[SOLVED] Can't figure why my form is not inserting...

[pneudralics]

<?php
$cookieid = $_COOKIE['id'];
$postcomment = $_POST['postcomment'];

if (isset($_POST['submitpostcomment'])) {

if (!empty($_POST['postcomment'])){

	$postcommentq = "INSERT INTO usercomments (from, to, comment, date, approve) VALUES ('$cookieid', '$getid', '$postcomment', NOW(), 'N')";
	if (mysql_query($postcommentq)) {
	echo "<font color=\"red\">Comment posted. Waiting for approval.</font>";
	}
	else {
	echo "<font color=\"red\">There was an issue posting the comment.</font>";
	}

}
else {
echo "<font color=\"red\">Post comment is empty.</font>";
}

}//End isset submitpostcomment

?>
<form action="comment.php" method="post">
Post Comment
<br />
<textarea name="postcomment" cols="30" rows="10">
</textarea>
<br />
<input type="submit" name="submitpostcomment" value="Submit" />
</form>

[dennismonsewicz]

are you getting any errors?

 

Also, where is $getid being set at?

[pneudralics]

No errors I just keep getting: There was an issue posting the comment.

It also does not insert into the database.

 

$getid = $_GET['id']; //Top of page

 

[dennismonsewicz]

try echoing out your SQL command

 

echo $postcommentq; 

[jackpf]

And put or die(mysql_error()); after your query/connections.

[pneudralics]

No errors.

echo shows data are in the value section.

[BobcatM]

The only thing I can see just make sure the usercomments table is the right DB table, and all the fields are correct.

[dennismonsewicz]

try doing this to your postcomment var

 

$postcomment = mysql_real_escape_string($_POST['postcomment']);

[jackpf]

If you put or die(mysql_error()); after your query and didn't get any errors then it must be inserting.

[pneudralics]

Tried the escape string nothing happened.

 

This is how I displayed the error:

	if (mysql_query($postcommentq).mysql_error()) {
	echo "<font color=\"red\">Comment posted. Waiting for approval.</font>";
	}

[dennismonsewicz]

Try this

 

 $postcommentq = mysql_query("INSERT INTO usercomments (from, to, comment, date, approve) VALUES ('$cookieid', '$getid', '$postcomment', NOW(), 'N')")or die(mysql_error());
      if ($postcommentq) {

[pneudralics]

I get the following:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'from, to, comment, date, approve) VALUES ('31', '31', 'testtexttesttext' at line 1

 

Oh and I'm using the latest XAMPP

[Philip]

from and to are reserved words that needs to be surrounded by backticks. (I find it good practice to put backticks around all of the col/table names)

 

 

$postcommentq = mysql_query("INSERT INTO `usercomments` (`from`, `to`, `comment`, `date`, `approve`) VALUES ('$cookieid', '$getid', '$postcomment', NOW(), 'N')")or die(mysql_error());

 

Also, make sure to clean the $cookieid/$getid variables

[dennismonsewicz]

XAMPP is a great program.

 

Kudos KingPhillip -> I bow before thee

[pneudralics]

from and to are reserved words that needs to be surrounded by backticks. (I find it good practice to put backticks around all of the col/table names)

 

 

$postcommentq = mysql_query("INSERT INTO `usercomments` (`from`, `to`, `comment`, `date`, `approve`) VALUES ('$cookieid', '$getid', '$postcomment', NOW(), 'N')")or die(mysql_error());

 

Also, make sure to clean the $cookieid/$getid variables

 

Thanks. The backtick solved the problem.