inserting mistake in sql

Question

I want to insert 2 values in the database, I want to give each $userid a $teamid so that users are connected to a team.

but whenever I run it, I won't see it in the database.

<?php
if(isset($_POST['submit']))

{
$db = mysql_connect("localhost","root","usbw");
mysql_select_db("the red socks",$db) or die ("fout: openen db mislukt");

$userid = $_POST['userid'];
$teamid = $_POST['teamid'];

 $query = "INSERT INTO team_users (userid, teamid, ) VALUES ('$userid', '$teamid')";
 echo $query;
 $result = mysql_query($query);
 echo $result;
}
else
{
?>
<form method='post' action=''>
<table>
<tr><td>invoegen<br></td></tr>
<tr><td>User ID</td></tr>
<tr><td><input name='userid'></td></tr>
<tr><td>Team ID</td></tr>
<tr><td><input name='teamid'></td></tr>
<tr><td><input name='submit' type='submit' value='inloggen'>
<input type='reset' name='reset'value='wissen'></td></tr>
</table>
</form>
<?php
}
?>

the echo I get from the query when I type 7 into userid and 2 in teamid:

INSERT INTO team_users (userid, teamid, ) VALUES ('7', '2')

You have a , before your ) ... If you'd done something like `mysql_query(...) or die(mysql_error())` you might have seen an error message. — ta.speot.is, Jun 19 '14 at 13:42
Add error reporting to the top of your file(s) `error_reporting(E_ALL); ini_set('display_errors', 1);` — Funk Forty Niner, Jun 19 '14 at 13:43
@ta.speot.is Thank you very much. I already thought it would have been a stupid mistake. — Chielarts, Jun 19 '14 at 13:45
Your present code is open to [**SQL injection**](http://stackoverflow.com/q/60174/). Use [**prepared statements**](http://www.php.net/manual/en/mysqli.quickstart.prepared-statements.php), or [**PDO**](http://php.net/pdo) with [**prepared statements**](http://php.net/pdo.prepared-statements). — Funk Forty Niner, Jun 19 '14 at 13:49
You're also inserting your IDs as strings. No need for quotes. — Dan, Jun 19 '14 at 13:50

score 1 · Accepted Answer · edited Jun 19 '14 at 13:47

1

change this

$query ="INSERT INTO team_users (userid, teamid, ) VALUES ('$userid','$teamid')";
                                               ^----remove the , here

to

$query ="INSERT INTO team_users (userid, teamid ) VALUES ('$userid', '$teamid')";

edited Jun 19 '14 at 13:47

Funk Forty Niner

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answered Jun 19 '14 at 13:44

echo_Me

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inserting mistake in sql

1 Answers1