I have this JavaScript code:
$(document).ready(function(){
$('#sel').change(function(){
$.ajax({
type: "POST",
url: "modules.php?name=TransProject_Management&file=index",
data: "&op=index_stat&stat="+$(this).val(),
cache: false,
success: function(data) {
//alert(data);
$("#ajax_results").html(data);
}
});
});
});
On status change i need to refresh a div without page reload. But it returns blank page. If i try alert the result on success, i get the response, also i checked with inspect element, its ok. The problem is that it returns blank page.
The file i'm working on, is the same( modules.php?name=TransProject_Management&file=index ) i called in ajax.
the html:
<body>
//...
<div id="ajax_results">
//.....
//somewhere here is the select option <select id="sel">......</select>
//.....
</div>
</body>
Any help, would be very appreciated.
use the following code to return your response html:
echo json_encode(array($your_response));
Then in your javascript, you will need to reference the data as:
success: function(data) {
$("#ajax_results").html(data[0]);
}
since it is now an array.
this in your ajax function refers to the jQuery XHR object, NOT the $('#sel') object. Just assign it to a variable before the ajax function like var sel = $(this) then use it later inside the function. Try this:
$('#sel').change(function(){
var sel = $(this);
$.ajax({
type: "POST",
url: "modules.php?name=TransProject_Management&file=index",
data: "&op=index_stat&stat="+sel.val(),
cache: false,
success: function(data) {
//alert(data);
$("#ajax_results").html(data);
}
});
});
});
Hmm, first glance the code looks good. Have you tried using Chrome debug tools? Hit F12 and check the Network tab, this will show you what is being returned. You can also debug without using an alert so you can step through to see what exactly the properties are.
Just thought, you might need to add 'd' to the data returned. Anyway, if you do what I suggested above, put a pause break on the line and run the code you will see what you need.
Based on your comments below the question, it seems that you are using the same script to display your page and to call in the javascript. This script seems to return a complete html page, starting with the <html> tag.
A page can only have one <html> tag and when you try to dump a complete html page inside an element in another page, that will lead to invalid html and unpredictable results.
The solution is to have your ajax script only return the necessary elements / html that needs to be inserted in #ajax_results, nothing more.
Related
Something weird - I have the script inside the bootstrap modal.
sometimes the script is loaded and works and sometimes it doesn't.
here is a URL for example:
https://ns6.clubweb.co.il/~israelig/sites/followmyroutes/test_sec.php
Click on the button and see the modal, then close the modal and open it again. After the couple of times, the scripts inside the modal stops working (scripts like form validation [when you submit it], image browser)
How can I fix it so all the script will work every time?
The way you populate the html is right or not suggested to be advised by coders. Check again from where you got the documentation.
your existing code from backend call
$(document).on('ready', function() {
$("#input-8").fileinput({
});
});
Try changing like
$(document).on('ready', function() {
setImageUploader();
});
function setImageUploader(){
$(document).find("#input-8").fileinput({
});
}
And also
$.ajax({
cache: false,
type: 'GET',
url: 'itinPage-secManage.view.php',
data: info,
success: function(data) {
$modal.find('.modal-body').html(data);
setTimeout(function(){ //added this line.
setImageUploader()
})
}
});
i think the problem in the syncronisation of the request, you can use Ajax reque
It looks like your common libraries like jquery bootstrap-datepicker, fileinput theme etc are being get every time the modal is launched.
This may cause a sort of namespace corruption or some weird side effect of the kind you seem to see.
You could put all the common libraries outside of the modal to prevent this from happening.
the problem in the ajax request, I means you maste waite while the request it's done.
var request = $.ajax({
cache: false,
type: 'GET',
url: 'itinPage-secManage.view.php',
data: info
});
request.done(function(data) {
$modal.find('.modal-body').html(data);
});
I've this script:
<script>
$(document).ready(function() {
$('#signUpForm').submit(function() {
$.ajax({
type: 'POST',
url: "process.php",
data: $("#signUpForm").serialize(),
success: function(data)
{
$('#errors').show();
$('#errors').append(data);
}
});
return false;
});
});
</script>
It displays errors from php script. And right now every time I click submit button, new errors are added to old ones. I wonder, is it possible to reset old errors and show just new, after I click submit button?
