PHP / Ajax cannot send value

Question

I have problem with ajax and php.I have next code

$("form").submit(function() {
 var poruka = $("#poruka").val();
  $.ajax({
        type: "GET",
        url: "/ajax/gather/send/"+ poruka + "/'.$id.'",
        success: function(){
         loaduj('.$id.');
         alert("Your message is "+poruka);
        },
        error: function(){
            alert("fail");
        }
    });
});

"poruka" is the message and it's taken from form

<form action='javascript:;'>
<input type='text' name='poruka' id='poruka' placeholder='Poruka..' style='width:100%;'>
<input type='button' name='button' hidden>
</form>

And when i try to send to php it won't insert that into database :

Php code :

$text = $_POST['text'];
$id = $_POST['id'];
if($text != ""){
  mysql_query("INSERT INTO messages (text,id2) VALUES ('$text','$id')") or     die(mysql_error());
}

And it can't work..How can i fix it ? Thanks

lukas.pukenis · Accepted Answer · 2013-03-07 17:13:15Z

2

You use POST instead of GET. In your AJAX form you specify that you use GET, but in PHP you use $_POST superglobal.

Change $_POST into $_GET in your PHP code and it should work :)

Also if you do not specify method in your tag, the default is GET.

From experience, in case if the data from the form is not written/updated in/to the database use

echo mysql_error();

after the MySQL query and also open the developer tools on your browser and inspect in the network panel. Developer tools can be opened using F12 and with CTRL+SHIFT+I if your F12 is not working due to the coffee spilled on it. Network panel shows the data being sent, the method used and every other detail of a network call. Also there you can see the response of your server side script which in this case would return a mysql error message because you echo'ed it out.

edited Mar 7, 2013 at 17:13

answered Mar 11, 2012 at 16:48

lukas.pukenis

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1 Comment

lukas.pukenis Over a year ago

No problem, hope that helped! :)

Grigorash Vasilij · Accepted Answer · 2012-03-11 16:54:09Z

0

The usual deal is to first locate the problem, then post:

Form not calling JS
JS not making the request
PHP not getting the variable

SQL injection detected

answered Mar 11, 2012 at 16:54

Grigorash Vasilij

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5 Comments

lukas.pukenis Over a year ago

SQL injection is not the case at this point. He just wants to have a simple code working, not to implement full functionality when even a simple thing does not work as expected

Grigorash Vasilij Over a year ago

i can not be speaking on the part of the author - what he does or does not want, but usually you don't write code to keep it in a sealed box with you being the only person to ever open it. There can be trouble with the given SQL; it's a mere warning.

lukas.pukenis Over a year ago

That's good, however I believe your answer does not suit the question here :)

Grigorash Vasilij Over a year ago

SQL injection detected is not answer. The answer is debug first O-)

lukas.pukenis Over a year ago

Yes, debugging is one of the methods every programmer uses when something does not work. I believe the question author is fully aware of it :)

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