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问题描述
我收到一条警告消息,我不明白为什么并且无法解决,(见下文)
I am getting a warning message which I don't understand why and unable to resolve, (see below)
Warning: Supplied argument is not a valid MySQL result resource in /detail.php on line 34
这是我的代码:
$rs = mysql_query($strSQL);
$strSQL = "SELECT * FROM <tablename> WHERE id=" . $_GET["serviceName"];
// Loop the recordset $rs
while($row = mysql_fetch_array($rs)) **(line 34) here ***
{
echo $row['ID']."<br />";
echo $row['serviceName']."<br />";
// Close the database connection
mysql_close();
?>
</dl>
<p><a href="li.php">Return to the list</a></p>
</body>
</html>
提前谢谢,我也没有得到这个网页上的任何数据,谢谢...singhy
thanks in advance, I am not getting any of the data on this webpage either,thanks...singhy
推荐答案
查询失败 - 您需要在 MySQL 中用引号将字符串括起来:
The query is failing - you need to wrap quotes around strings in MySQL:
$strSQL = "SELECT * FROM gu_service_cat WHERE id = '" .$_GET["serviceName"] ."'";
另外,$rs 应该低于 $strSQL...
plus, the $rs should be BELOW the $strSQL...
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