我无法弄明白,这不是错误,但是当我试图回应查询的结果时:
echo "SELECT * FROM `active_customers` WHERE `cus_email`='".$k."' AND `cus_product`='".$_GET['product']."'" . "<br />";
$q = mysql_query("SELECT * FROM `active_customers` WHERE `cus_email`='".$k."' AND `cus_product`='".$_GET['product']."'");
$a = mysql_fetch_array($r);
$n = mysql_num_rows($q);
echo "Number of rows:" . $n;
echo $a['cus_id'];
echo $a['cus_email'];
帖子数量返回正确的数字,即1,这些值 $ a [&#39; cus_id&#39;]; 和 $ a [&#39; cus_email& #39;]; 根本没有通过,我甚至在phpmyadmin中做了一个mysql查询,它在那里工作。
谁能看到我做错了什么?
答案 0 :(得分:0)
正确的代码可以是
echo "SELECT * FROM `active_customers` WHERE `cus_email`='".$k."' AND `cus_product`='".$_GET['product']."'" . "<br />";
$q = mysql_query("SELECT * FROM `active_customers` WHERE `cus_email`='".$k."' AND `cus_product`='".$_GET['product']."'");
$n = mysql_num_rows($q);
echo "Number of rows:" . $n;
while($row = mysql_fetch_array($q, MYSQL_ASSOC)){
echo $row['cus_id'];
echo $row['cus_email'];
}
虽然我建议将mysqli或PDO用于sql数据库