本文共 811 字，大约阅读时间需要 2 分钟。

题目：（中等）

标签：SQL

解法一：

SELECT I.invoice_id,       C.customer_name,       I.price,       IFNULL(U1.contacts_cnt, 0) AS contacts_cnt,       IFNULL(U2.trusted_contacts_cnt, 0) AS trusted_contacts_cntFROM Invoices AS I         LEFT JOIN     Customers AS C ON I.user_id = C.customer_id         LEFT JOIN     (SELECT user_id,             contact_name AS customer_name,             COUNT(contact_name) AS contacts_cnt      FROM Contacts      GROUP BY user_id) AS U1 ON I.user_id = U1.user_id         LEFT JOIN     (SELECT user_id,             COUNT(contact_name) AS trusted_contacts_cnt      FROM Contacts      WHERE contact_email IN (SELECT email FROM Customers)      GROUP BY user_id) AS U2 ON I.user_id = U2.user_idORDER BY I.invoice_id;

转载地址：https://dataartist.blog.csdn.net/article/details/110421387 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！