Passing PHP variable with jQuery

Question

I need an assistance on how to transfer this value $row[COMPONENT] in my class1.php into another page process_class.php with jQuery using post method.

i did this but it seems doesnt work ,

 $('.cuttingCheckbox').change(function() { 
 if (this.checked) { 
 $.post('process_class.php', 
 { comp : $($row[COMPONENT]).val(), comp_id : $($row[ID]).val() },
 function(response) { 
 this.setAttribute("disabled", true), alert(comp,comp_id); }); 
 } 
});

Does anyone willing to help me ?

• rockStar: you would need ajax for that. api.jquery.com/jQuery.ajax
• doniyor: You can also save it into session and catch it back in another php file. it doesnot have to be form..
• Chriskonz: i did this but it seems doesnt work , $('.cuttingCheckbox').change(function() { if (this.checked) { $.post('process_class.php', { comp : $($row[COMPONENT]).val(), comp_id : $($row[ID]).val() }, function(response){ this.setAttribute("disabled", true), alert(comp,comp_id); }); } });
• Chriskonz: @doniyor I already have the variable like this, $row[COMPONENT]. would that possible to save this in a session ? This variable dynamically changes as the database changes
• doniyor: @Chriskonz, sure this is possible. i will show you in a second how to do this

Answer 1

you can save that $row[COMPONENT] into session like this:

$_SESSION['row_component'] = $row[COMPONENT];

and in your next page, you just retrieve it:

$row = $_SESSION['row_component'];

As Jonast said (thanks dude), you should initiate the session first at the top of your php file: session_start();

• Jonast92: Message to OP: Make sure you got session_start(); at the top of every PHP page so that sessions will work. (It must appear before any output is sent to the browser from the server). That is, if you use this approach, which you should.
• Chriskonz: @doniyor This is my implementation. and it still doesnt work even the alert... pastebin.com/mSVbF0fv pastebin.com/RS6Y1Ac3