here is my input
<textarea id="mytext" class="txtarea" name="in_content" cols="120" rows="15"><?php echo $term;?></textarea>
here is my ajax code that take value of the textarea above
$('#spdf-form').submit(function() {
$.ajax({
type: 'POST',
url: $(this).attr('action'),
data: $(this).serialize(),
success: function(data) {
$('#spdf_results').html(data);
$('#spdf-form').fadeOut('slow');
var textAreaValue = $("#mytext").text();
alert(textAreaValue);
}
})
return false;
});
it's works, and show the value in the alert popup. Now i want to show the result as php code. i want to insert the value into tinymce editor, and the editor will be called like this
<?php the_editor(''); ?>
so i think i should do this
$myvalues = something to get the results from ajax function
and then i call the editor like this
<?php the_editor($myvalues); ?>
but i dont know how to do that, can someone help me please? i tried to get the value directly but it didnt work also.
PHP is server-side. JavaScript/jQuery/AJAX is client-side. You can't run PHP on the same page after it's been sent to the client.
You'll need to use AJAX to request your editor code from the server, then take the editor code and insert it into the page.
Related
To sum up what I'm trying to achieve here:
Inside of index.php, when selecting an option in a dropdown list, a function is called with the onchange="displayData(this) event inside of <select>
This function performs an AJAX POST request to a PHP page (target.php) with the value of the selected option inside of data
The PHP page is displayed inside a div on the page
Here is the function, using jQuery:
function displayData(str){
$.ajax({
url: "target.php",
type: "POST",
data: {"value": str.value},
success: function(data){
console.log(data);
}
});
$('#my-div').load('target.php');
}
To make things easier, here is what the PHP page looks like: <?php echo $_POST['value']; ?>
When logging data on success, everything seems to work fine, the value of $_POST['value'] is displayed in the console correctly. In the page itself though, I get an error:
Notice: Undefined index: value
Sorry if it seems kind of dumb, but I can't figure out what I'm doing wrong... So I thought of asking the community. Thank you for your help guys! Cheers.
Please try this.
and you need to return the result from target.php page
function displayData(str){
$.ajax({
url: "target.php",
type: "POST",
data: {"value": str.value},
success: function(data){
$('#my-div').html(data);
}
});
}
If you use .load() your browser will make another GET request to target.php and display it, so your $_POST will be empty.
You can use $('#my-div').html(data) inside success: of ajax or you can use $_GET instead of $_POST in your php and pass variable in url like this
$('#my-div').load('target.php?value='str.value);
hey the problem is quite simple, if you are posting the data value getting from ajax to target.php then the $('#my-div').load('target.php') should be inside the ajax success function and you have to put the data using html function like this $("#my-div").html(data). it will directly load the data in html format inside the div.
I would like to submit some data. Weather this be using a form, or onClick execute some AJAX, I'm not sure.
For example, I have this input
<input id='inpamount' type="text" name="amount" value="2.00" onkeyup="pad();validatemin();product()">
Now say if I wanted to send this data to a PHP file, using post ( i could just add a form). But then, without reloading the page (I can do this), how could I fetch a response (using AJAX).
Essentially I would like a user to be able to press a button, then to submit the inputs to a php file, and then get the output and assign it to a variable (I know how to do this, I just want to be able to get the data.
The php execute takes a few centiseconds because it comunicates with SQL. (If this matters).
I have considered using invisible forms but it didn't seem to work
If anyone could point me in the right direction, that would be great.
Something like this?
$(document).ready(function(e) {
$('#inpamount').click(function() {
var data = 'myValue=' + $(this).val();
$.ajax({
url: "yourscript.php",
type: "POST",
data: data,
cache: false,
success: function(scriptOutput) {
//handle the result
}
}
});
return false;
});
});
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 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.
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.