[英]MySQL query, subquery, distinct, count,
Employees
表名称。 | 新的。 |
---|---|
沙达布。 | 是的 |
阿里。 | 不 |
沙达布。 | 是的 |
名称。 | 新的(是)。 | 新品(否) |
---|---|---|
沙达布。 | 2. | 0 |
阿里。 | . | 1 |
如何从Employees
表中得到这个结果? 使用 MySQL。
只需计算New
列的Yes
和No
。
SELECT DISTINCT
a.name,
coalesce(yes.numbers, 0),
coalesce(no.numbers, 0)
FROM employees AS a
LEFT JOIN (SELECT name, count(name) AS numbers FROM employees WHERE new = 'Yes' GROUP BY name) AS yes ON (a.name = yes.name)
LEFT JOIN (SELECT name, count(name) AS numbers FROM employees WHERE new = 'No' GROUP BY name) AS no ON (a.name = no.name)
或使用 SUM 和 IF:
SELECT
name,
SUM(IF(new = 'Yes', 1, 0)) as yes_num,
SUM(IF(new = 'No', 1, 0)) as no_num
FROM employees
GROUP BY name
这是您需要的查询:
SELECT
Name,
SUM(CASE WHEN New='Yes' THEN 1 ELSE 0 END) as NEW_Yes,
SUM(CASE WHEN New!='Yes' THEN 1 ELSE 0 END) as NEW_No
FROM `employees`
GROUP BY Name
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