Use $('#errors').html(data); instead of $('#errors').append(data);, When you use .append(), you add to the content you already had. .html() replaces the content.
Another alternative might be to empty first and then append. Like $('#errors').empty().append(data);
I wonder why do you have $('#errors').show();? If it was hidden first, maybe better to put .show() after the .html() / .append()`
Use $('#errors').empty().append(data);
I know there are a few topics on this subject, but after I spent 2 or 3 hours trying to get something good out of them, I just decided to ask this question on a specific point.
So here is my problem : I have got a table and I am using a jQuery function to select a row of this table. Now what i actually want to do is getting the text content of the div contained in the first td of the row.
I already used a getter on it and I am checking the getted value with an alert as you can see in th following code :
$("#myRow").click(function() {
$(".selectedRow").removeClass("selectedRow").addClass("unselected");
$(this).addClass("selectedRow").removeClass("unselected");
var myValue = $(".selectedRow .firstTd div").text();
alert('myValue');
});
So now, what I am trying to do is to send the myValue variable through an ajax request by replacing my alert by this piece of code :
$.ajax({
type: 'get',
url: 'index.php',
data: {"myValue" : myValue},
success: function(rs)
{
alert(myValue);
}
});
Then, back to my php code, I am tring to observe the obtained variable by using an echo, just like this :
<?php echo $_GET['myValue']; ?>
But there is just no way for me to know if my page got it beacause the echo just prints nothing... So i was wondering if someone could do something for me. Thanks.
PS : Oh, by the way ; I don't really know if this can matter, but my page index.php already receives data by a post.
You can't, but read this, php is on the server, while js usually runs on the client, but your ajax trick can work. Just do some processing in the recieving php.
I usually put my ajax recieving end in a different file, and process the rest by the variables posted.
Just try to put the $_GET['myValue']; into an if, or a switch.
Do a var dump of the request var to see if anything is coming through:
<?php
var_dump($_REQUEST);
If not, do a console.log() on 'myValue' to make sure it exists before sending the ajax request - the issue may lie in your js rather than you php.
If you are POSTing data then adjust accordingly - e.g.
$.ajax({
type: 'post',
url: 'index.php',
data: {"myValue" : myValue},
success: function(data)
{
console.log('successfuly posted:');
console.log(data);
}
});
then:
<?php echo $_POST['myValue']; ?>
If you were using GET your data would be in the url, e.g:
index.php?myValue=something
I'm not sure if you are aware of that, but you should wrap you function in document ready statement as below.
Next, call the AJAX request on some action, in this case we can use a click on the row in table.
$(document).ready(function () {
$("#myRow").click(function() {
$(".selectedRow").removeClass("selectedRow").addClass("unselected");
$(this).addClass("selectedRow").removeClass("unselected");
var myValue = $(".selectedRow .firstTd div").text();
alert('myValue');
$.ajax({
type: 'get',
url: 'index.php',
data: {"myValue" : myValue},
success: function(data)
{
console.log('you have posted:' + data.myValue);
}
});
});
});
Okay so it seems that i totally misunderstanded on the way that the $.ajax function works.
I now do use the $.post function (which is actually the same), this way :
$.post('pageElement.php', { myValue : $(".selectedRow .firstTd div").text() },
function(data) { $("#test").html(data); }
);
The url "pageElement.php" refers to a page containing this code :
<div><?php echo $_POST['myValue']; ?></div>
The function called at the end of the process just puts this code into a div of my original page, so i can use it as a php variable now and then send it to another page through a form.
I am doing a basic jquery ajax call on a php file and can't seemsto figure out why it isn't working. Any help is appreciated. Fiebug does not seem to show any ajax or XHR action going on. I want to not to refresh the page and just execute the ajax call. Thanks.
