-4
我想在數據庫中插入2個值,我想給每個$ userid一個$ teamid,以便用戶連接到一個團隊。在sql中插入錯誤
但是每當我運行它,我都不會在數據庫中看到它。
<?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
}
?>
回波餘從查詢得到當我在teamid鍵入7成用戶ID和2:
INSERT INTO team_users(用戶ID,teamid,)VALUES( '7', '2')
爲什麼SQL-Server標記? – Frazz
你有一個,在你之前)...如果你做了'mysql_query(...)或死(mysql_error())'''你可能看到了錯誤信息。 –
刪除最後一個逗號'(userid,teamid,)' –