mysql_fetch_array
不起作用。一切都很美好。我不知道我在哪里做错了什么。执行mysql_fetch_array时出错
$sql = "SELECT * FROM `$tbl_name` limit $start,$limit";//if echo gives o/p Resource id #14
$resultw = mysql_query($sql);
while($gup=mysql_fetch_array($resultw))//if echo gives o/p Array
{
//if echo $gup['to']; gives o/p [email protected];
$anusha=mysql_query("select * from users where email='$gup[to]'");//if echo gives o/p Resource id #15
while($resulter = mysql_fetch_array('$anusha'))//here is what iam getting error Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in
{
}
}
有人能帮我找出我的错误在我的mysql_fetch_array
?
** mysql_fetch_array($ anusha)** – HoangHieu
由于某种原因,你包裹在单引号'$ anusha'。不要这样做。 – ElefantPhace
我还没有使用 –