How to echo a variable from jquery POST to another page

Question

How to use the a variable that is posted from page1 to page2 from jquery.I am using the following scenario

page1.php

    if ( this.checked ) { 
            $.ajax({
                url: 'page2.php',
                type: "POST",
                data: ({name: 145}),
                success: function(data){
               //$("#result").html(data);
                }
            });        

        }

This is not working why,if i separatly opens a page2.php and wants further usage of the post data

    Page2.php
    <?php
    $userAnswer = $_POST['name'];    
    if ($_POST['submit']){
         echo $userAnswer; 
    }       

    ?>
    <form method="post" action="page2.php">
<input type="submit" value="submit" name="submit" />
</form>

if i open page2.php and presses the submit button,the posted value from page1 is not displaying

Do var_dump($_POST) to see the contents. If the page reloads then there's some Javascript error, try debugging it (Ctrl + Shift + J). — ᴍᴇʜᴏᴠ
– ᴍᴇʜᴏᴠ, Commented Dec 21, 2012 at 19:29

gpojd · Accepted Answer · 2012-12-21 19:29:37Z

1

You aren't sending the "submit" data.

if ( this.checked ) { 
    $.ajax({
        url: 'page2.php',
        type: "POST",
        data: ({name: 145, submit: "submit"}), // note the submit property
        success: function(data){
            //$("#result").html(data);
        }
    });        
}

I just noticed this:

if i open page2.php and presses the submit button,the posted value from page1 is not displaying

That will never happen. The value that you post from an ajax request will not persist to a second request coming from a different page. You want to store that in some sort of persistant storage (e.g. a database), not a variable that will be lost once the request is finished.

answered Dec 21, 2012 at 19:29

gpojd

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5 Comments

OpenSorceress Over a year ago

"Not working dear" is not useful feedback. Please explain what is not working or what the outcome was.

Creative Solution Over a year ago

any other alternative i just need to store data temporarily

ᴍᴇʜᴏᴠ Over a year ago

try using php sessions? so the data is kept server-side between requests

Creative Solution Over a year ago

so if i need to display the data,how i do,echo is not working??

ᴍᴇʜᴏᴠ Over a year ago

Start a session by calling sesion_start() at the beginning of your scripts. Then assign any variable inside of it, e.g. $_SESSION['my_variable'] = 'some value';. You can now call that variable anywhere with echo $_SESSION['my_variable']. You can also see the $_SESSION contents by calling print_r($_SESSION);. Hope this helps

rachids · Accepted Answer · 2012-12-22 16:01:53Z

0

Why don't you use ".load()" function instead of post ?

$('#resultat').load('page2.php',{name: 145});

edited Dec 22, 2012 at 16:01

answered Dec 22, 2012 at 15:53

rachids

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Comments

Arunkumar Srisailapathi · Accepted Answer · 2012-12-22 16:04:51Z

0

if ( this.checked ) {

$.post("page2.php", {name1: variable1, name2: variable2, ...}, function (result) {

});
}

page2.php

$storeVariable1 = $_POST['name1'];
$storeVariable2 = $_POST['name2'];

answered Dec 22, 2012 at 16:04

Arunkumar Srisailapathi

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