Question

ticket                      
+----------+--------+
| ticketID | assign |
+----------+--------+
|     1015 | NULL   |
|     1020 | James  |
|     1021 | Nick   |
+----------+--------+

staffinfo
+---------+-------+
| staffID | staff |
+---------+-------+
|       1 | Jane  |
|       2 | James |
|       3 | Nick  |
|       4 | Cole  |
+---------+-------+

SELECT staff,COUNT(*) as count FROM staffinfo,ticket
WHERE ticket.assign = staffinfo.staff
GROUP BY staff

result:
    +-------+-------+
    | staff | count |
    +-------+-------+
    | James |     1 |
    | Nick  |     1 |
    +-------+-------+

工作正常，但事实上我需要像：

+-------+-------+
| staff | count |
+-------+-------+
| James |     1 |
| Nick  |     1 |
| Jane  |     0 |
| Cole  |     0 |
+-------+-------+

COUNT不计算表中不存在的记录，因为我刚刚开始学习SQL，我想问一下是否可以计算上述结果？

Answer 1

您应该使用LEFT JOIN

SELECT  a.staff,    COUNT(b.assign) as count 
FROM    staffinfo a
        LEFT JOIN ticket b
            ON b.assign = a.staff
GROUP   BY a.staff

SQLFiddle Demo

要充分了解联接知识，请访问以下链接：

Visual Representation of SQL Joins

Answer 2

使用LEFT JOIN

LEFT JOIN关键字返回左表（table_name1）中的所有行，即使右表（table_name2）中没有匹配项也是如此。

SELECT staffinfo.staff, count(ticket.assign)
FROM staffinfo
LEFT JOIN ticket
ON ticket.assign =staffinfo.staff
GROUP BY staffinfo.staff

LEFT JOIN关键字返回左表（staffinfo）中的所有行，即使右表（票证）中没有匹配项也是如此。

如何连接两个表，保持不符合JOIN条件的行？

2 个答案: