Question

我想在数据库中插入2个值，我想给每个$ userid一个$ teamid，以便用户连接到一个团队。

但每当我运行它时，我都不会在数据库中看到它。

<?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
}
?>

当我输入7到userid时，我从查询中得到的回声，在teamid中输入2：

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

Answer 1

更改此

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

到

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

在sql中插入错误

1 个答案: