Question

唯一的问题是，我的代码仅在wp_realcustomers表中至少有一个列表可用时返回结果，但我希望它返回结果，无论它是否在wp_realcustomers中有列表/数据，使它成为一种可选的。

这是我的代码：

"SELECT 
    usermeta1.user_id,
    usermeta2.meta_value AS firstname,
    usermeta3.meta_value AS lastname,
    usermeta4.meta_value AS phone1,
    usermeta5.meta_value AS phone2,
    mainuser.user_email AS email,
    COUNT(customers.id) AS numberofcustomers


FROM wp_usermeta usermeta1
JOIN wp_usermeta usermeta2 ON(usermeta1.user_id = usermeta2.user_id AND usermeta2.meta_key = 'first_name' )
JOIN wp_usermeta usermeta3 ON(usermeta1.user_id = usermeta3.user_id AND usermeta3.meta_key = 'last_name' )
JOIN wp_usermeta usermeta4 ON(usermeta1.user_id = usermeta4.user_id AND usermeta4.meta_key = 'tm_phone_1' )

JOIN wp_usermeta usermeta5 ON(usermeta1.user_id = usermeta5.user_id AND usermeta5.meta_key = 'tm_phone_2' )
JOIN wp_users mainuser ON(usermeta1.user_id = mainuser.ID )
JOIN wp_realcustomers customers ON(usermeta1.user_id = customers.team_member_id )


WHERE usermeta1.meta_key='teamleader_id' && usermeta1.meta_value='1'

Answer 1

在wp_realcustomers上使用LEFT JOIN

"SELECT 
    usermeta1.user_id,
    usermeta2.meta_value AS firstname,
    usermeta3.meta_value AS lastname,
    usermeta4.meta_value AS phone1,
    usermeta5.meta_value AS phone2,
    mainuser.user_email AS email,
    COUNT(customers.id) AS numberofcustomers


FROM wp_usermeta usermeta1
JOIN wp_usermeta usermeta2 ON(usermeta1.user_id = usermeta2.user_id AND usermeta2.meta_key = 'first_name' )
JOIN wp_usermeta usermeta3 ON(usermeta1.user_id = usermeta3.user_id AND usermeta3.meta_key = 'last_name' )
JOIN wp_usermeta usermeta4 ON(usermeta1.user_id = usermeta4.user_id AND usermeta4.meta_key = 'tm_phone_1' )

JOIN wp_usermeta usermeta5 ON(usermeta1.user_id = usermeta5.user_id AND usermeta5.meta_key = 'tm_phone_2' )
JOIN wp_users mainuser ON(usermeta1.user_id = mainuser.ID )
LEFT JOIN wp_realcustomers customers ON(usermeta1.user_id = customers.team_member_id )


WHERE usermeta1.meta_key='teamleader_id' && usermeta1.meta_value='1'"

如果你有数量，你必须有分组

"SELECT 
    usermeta1.user_id,
    usermeta2.meta_value AS firstname,
    usermeta3.meta_value AS lastname,
    usermeta4.meta_value AS phone1,
    usermeta5.meta_value AS phone2,
    mainuser.user_email AS email,
    COUNT(customers.id) AS numberofcustomers


FROM wp_usermeta usermeta1
JOIN wp_usermeta usermeta2 ON(usermeta1.user_id = usermeta2.user_id AND usermeta2.meta_key = 'first_name' )
JOIN wp_usermeta usermeta3 ON(usermeta1.user_id = usermeta3.user_id AND usermeta3.meta_key = 'last_name' )
JOIN wp_usermeta usermeta4 ON(usermeta1.user_id = usermeta4.user_id AND usermeta4.meta_key = 'tm_phone_1' )

JOIN wp_usermeta usermeta5 ON(usermeta1.user_id = usermeta5.user_id AND usermeta5.meta_key = 'tm_phone_2' )
JOIN wp_users mainuser ON(usermeta1.user_id = mainuser.ID )
LEFT JOIN wp_realcustomers customers ON(usermeta1.user_id = customers.team_member_id )


WHERE usermeta1.meta_key='teamleader_id' && usermeta1.meta_value='1'
group by  usermeta1.user_id, usermeta2.meta_value , 
          usermeta3.meta_value , usermeta4.meta_value ,
          usermeta5.meta_value , mainuser.user_email "

Answer 2

使用left outer join代替join - 更多详情here

如何JOIN表可选？

2 个答案: