PHP and JQuery Post communication errors

Question

Ok, so i tried multiple of solutions but nothing seems to work for me, (or i'm just too stupid to get it).

I have such jquery code:

Jquery:

var current_id = $(this).attr("id");
var dataString = {"id" : current_id};
$.post("popup.php" , {"id" : $(this).attr("id")}, function(msg) { alert(msg);} );

I know few lines are useless here but i tried all options,

PHP:

$tmp = $_POST['id'];
if($tmp == "kontakt"){
echo '<p>UDANE PRZESŁANIE</p>';}
else{
echo '<p>NIEUDANE PRZESŁANIE</p>';
}

When i click on id=kontakt php returns error :

 Notice: Undefined index: id in D:\***\popup.php on line 2

but alert sends back

<p>UDANE PRZESŁANIE</p>

Edit:

Ok, Maybe i haven't make my question clear. I don't care if php error is shown or not, I just want to know why it shows up in the first place.

I wonder why do you first assign a dataString, but then essentially throw it away - and calculate the data again? Why is {"id": $(this).attr("id")} code used twice? — raina77ow
– raina77ow, Commented May 21, 2012 at 22:15
@raina77ow: I think these are relicts from what the asker has tried... — Lukx
– Lukx, Commented May 21, 2012 at 22:20
But why try this instead of just logging the current_id value, I wonder? With console.log(current_id) or something like that. ) — raina77ow
– raina77ow, Commented May 21, 2012 at 22:21

LZNPDE · Accepted Answer · 2012-05-21 23:44:50Z

1

It's just a PHP warning - it's basically saying that $_POST has no index called 'id'.

You can get rid of it in several ways...

A shorted version of Austin Allover's answer:

if(!isset($_POST['id')) $tmp = '';
Turn of warnings in your script or your php config file, see: http://php.net/manual/en/function.error-reporting.php (Look for E_WARNING)
You might also want try $.post("popup.php" , {"id" : "BLAHBLAHBLAH"}, function(msg) { alert(msg);} ); - just in case $(this).attr("id") doesn't contain anything...

edited May 21, 2012 at 23:44

answered May 21, 2012 at 23:15

LZNPDE

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score 0 · Accepted Answer · 2012-05-21 22:57:11Z

0

Declare $tmp var this way:

$tmp = (isset($_POST['id'])) ? $_POST['id'] : "";

Or ...

if(isset($_POST['id'])) { $tmp = $_POST['id']; } else { $tmp = ""; }

If you don't understand what this code does...

if post id is set, then declare $tmp as the post id, else set $tmp as "". This will take away the PHP notice.

edited May 21, 2012 at 22:57

answered May 21, 2012 at 22:12

user1403947

1 Comment

Mateutek Over a year ago

Ok, I got rid of the error but the echo still returns $tmp="" i don't know why the $_POST['id'] isn't working

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