Mysql 为派生表选择数据时获取未知列错误
我需要一个查询来创建一个表,计算给定字段中每个值的数量,但由于此表中的数据不足,我不得不加入另一个表以获得一个附加值NName:Mysql 为派生表选择数据时获取未知列错误,mysql,subquery,derived-table,Mysql,Subquery,Derived Table,我需要一个查询来创建一个表,计算给定字段中每个值的数量,但由于此表中的数据不足,我不得不加入另一个表以获得一个附加值NName: Records_table-------------- Name_table--------- Ref Score Iteration Ref NName 1 High 1 1 Falcon 1 Middle 2
Records_table-------------- Name_table---------
Ref Score Iteration Ref NName
1 High 1 1 Falcon
1 Middle 2 2 Willow
2 Middle 1 3 Lance
2 Middle 2 4 Ranger
2 Low 3
3 Low 1
4 High 1
4 High 2
4 High 3
所需输出:
NName High Middle Low
Falcon 1 1 0
Willow 0 2 1
Lance 0 0 1
Ranger 3 0 0
这就是我所拥有的:
编辑,因为我注意到我在选择字段中留下了我在我的版本中删除的参考
SELECT
tc.nname AS NName,
COUNT(High) High,
COUNT(Middle) Middle,
COUNT(Low) Low
FROM
(SELECT
NName, 'High' High, NULL Middle, NULL Low
FROM
records_table tr
JOIN name_table tc ON tc.ref = tr.ref
WHERE
tr.score = 'High'
UNION ALL
SELECT
NName, NULL, 'Middle', NULL
FROM
records_table tr
JOIN name_table tc ON tc.ref = tr.ref
WHERE
tr.score = 'Middle'
UNION ALL
SELECT
NName, NULL, NULL, 'Low'
FROM
records_table tr
JOIN name_table tc ON tc.ref = tr.ref
WHERE
tr.score = 'Low' ) T
GROUP BY NName;
此操作失败,错误为1054 42S22:“字段列表”中的未知列“NName”
我做错了什么,我需要做什么才能使这项工作正常?您的表别名不是tc,那么您应该在主选择中使用T.nname
SELECT
T.`NNAME` AS NName,
COUNT(High) High,
COUNT(Middle) Middle,
COUNT(Low) Low
FROM (
SELECT
Ref,`NNAME`, 'High' High, NULL Middle, NULL Low
FROM records_table tr
JOIN name_table tc ON tc.ref = tr.ref
WHERE tr.score = 'High'
UNION ALL
SELECT
Ref,`NNAME`, NULL, 'Middle', NULL
FROM records_table tr
JOIN name_table tc ON tc.ref = tr.ref
WHERE tr.score = 'Middle'
UNION ALL
SELECT
Ref,`NNAME`, NULL, NULL, 'Low'
FROM records_table tr
JOIN name_table tc ON tc.ref = tr.ref
WHERE tr.score = 'Low'
) T
GROUP BY T.`NNAME`;
虽然不是严格意义上的问题是什么,这是一个封闭的问题,@scaisEdge给出了正确的答案,但我不认为工会是最好的方式。考虑条件聚集< /P>
select n.ref,n.nname,
sum(case when score = 'high' then 1 else 0 end) High,
sum(case when score = 'middle' then 1 else 0 end) middle,
sum(case when score = 'low' then 1 else 0 end) low
from n
join r on r.ref = n.ref
group by n.ref,n.nname;
+------+--------+------+--------+------+
| ref | nname | High | middle | low |
+------+--------+------+--------+------+
| 1 | Falcon | 1 | 1 | 0 |
| 2 | Willow | 0 | 2 | 1 |
| 3 | Lance | 0 | 0 | 1 |
| 4 | Ranger | 3 | 0 | 0 |
+------+--------+------+--------+------+
4 rows in set (0.00 sec)
您是否可以执行内部SELECT,即带有联合的SELECT,并查看您在输出中看到的列名?我想我会使用条件聚合进行不同的操作。@Darshan Mehta,我只执行了第一个,因为我在开始时不再有SELECT语句,我不得不用NName代替现场名称tc.NName,这很有效。但是,当我再次运行整个程序时,它仍然失败:从SELECT tc.nname中选择tc.nname作为nname,COUNTHigh High,COUNTMiddle Middle,COUNTLow Low,“High”High,NULL Middle,tc.ref=tr.ref,其中tr.score='High'UNION ALL上的记录\u table tr JOIN name\u table tc为NULL Low。。。。。。。其中tr.score=根据tc.nname划分的“低”T组;错误1054 42S22:未知列“cases.project”在“field list”中@P Salmon,您是明星,工作过一次,将在下面发布完整的解决方案,谢谢。您的查询中可能有一些垃圾字符,请尝试重新键入与NNAME相关的代码。。删除逗号之间的所有字符或用bactik包装colname..谢谢你-根据你上面的问题,我用COUNT而不是SUM得出了相同的结论,但本质上是一样的-非常感谢!