Question

我在MySQL表中有以下数据。

+--------+----------+------+
| job_id | query_id | done |
+--------+----------+------+
|  15145 | a002     |    1 |
|  15146 | a002     |    1 |
|  15148 | a002     |    1 |
|  15150 | a002     |    1 |
|  15314 | a003     |    0 |
|  15315 | a003     |    1 |
|  15316 | a003     |    0 |
|  15317 | a003     |    0 |
|  15318 | a003     |    1 |
|  15319 | a003     |    0 |
+--------+----------+------+

我想知道是否有可能有一个SQL查询，可以按query_id分组如果所有'完成'字段标记为1 。我想象的可能输出是：

+----------+------+
| query_id | done |
+----------+------+
|  a002    |  1   |
|  a003    |  0   |
+----------+------+

我尝试过以下SQL查询：

select job_id, query_id, done from job_table group by done having done = 1 ;

但没有运气。我真的很感激你的帮助！

Answer 1

我并不为这个解决方案感到特别自豪，因为它不是很清楚，但至少它快速而简单。如果所有项目都“done”= 1 ，则总和将等于计数SUM = COUNT

SELECT query_id, SUM(done) AS doneSum, COUNT(done) AS doneCnt 
FROM tbl 
GROUP BY query_id

如果你添加一个having子句，你会得到“完成”的项目。

HAVING doneSum = doneCnt

我会让你正确地格式化解决方案，你可以做一个分歧来获得“未完成”的项目或doneSum＆lt;＆gt; doneCnt。

顺便说一句，SQL小提琴here。

Answer 2

所有“完成”工作：

<body>
  <ul>
    <li>Foo</li>
    <li>Bar</li>
    <li>Baz</li>
    <li>Qux</li>
  </ul>
</body>

所有“未完成”的工作：

SELECT * FROM job_table GROUP BY done HAVING MIN(done) = 1;

所有具有相同值的作业（“完成”或“撤消”）：

SELECT * FROM job_table GROUP BY done HAVING MAX(done) = 0;

Answer 3

您的示例中的输出可以通过以下方式获得：

SELECT query_id, min(done) as done
FROM job_table
GROUP by done

Answer 4

我认为你应该看看这个

SELECT column_name, aggregate_function(column_name)
FROM table_name
WHERE column_name operator value
GROUP BY column_name

Answer 5

尝试类似

的内容

SELECT qry_id, 1
FROM (
    SELECT qry_id, done
    FROM tbl
    GROUP BY qry_id, done
) AS t
GROUP BY qry_id
HAVING COUNT(*)=1
UNION
SELECT qry_id, 0
FROM (
    SELECT qry_id, done
    FROM tbl
    GROUP BY qry_id, done
) AS t
GROUP BY qry_id
HAVING COUNT(*)>1

将MySQL中特定字段中具有相同值的行分组

5 个答案: