Ajax, PHP, MySQL returning sql table

Question

I have the following Ajax code to send information from an HTML form to a PHP file.

   $(document).ready(function(){
         $('#txt').load( '../../do_comment.php' );
     });
     $(function(){
        $("#submit").click(function(e) {
            e.preventDefault();
            var name = $("#user_name").val();
            var comment = $("#user_comment").val();
            var ID = '2'; //must change for each post
            $.ajax({
            type: "POST",
            url: "../../do_comment.php",
            data: {user_name:name, user_comment:comment, ID:ID},
            success: function(){
                $('#txt').load( '../../do_comment.php' );
            },
            error:function(e){alert("it failed");}
            });
        }); 
});

In my PHP file I declare the variables like this:

$name = $_POST[user_name];
$comment = $_POST[user_comment];
$ID = $_POST[ID];

And correctly populate my database with this:

 if($_POST[user_comment] != Null) {
    $sql = "INSERT INTO $table_name (post_ID, user_name, comments)
    VALUES ('$ID','$name', '$comment')";

    $result = @mysql_query($sql,$connection) or die(mysql_error());

 }

The problem is none of the variables will echo any sort of value, and when I try to query the database it only works if I hard code the ID value in instead of using the variable.

$data = mysql_query("SELECT * FROM $table_name WHERE post_ID = 
 '".mysql_real_escape_string($ID)."'") or
    die(mysql_error());

In your $_POST invocations you should do $_POST['user_name']; instead of $_POST[user_name]; Notice the quotes. — Jon Rubins
– Jon Rubins, Commented Mar 8, 2013 at 20:39
Also, don't user mysql_query. Use PDO. Search PDO php and use it. — Jon Rubins
– Jon Rubins, Commented Mar 8, 2013 at 20:41
You should include in your post what is it your getting echoed back — rantsh
– rantsh, Commented Mar 8, 2013 at 20:47
that variables will echo back nothing, meaning they must be null. But if they are null, how come they are populating the database properly? — user2138541
– user2138541, Commented Mar 9, 2013 at 0:37

UnholyRanger · Accepted Answer · 2013-03-08 20:42:37Z

2

Use the following when gathering from $_GET/$_POST/$_REQUEST:

$name = $_POST['user_name'];
$comment = $_POST['user_comment'];
$ID = $_POST['ID'];

Notice the tics. Proper syntax is $_POST[''].

Have you checked the database to make sure the proper values are being inserted?

Also, if the post_id is an integer, don't use tics

SELECT * FROM table WHERE post_ID = 1234

NOTICE: do not use MySQL_*, it has been deprecated in PHP 5.5. Use MySQLi or PDO. Watch out for SQL injections as well, especially when using MySQL_*.

answered Mar 8, 2013 at 20:42

UnholyRanger

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14 Comments

user2138541 Over a year ago

I added the tics into the declaration. No luck. My database is being filled with the proper values, which is why this does not make sense to me. Also if I do not use tics around the variable name in the select statement when using a variable, I receive an error.

UnholyRanger Over a year ago

what's your error? Integers do not use tics and may cause the query to return nothing

UnholyRanger Over a year ago

also, try echo $data; and see how the query was built

user2138541 Over a year ago

I can hardcode the integers in with no error. But when I try to use the ID variable with the select statement I receive a syntax error.

user2138541 Over a year ago

when I echo $data I get Resource id #4

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