此段sql报错如下:Column 'user_worker.Title' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
时间: 2024-04-13 10:31:05 浏览: 12
这个错误是因为在你的查询中,选择列表中的列 'user_worker.Title' 未在聚合函数或 GROUP BY 子句中包含,而你的查询中使用了聚合函数。
根据你的需求,你可以选择以下两种解决方案之一:
1. 将 'user_worker.Title' 列添加到 GROUP BY 子句中:
```sql
SELECT user_worker.Title, COUNT(*) AS Count
FROM YourTable
GROUP BY user_worker.Title
```
在上述例子中,将 'YourTable' 替换为你要查询的表名。通过将 'user_worker.Title' 列添加到 GROUP BY 子句中,可以确保在使用聚合函数时不会出现错误。
2. 如果你不需要按照 'user_worker.Title' 列进行分组,可以使用聚合函数进行计算,而不选择 'user_worker.Title' 列:
```sql
SELECT COUNT(*) AS Count
FROM YourTable
```
在上述例子中,将 'YourTable' 替换为你要查询的表名。这样做将返回整个表中的记录计数,而不需要按照 'user_worker.Title' 列进行分组。
请选择适合你需求的解决方案,并相应地修改查询语句。
相关问题
with user_worker as (select Title,ID from [biUser]) select user_worker.Title as worker, (COUNT(CASE WHEN (CASE WHEN bl.Checker=0 THEN '' WHEN bl.Checker>0 and bl.Audited=-1 THEN 'NG' WHEN bl.Checker>0 and bl.Audited=1 THEN 'OK' END) = 'NG' THEN 1 END) / NULLIF(COUNT(CASE WHEN (CASE WHEN bl.Checker=0 THEN '' WHEN bl.Checker>0 and bl.Audited=-1 THEN 'NG' WHEN bl.Checker>0 and bl.Audited=1 THEN 'OK' END) = 'OK' THEN 1 END), 0)+ COUNT(CASE WHEN (CASE WHEN bl.Checker=0 THEN '' WHEN bl.Checker>0 and bl.Audited=-1 THEN 'NG' WHEN bl.Checker>0 and bl.Audited=1 THEN 'OK' END) IS NULL THEN 1 END)) AS ResultRatio from [bdWorkOrderList] bl left join [bdWorkOrder] bd on bl.OrderID=bd.ID left join [reportMJWXJL] rwx on bl.OrderID=rwx.OrderID left join [reportMJQHLHJL] rqh on bl.OrderID=rqh.OrderID left join user_worker on user_worker.ID=bl.Worker where bd.IsDeleted=0 and bl.StepName like 'Maker%' and bl.StepName in('Maker修模','maker切换料号','Maker模具保养') and DATEPART(MONTH, bl.StartTime) = DATEPART(MONTH, GETDATE()) order by user_worker.Title 这段代码出现如下报错:Column 'user_worker.Title' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
这个错误是因为在你的查询中,选择列表中的列 'user_worker.Title' 未在聚合函数或 GROUP BY 子句中包含,而你的查询中使用了聚合函数。
根据你的需求,你可以将查询修改如下:
```sql
WITH user_worker AS (
SELECT Title, ID
FROM [biUser]
)
SELECT
user_worker.Title AS worker,
(COUNT(CASE WHEN (CASE WHEN bl.Checker = 0 THEN ''
WHEN bl.Checker > 0 AND bl.Audited = -1 THEN 'NG'
WHEN bl.Checker > 0 AND bl.Audited = 1 THEN 'OK' END) = 'NG' THEN 1 END)
/ NULLIF(COUNT(CASE WHEN (CASE WHEN bl.Checker = 0 THEN ''
WHEN bl.Checker > 0 AND bl.Audited = -1 THEN 'NG'
WHEN bl.Checker > 0 AND bl.Audited = 1 THEN 'OK' END) = 'OK' THEN 1 END), 0)
+ COUNT(CASE WHEN (CASE WHEN bl.Checker = 0 THEN ''
WHEN bl.Checker > 0 AND bl.Audited = -1 THEN 'NG'
WHEN bl.Checker > 0 AND bl.Audited = 1 THEN 'OK' END) IS NULL THEN 1 END)) AS ResultRatio
FROM [bdWorkOrderList] bl
LEFT JOIN [bdWorkOrder] bd ON bl.OrderID = bd.ID
LEFT JOIN [reportMJWXJL] rwx ON bl.OrderID = rwx.OrderID
LEFT JOIN [reportMJQHLHJL] rqh ON bl.OrderID = rqh.OrderID
LEFT JOIN user_worker ON user_worker.ID = bl.Worker
WHERE bd.IsDeleted = 0
AND bl.StepName LIKE 'Maker%'
AND bl.StepName IN ('Maker修模', 'maker切换料号', 'Maker模具保养')
AND DATEPART(MONTH, bl.StartTime) = DATEPART(MONTH, GETDATE())
GROUP BY user_worker.Title
ORDER BY user_worker.Title
```
在上述修改中,将聚合函数 `COUNT` 和条件语句都放在了 `GROUP BY` 子句中,并且将查询结果按照 `user_worker.Title` 进行分组。
请注意,根据你的具体需求,你可能还需要对其他列进行适当的聚合或添加到 `GROUP BY` 子句中。请根据实际情况进行相应的调整。
Cause: java.sql.SQLSyntaxErrorException: Expression #2 of SELECT list is not in GROUP BY clause and contains nonaggregated column 'water.w.worker_image' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by
这个错误是因为MySQL的sql_mode设置了only_full_group_by,导致在SELECT语句中的非聚合列必须在GROUP BY子句中出现。
你可以通过以下两种方式来解决这个问题:
1. 修改sql_mode
你可以通过修改MySQL的sql_mode来解决这个问题,将only_full_group_by从sql_mode中删除即可。可以通过以下命令查看当前的sql_mode:
```
SELECT @@sql_mode;
```
然后可以通过以下命令将only_full_group_by从sql_mode中删除:
```
SET sql_mode=(SELECT REPLACE(@@sql_mode,'ONLY_FULL_GROUP_BY',''));
```
2. GROUP BY所有非聚合列
另一种解决方法是将SELECT语句中的所有非聚合列都添加到GROUP BY子句中,例如:
```
SELECT w.worker_name, w.worker_image
FROM worker w
GROUP BY w.worker_name, w.worker_image;
```
以上是两种解决这个问题的方法,你可以根据实际情况选择其中之一来解决这个问题。