You can have the callback function do all the work for you.
Add a unique css class to all divs that need replacing. This will allow you to select all of them in one shot. Remember that html elements can have more than one css class (that's what the "c" in CSS means). So, they'd all be <div id="[some unique id]" class="replace_me"... Then, if you have a variable set to $("div.replace_me"), this will be a collection of all divs with the replace_me style.
Whatever elements that come from the ajax call (whether they're another div container or just a single "a" element) should have a unique id similar to the container they're to be inserted into. For example, div_replace1 would be the id of a container and div_replace1_insert would be the id of the element to be inserted
Inside the callback function, iterate over the replacements using $("div.replace_me").each(function(){ ...
Inside each iteration the "this" keyword refers to the current item. You can grab the id of this item, have a variable like var replacement_id = this.id + "_insert"; (as in the example above) which is now the unique id of the element you'd like to insert. $("#" + replacement_id) will now give you a reference to the element you want to insert. You can do the insertion something like this: this.html( $("#" + replacement_id) );
You may have to edit the code above (it's not tested), but this would be the general idea. You can use naming conventions to relate elements in the ajax return data to elements on the page, iterate the elements on the page with "each", and replace them with this.html()
did you really mean to declare your ajax success return function as
function(html)
? .. i think maybe you mean for the param to be 'data' ?
Since your php script is returning json you should set the dataType to json. Note that in your posted code sample, the success function() was outside of the $.ajax() and it needs to be inside.
$.ajax({
url: "ajaxPostOnline.php?b=" + bizID + "&p=" + shouldPost,
dataType: "json",
success: function(json){
$("#onlineStatus" + bizID).html(json.return);
}
});
You might want to check out the getJSON method since it's more concise for this particular situation.
$.getJSON("ajaxPostOnline.php", {b:bizID, p:shouldPost}, function(json) {
$("#onlineStatus" + bizID).html(json.return);
});
EDIT: Original question was edited and the provided sample changed significantly. I would still recommend the $.getJSON method.
Unless I am mistaken, it seems you have an error mixing AJAX and server-side scripting.
That depends on whether $return is PHP parsed anywhere after assignment snippet in ajaxPostOnline.php (hardly, if it is called from AJAX!).
$return = "<a class='onlineStatus' href='#' onchange='doAjaxPostOnline( 1, <?php echo $b_id ?> ); return false;' >Post Online</a>";
Surely this should be:
$return = "<a class='onlineStatus' href='#' onchange='doAjaxPostOnline( 1, ".$id." ); return false;' >Post Online</a>";