这是我的CODE,我需要来自2个表的数据。第一个:ps_customer(id_customer,firstname,lastname,email)
和第二个:ps_adress(phone_mobile)
。
我收到了错误:
警告:mysql_query()期望参数2是资源,第19行/home/domain/public_html/k2.php中给出的对象
这是我的代码:
<?php
$conn=mysqli_connect("localhost","login","pass","dbNAME");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$gender = 6;
$sql =
"SELECT
ps_customer.id_customer, ps_customer.firstname, ps_customer.lastname, ps_customer.email, ps_adress.phone_mobile
FROM
ps_customer, ps_adress
ps_customer.id_gender = '$gender' ";
$rs=mysql_query($sql,$conn) or die(mysql_error());
echo '<table width="100%" border="0" cellspacing="5" cellpadding="5">';
while($result=mysql_fetch_array($rs))
{
echo '<tr>
<td>'.$result["id_customer"].'</td>
<td>'.$result["firstname"].'</td>
<td>'.$result["lastname"].'</td>
<td>'.$result["email"].'</td>
<td>'.$result["phone_mobile"].'</td>
</tr>';
}
echo '</table>';
?>
答案 0 :(得分:1)
首先,你必须在两个表之间建立一个关系,比如有一个关键字段,第二个你应该加入表,然后你可以继续你的SELECT查询。