Question

两个查询都使用相同的表（此处为：test）

首先查询：

SELECT `test` as t1, COUNT( * ) as t2
FROM  `test_table`
WHERE  `test` > 5
GROUP BY `test`

以表格

返回结果

--------------------------------------------------
| t1 | t2 |
--------------------------------------------------
|  6 |  2 |
--------------------------------------------------
|  8 |  7 |
--------------------------------------------------

第二次查询：

SELECT TRUNCATE(((num1/num2) * 100),3) as t3
FROM
(SELECT COUNT( * ) as num1
FROM `test_table`
WHERE test > 5
group by `test`) a,
(SELECT COUNT( * ) as num2
FROM `test_table`
WHERE test > 5) b

以表格

返回结果

--------------------------------------------------
| t3      |
--------------------------------------------------
| 40.456% |
--------------------------------------------------
| 59.544% |
--------------------------------------------------

我想要的是：

--------------------------------------------------
| t1 | t2 | t3 |
--------------------------------------------------
|  6 |  2 | 40.456% |
--------------------------------------------------
|  8 |  7 | 59.544% |
--------------------------------------------------

我该怎么办？联盟将t3结果置于t1 t2以下，或许加入声明？但加入什么？或者我可以从这两个选择所有数据的一个查询？

Answer 1

在select中使用子查询使其非常易读

   DROP TABLE IF EXISTS TEST_TABLE;
    CREATE TABLE TEST_TABLE(TEST INT);
    INSERT INTO TEST_TABLE VALUES
    (6),(6),
    (8),(8),(8),(8),(8),(8),(8);

    SELECT `test` as t1, COUNT(*) as t2, 
            TRUNCATE(COUNT(*) / (SELECT COUNT(*) FROM TEST_TABLE WHERE TEST > 5) * 100,3) AS T3
    FROM  `test_table`
    WHERE  `test` > 5
    GROUP BY `test`

结果

+------+----+--------+
| t1   | t2 | T3     |
+------+----+--------+
|    6 |  2 | 22.222 |
|    8 |  7 | 77.777 |
+------+----+--------+
2 rows in set (0.00 sec)

Answer 2

SELECT 
    Z.t1, Z.t2, Z1.t3
FROM
    (SELECT 
        `test` AS t1, COUNT(*) AS t2
    FROM
        `test_table`
    WHERE
        `test` > 5
    GROUP BY `test`) Z
        INNER JOIN
    (SELECT 
        t1, TRUNCATE(((num1 / num2) * 100), 3) AS t3
    FROM
        (SELECT 
        test t1, COUNT(*) AS num1
    FROM
        `test_table`
    WHERE
        test > 5
    GROUP BY `test`) a, (SELECT 
        COUNT(*) AS num2
    FROM
        `test_table`
    WHERE
        test > 5) b) Z1 ON Z.t1 = Z1.t1;

尝试以上代码。

希望这会对你有所帮助。

Answer 3

这可以通过对第二个查询进行一些修改来完成：

SELECT a.t1, num1 as t2, TRUNCATE(((num1/num2) * 100),3) as t3
FROM
(SELECT `test` as t1, COUNT( * ) as num1
FROM `test_table`
WHERE test > 5
group by `test`) a,
(SELECT COUNT( * ) as num2
FROM `test_table`
WHERE test > 5) b

Answer 4

在您的示例中构建它的方式您应该能够在test_table.test（您正在使用Group By的字段）上加入它们

SQL合并两个查询的结果

4 个答案: