参数确实是true，但是还是有警告mysql_fetch_row() expects parameter 1 to be resource？

function checkCancelUserStop($phoneNum){	$con = mysql_connect("localhost", "root", "");	if(!$con)	{		echo "";		echo "";		return;	}		$result = mysql_select_db ("teleservice",$con);	if(!$result)	{		echo "";		echo "";		return;	}	$result = mysql_query("SET NAMES 'GBK'");	if(!$result)	{		echo "";		echo "";		return;	}	$query = "select state			  from phoneDetail			  where phoneDetail.phoneNum = $phoneNum";	if($result) 	{		echo "";		echo "";			} else {		echo "";		echo "";		return;	}	$oldState = "正常开通";	while($row = mysql_fetch_row($result))	{		echo "in while loop...";		$oldState = $row[0];		echo $oldState;	}	if($oldState == "主动停机") {		echo "yes";		return true;	} else {		echo "no";		return false;	}	mysql_close($con);}

问题：通过调试，已经可以成功查询到号码的原状态。此时$result = 1；但是后面fetch_row的时候，不能进入到while循环中，出现了警告：Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in H:\wamp\www\frr_cancelUserStop.php on line 58
请问：$result = 1确实是1（true）的啊，那为什么进不了循环？
怎么才能取得结果呢？
谢谢各位！

回复讨论(解决方案)

我知道怎么错了，少写了一句话。不好意思。

SQL 指令为执行

我也遇到了这种错误，能不能帮忙看下？
Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in D:\WWW\blog\show_pub.php on line 7
代码：
error_reporting(E_ALL^E_NOTICE);
session_start();
include "conn/conn.php";
$pubsql = "select * from tb_public where id= ".$_GET['id'];
$pubrst = mysql_query($pubsql,$link);
$pubrow = mysql_fetch_row($pubrst);
echo "

".$pubrow[2]."

";
?>