Question

是的，现在我知道JOIN，但仍然没有解决这个问题。

$ gender = 6;

$ result = mysqli_query（$ con，＆＃34; SELECT * FROM ps_customer c LEFT JOIN   ps_adress a ON a.customer_id = c.customer_id WHERE   c.id_gender =＆＃39; $性别＆＃39;＆＃34）;

的var_dump（$结果）;

mysqli_close（$ CON）;

这是我的代码，但结果为：

我尝试了var_dump来检查为什么我无法打印数据。

我的代码的数据打印是否正确？

 echo "<h1>People</h1>";
echo "<table border='1'>
<tr>
<th>ID</th>
<th>Name</th>
<th>Lastname</th>
<th>Email</th>
<th>Company</th>
<th>Phone</th>
</tr>";



$result = mysql_query($query) or die(mysql_error());
echo var_dump($result);
while($row = mysql_fetch_array($result)){
  echo "<tr>";
  echo "<td>" . $row['id_customer'] . "</td>";
  echo "<td>" . $row['firstname'] . "</td>";
  echo "<td>" . $row['lastname'] . "</td>";
  echo "<td>" . $row['email'] . "</td>";
  echo "<td>" . $row['company'] . "</td>";
  echo "<td>" . $row['mobile_phone'] . "</td>";
  echo "</tr>";
}

Answer 1

简单加入即可。

SELECT *
FROM ps_customer c
    LEFT JOIN ps_address a ON a.customer_id=c.customer_id
WHERE c.id_gender = '$gender'

Answer 2

SELECT *
FROM adress a
INNER JOIN customer c ON a.customer_id = c.id
WHERE c.id_gender = $gender

Answer 3

您需要LEFT JOIN（如果客户没有地址记录）。假设两个表都具有相同的列名称＆＃34; customer_id＆＃34;识别客户，您可以使用：

SELECT ps_customer.name,
       ps_customer.lastname,
       ps_customer.company,
       ps_address.phone,
       ps_address.mobile
FROM ps_customer
LEFT JOIN ps_address ON ps_address.customer_id = ps_customer.customer_id
WHERE ps_customer.id_gender = '$gender'

但要小心点。如果每个客户有多个地址记录，您将获得包含每个地址记录的相同客户数据的多行。如果是这种情况，则需要更复杂的查询：）

Answer 4

您的查询应该是

"select cus.name, cus.lastname, cus.company from ps_customer cus, ps_address add where id_gender='".$gender."' and cus.customer_id=add.customer_id"

来自2表的mysql数据

4 个答案: