how to use JavaScript and and php to store a variable?

Question

i have a success function that has some data stored inside it:

function(receiverUserIds) {
console.log("IDS : " + receiverUserIds.request_ids);
}

this will log: IDS :123123213, 4645646654, 7897987989, ....

what i want to do is grab all those id's and store them in the database.

one way i was thinking to do it is by using ajax:

    function(receiverUserIds) {
        console.log("IDS : " + receiverUserIds.request_ids);
        $.ajax({
            type: "POST",
            url: "<?php echo $_SERVER['PHP_SELF']; ?>",
            friends_invite: receiverUserIds.request_ids,
            success: function(msg){
                /* alert( "Data Saved: " + msg ); */
            }
        });
}

and on the same page:

if(isset($_POST['friends_invite'])){
print_r($_POST['friends_invite']);
}

but it doesn't seem to work.

something might be wrong with either the ajax or i don't know. maybe you guys can suggest another way of doing this..??

any ideas?

Thanks

edit: if i enable the alert lert( "Data Saved: " + msg ); i get an alert so i know that the ajax is successful, but i don't see my $_POST being echoed out

This method should work. Define doesn't seem to work. What happens? Can you post a snippet of what the code looks like when you view source in your browser after it has been generated? — mellamokb
– mellamokb, Commented Oct 4, 2011 at 18:18
As the javascript being written out with php? If not, _SERVER[PHP_SELF] is unavailable. — Explosion Pills
– Explosion Pills, Commented Oct 4, 2011 at 18:20
if i enable the alert lert( "Data Saved: " + msg ); i get an alert so i know that the ajax is successful, but i don't see my $_POST being echoed out — Patrioticcow
– Patrioticcow, Commented Oct 4, 2011 at 18:24
possible duplicate of sending information back and forth with AJAX — hakre
– hakre, Commented Oct 4, 2011 at 18:28

mellamokb · Accepted Answer · 2011-10-04 18:20:43Z

3

This method should work fine. First thing that sticks out is that you should be using the data option to pass the data. See the specs of the documentation for more info.

$.ajax({
    ...
    data: { friends_invite: receiverUserIds.request_ids },
    ...
});

answered Oct 4, 2011 at 18:20

mellamokb

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8 Comments

mellamokb Over a year ago

Use fiddler (fiddler2.com/fiddler2) or something to track what data is flowing back and forth instead of just saying "doesn't work". What doesn't work? What's happening? What's not happening? Why? Go through the process step-by-step systematically and see how far the data is going before something fails.

Patrioticcow Over a year ago

if i enable the alert 'alert( "Data Saved: " + msg );' i get an alert so i know that the ajax is successful, but i don't see my '$_POST' being echoed out

mellamokb Over a year ago

This happens with or without the change I suggested in my post (or both)?

Patrioticcow Over a year ago

actually with your change i get this in the alert Data Saved: Array( [0]=>123123123 )

mellamokb Over a year ago

So the data is now being passed. That's what you wanted right?

|

dqhendricks · Accepted Answer · 2011-10-04 18:21:09Z

1

friends_invite: receiverUserIds.request_ids

I don't belive JQuery works this way. It would have to be changed to something like:

data: 'friends_invite=' + receiverUserIds.request_ids

answered Oct 4, 2011 at 18:21

dqhendricks

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Comments

Vaibhav Naranje · Accepted Answer · 2011-10-04 18:25:15Z

1

Use this-

       $.ajax({
            type: "POST",
            url: "<?php echo $_SERVER['PHP_SELF']; ?>",
            'data':{friends_invite: receiverUserIds.request_ids},
            success: function(msg){
                /* alert( "Data Saved: " + msg ); */
            }
        });

answered Oct 4, 2011 at 18:25

Vaibhav Naranje

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