mysql - 返回 mysql 所在的所有行

Question

我有这个表:

tbl_1
+---+--------+
|ID | NAME   |
+---+--------+
| 1 | Name_1 |
| 2 | Name_2 |
| 3 | Name_3 |
+---+--------+

tbl_2
+---+----------+---------+
|ID | MENTIUNI | ID_NAME |
+---+----------+---------+
| 1 | abc      |   1     |
| 2 | abcd     |   2     |
| 3 | abc      |   1     |
| 4 | abcd     |   1     |
| 5 | abc      |   2     |
| 6 | abcd     |   3     |
+---+----------+---------+

SELECT
	tbl_1.NAME,
	count(tbl_2.MENTIUNI) AS NR_MENTIUNI

FROM tbl_1

LEFT JOIN tbl_2 ON tbl_1
	tbl_1.ID = tbl_2.ID_NAME

WHERE tbl_2.MENTIUNI = 'abc'

查询 ul de mai sus imi selecteaza din tbl_1.ID，numai cele care se regasesc in tbl_2 sub conditia tbl_2.MENTIUNI = 'abc'

Cum pot afisa toate id-urile din tbl_1 chiar daca am conditie 在哪里？

上面的查询从 tbl_1.ID 中选择我，仅在 tbl_2.MENTIUNI = 'abc' 条件下在 tbl_2 中找到的内容

即使我在 where 中有条件，如何显示 tbl_1 中的所有 id？

+------+-------------+
| NAME | NR_MENTIUNI |
+------+-------------+
|Name_1|      2      |
|Name_2|      1      |
|Name_3|      0      |
+------+-------------+

谢谢!

Answer 1

将条件从整个查询移至 JOIN 子句，因此它仅过滤要连接的第二个表中的记录。像这样的事情:

SELECT
    tbl_1.NAME,
    count(tbl_2.MENTIUNI) AS NR_MENTIUNI
FROM tbl_1
    LEFT JOIN tbl_2
        ON tbl_1.ID = tbl_2.ID_NAME
        AND tbl_2.MENTIUNI = 'abc'