$cmd= "update..........";
$query = mysql_query( $cmd );
if( $query )
{
echo "success";
}
else
{
echo "false";
}
$query = mysql_query( $cmd );
if( $query )
{
echo "success";
}
else
{
echo "false";
}
你的查询连接结果标示都没有,怎么判断的?
http://www.52codes.com/code.php?id=21888
update 代码
$result= mysql_query( $cmd );
if( !$result)
{
echo "执行失败";
exit();
}
我就是这么判断的.为什么执行成功了,还返回执行失败呢?
$cmd= "update..........";
$query = mysql_query( $cmd );
if( $query ){
$result = mysql_fetch_assoc($query);
var_dump($result);
}else{
echo "false";
} var_dump或者是print_r输出个...
$cmd= "update..........";
$query = mysql_query( $cmd );
var_dump($query);跟上面那个对比一下你就知道了....