Question

我有两个不同的表（wp_usermeta＆amp; my cutom table-sia_licence）：

wp_usermeta表： - enter image description here sia_licence表（自定义）： - 我希望最终数据表是这样的： -

我们尝试使用

$data = $wpdb->get_results("SELECT * FROM wp_usermeta, sia_licence WHERE  wp_usermeta.user_id = sia_user_id  
        AND wp_usermeta.meta_key = 'first_name'
        AND wp_usermeta.meta_key = 'last_name'
        AND wp_usermeta.meta_key = 'phone_number'
        AND wp_usermeta.meta_key = 'email'
        AND wp_usermeta.meta_key = 'address'");

但得到空白数组..

Answer 1

试试这个：

SELECT * FROM wp_usermeta wu
 INNER JOIN sia_licence sl ON wu.user_id = sl.sia_user_id
 WHERE <you can specify your filter here>;

Answer 2

sla_user_id是否引用user_id？

如果是这样你想使用像这样的JOIN ......

SELECT um.*,sl.select_licence,sl.visa_expiry,sl.licence_number,sl.sia_licence,sl.driving_licence FROM `wp_usermeta`um JOIN `sia_licence`sl ON um.`user_id`=sl.`sia_user_id`;

你也可以像这样添加一个where子句

SELECT um.*,sl.select_licence,sl.visa_expiry,sl.licence_number,sl.sia_licence,sl.driving_licence FROM `wp_usermeta`um JOIN `sia_licence`sl ON um.`user_id`=sl.`sia_user_id` WHERE um.`user_id`=101;

如果需要的话。在上面的例子中，它只返回用户ID = 101。

在这种情况下，您也可以使用SELECT *，但通常这是一个坏主意，因为如果两个表之间有任何不明确的列（具有相同名称的列），它将给您一个错误，但在您的看起来你应该没问题：

SELECT * FROM `wp_usermeta`um JOIN `sia_licence`sl ON um.`user_id`=sl.`sia_user_id`;

Answer 3

您需要在wp_usermeta上执行条件聚合，以便按user_id转动所有属性（名字，姓氏......），然后JOIN转换sia_licence产生所需的输出

SELECT q.user_id, q.first_name, q.last_name, q.email, q.phone_number, q.address,
       l.select_licence, l.visa_expiry, l.licence_number, l.sia_licence, l.driving_licence
  FROM 
(
  SELECT user_id, 
         MAX(CASE WHEN meta_key = 'first_name'   THEN meta_value END) first_name,
         MAX(CASE WHEN meta_key = 'last_name'    THEN meta_value END) last_name,
         MAX(CASE WHEN meta_key = 'phone_number' THEN meta_value END) phone_number,
         MAX(CASE WHEN meta_key = 'email'        THEN meta_value END) email,
         MAX(CASE WHEN meta_key = 'address'      THEN meta_value END) address
    FROM wp_usermeta m
   WHERE EXISTS
  (
      SELECT *
        FROM sia_licence
       WHERE sia_user_id = m.user_id
  )
   GROUP BY user_id
) q JOIN sia_licence l
    ON q.user_id = l.sia_user_id

或先加入然后合并

SELECT q.user_id, 
       MAX(CASE WHEN meta_key = 'first_name'   THEN meta_value END) first_name,
       MAX(CASE WHEN meta_key = 'last_name'    THEN meta_value END) last_name,
       MAX(CASE WHEN meta_key = 'phone_number' THEN meta_value END) phone_number,
       MAX(CASE WHEN meta_key = 'email'        THEN meta_value END) email,
       MAX(CASE WHEN meta_key = 'address'      THEN meta_value END) address,
       MAX(select_licence)  select_licence,
       MAX(visa_expiry)     visa_expiry,
       MAX(licence_number)  licence_number,
       MAX(sia_licence)     sia_licence,
       MAX(driving_licence) driving_licence
  FROM wp_usermeta q JOIN sia_licence l
    ON q.user_id = l.sia_user_id
 GROUP BY q.user_id

Wordpress - 从多个表中获取数据

3 个答案: