Question

表中所有各行都有<代码>的领域,即0或1。

我需要一把零和一只一把一把。因此,这一结果应当看一看:

type0 | type1
------+------
1234  | 4211

如何执行?

Answer 1

select type, count(type) from tbl_table group by type;

Answer 2

Les...see

SELECT
    SUM(CASE type WHEN 0 THEN 1 ELSE 0 END) AS type0,
    SUM(CASE type WHEN 1 THEN 1 ELSE 0 END) AS type1
FROM
   tableX;

测试了not。

Answer 3

您不妨使用 rel=“nofollow noreferer”>。缩略语:

SELECT (SELECT COUNT(*) FROM table WHERE type = 0) AS type0,
       (SELECT COUNT(*) FROM table WHERE type = 1) AS type1;

在MySQL测试如下:

CREATE TABLE t (id INT NOT NULL AUTO_INCREMENT, type INT);

INSERT INTO t VALUES (NULL, 0);
INSERT INTO t VALUES (NULL, 0);
INSERT INTO t VALUES (NULL, 1);
INSERT INTO t VALUES (NULL, 1);
INSERT INTO t VALUES (NULL, 1);

SELECT (SELECT COUNT(*) FROM t WHERE type = 0) AS type0,
       (SELECT COUNT(*) FROM t WHERE type = 1) AS type1;

+-------+-------+
| type0 | type1 |
+-------+-------+
|     2 |     3 | 
+-------+-------+
1 row in set (0.00 sec)

Answer 4

这样的结果很容易实现:

Type  Count
-----------
type0 1234
type1 4221

你们可以使用这样的东西:

SELECT CONCAT( type , [type]) as Type, COUNT(*) as Count
FROM MyTable
GROUP BY Type

友情链接