将MySQL中特定字段中具有相同值的行分组

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，如果所有“完成”字段都标记为1，是否可以使用sql查询，该查询可以按查询id进行分组。我认为可能的结果是：

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

我尝试了以下SQL查询：

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

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但是没有运气。我真的非常感谢你的帮助

我想你应该看看这个

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

试试像这样的东西

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

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我并不特别为这个解决方案感到自豪，因为它不是很清楚，但至少它是快速和简单的。如果所有项目的“完成”=1，则总和将等于计数总和=计数

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

如果您添加having子句，那么您将得到“done”项

我会让你正确格式化解决方案，你可以做一个差异来获得“未完成”项或doneSum doneCnt

顺便说一句，SQL fiddle.

所有“完成”作业：

所有“撤消”作业：

具有相同值的所有作业（“完成”或“撤消”）：

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示例中的输出可通过以下方式获得：

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

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非常感谢。但是查询id都显示“0”，而不是“1”和“0”

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

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

SELECT * FROM job_table GROUP BY done HAVING MAX(done) = 0 OR MIN(done) = 1;

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