我的MySQL数据库中有两个表。
machine_tbl
|---------------------|------------------|
| section | machine |
|---------------------|------------------|
| 2400-TWO-001 | AT-TWB-001 |
|---------------------|------------------|
| 2400-TWO-001 | AT-TWB-002 |
|---------------------|------------------|
| 2400-TWO-001 | AT-TWB-003 |
|---------------------|------------------|
| 2400-TWO-001 | AT-TWB-004 |
|---------------------|------------------|
| 2400-TWO-002 | AT-TWB-005 |
|---------------------|------------------|
| 2400-TWO-002 | AT-TWB-006 |
|---------------------|------------------|
| 2400-TWO-003 | AT-TWB-007 |
|---------------------|------------------|
open_notification_tbl
|---------------------|------------------|
| user | machine |
|---------------------|------------------|
| a1 | AT-TWB-001 |
|---------------------|------------------|
| a1 | AT-TWB-001 |
|---------------------|------------------|
| a2 | AT-TWB-001 |
|---------------------|------------------|
| a3 | AT-TWB-002 |
|---------------------|------------------|
| a1 | AT-TWB-002 |
|---------------------|------------------|
| a4 | AT-TWB-003 |
|---------------------|------------------|
| a4 | AT-TWB-004 |
|---------------------|------------------|
我想要的是为特定的部分获取计算机数来比较这两个表。
例如,2400-TWO-001部分有3台AT-TWB-001机器,2400-TWO-001部分有2台AT-TWB-002机器。喜欢明智的。如果没有机器那么计数应为零。然后我根据使用php创建一个JSON。
查询...
SELECT DISTINCT a.section, a.machine, count(b.user) AS countX FROM machine_tbl AS a LEFT JOIN open_notification_tbl AS b ON a.machine= b.machine WHERE section = ?
答案 0 :(得分:0)
我想你想要:
SELECT COUNT(DISTINCT a.machine) AS countX
FROM machine_tbl m LEFT JOIN
open_notification_tbl o
ON m.machine = o.machine
WHERE section = ?;