JS
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"</script>
<script>
function getData(url_param){
$.ajax({
type: 'get',
url: 'data.php',
data: {url_param:url_param},
success: function(data) {
$('#data').html(data);
}
});
};
$('#clickMe').click(function(e){
e.preventDefault();
getData(2);
});
</script>
HTML:
<div><a id='clickMe' href='data.php?url_param=url_param'>CLICK ME TO RUN PHP</a></div>
<div id="data"></div> <!-- divto show result -->
PHP:
<?php
if($_GET['url_param']){
echo "simple ajax call";
}
?>
You have to bind the event inside an onload function. The most common practice is:
$(document).ready(function(){
$('#clickMe').click(function(e){
...
});
});
You should also add return false; in the last line of your event.
First, you have misspelled your function name (getGata != getData).
Secondly:
data: {url_param:url_param}
Are you setting the javascript variable url_param anywhere? The $.ajax data parameter is formatted as follows:
get/post variable name : get/post variable value
As you have it now, it doesn't seem that you are assigning a value to url_param.
you can simply use jQuery post function.
$.post('data.php',{param1:'your param 1', param2 : 'your param 2'}, function(response){
//do your operation here. response is what you get from data.php. 'json' spicifies that the response is json type
$("#data").html(response);
},'json');
The (amended?) JavaScript prevents your code from working, because you haven't closed the angle brackets on jQuery source, it should be:
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
One of the comments states you shouldn't have the href in the anchor, but because you've ignored defaults this isn't triggered (assuming JS is enabled in the user's browser).
Finally, I think that
return false;
should really be inside the function after
getData(2);
but since we're ignoring defaults, the anchor shouldn't make an attempt to go anywhere or reload anyway.
It is me again. I am getting so frustrated with this code it is not even funny. It's not that I am wanting to post it again. It is just that now I understand the where the problem was in the code and wanted to see if you guys can help me figure the last part out.
Basically I am trying to refresh a div without reloading the entire page. It's killing me. Here is some more information on it:
here is my js file first
$(function() {
$(".button").click(function() {
// validate and process form
// first hide any error messages
var email = $("input#email").val();
//var dataString = '&email=' + email; commented out
var dataString = email;
//try insted this //alert (dataString);return false;
$.ajax({ type: "POST", dataType:'HTML',
//or the appropiate type of data you are getting back
url: "http://www.edshaer.com/EdinburgCISD/Gorena/Gorena.php", data: {email:dataString},
//in the php file do $email = $_POST['email'];
//not a good practice but you can try with it and without it
success: function() {
$("#div").fadeOut($("#div").html());
$("#div").fadeIn($("#div").html());
$("#email").val('');
// Change the content of the message element
// Fade the element back in
} });
//ajax ends
return false; });
//click ends
});//document ready ends
Now the problem that I am running into with this code is on the Ajax part. After placing the alert(), I have relized that if I use the function() like this:
success: function(data)
Then the alert came out blank. The reason behind it is that my URL is going to my php file, but my div that I am trying to refresh is on my html file. Meaning if I do this:
success: function(data) {
$("#div").html(data)}
I am sending blank data because it's trying to get the div from my php file instead of my html file.
Now if I do this:
$("#div").html()
Then that gives me the div that is in my html file.
By knowing what is going on now, Can you guys please help me???
My dear you should generate some sort of html in your php file that you want to generate in your div. Then you will see that you are having some content in data in the success function. This is an easy approach.
But there is also some other approach that is more efficient but it needs some sort of search. This is the implementation of Client Side Scripting. You can do this with the help of a jquery plugin jquote2. I hope it will work for you.
You're using
$("#div").fadeOut($("#div").html());
$("#div").fadeIn($("#div").html());
Both are wrong, jQuery .fadeIn() and .fadeOut() arguments are either [duration,] [callback] or [duration,] [easing,] [callback]. None take HTML as input.
Try changing
$("#div").fadeOut($("#div").html());
to
$("#div").fadeOut();
and moving it outside the $.ajax call to hide the previously showed (if any) results before the post and also change
$("#div").fadeIn($("#div").html());
to
$("#div").html(result).fadeIn();
Also change
success: function()
to
success: function(result)
Hope it helps.
This might be a problem relating to the response from the php script. Jquery doesn't always correctly render the Ajax response as html.
Setting dataType: html in the $.ajax({ ... }) call can help. Also setting header("Content-Type: text/html"); at the top of your php ajax script.