if($_GET['confirm']){
$coupon_id = $_GET['confirm'];
$query = mysql_query("SELECT * FROM purchases WHERE coupon_id = '$coupon_id'");
while($rows = mysql_fetch_array($query)){
$user_id = $rows['user_id'];
$query = mysql_query("INSERT INTO purchases_confirm VALUES(NULL,'$coupon_id','$user_id' ");
if($query){
echo "inserting new values to database....done !";
}
}
exit;
}
它输出:Warning:mysql_fetch_array():提供的参数不是一个有效的mysql结果资源…奇怪的是,如果我从命令行或phpmyadmin执行查询,它就会工作!!
最佳答案
当使用$query
时,不能在内部重新分配mysql_fetch_array
变量。
应该是:
if($_GET['confirm']){
$coupon_id = $_GET['confirm'];
$query = mysql_query("SELECT * FROM purchases WHERE coupon_id = '$coupon_id'");
while($rows = mysql_fetch_array($query)){
$user_id = $rows['user_id'];
$query2 = mysql_query("INSERT INTO purchases_confirm VALUES(NULL,'$coupon_id','$user_id' ");
if($query2){
echo "inserting new values to database....done !";
}
}
exit;
}
关于php - 奇怪的mysql_fetch_array错误,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/5329865/