Sql server 2008 SQLServer2008中的减法

我有两个选择查询，如下所示：

SELECT COUNT(WARD_id) 
FROM IP_Admission 
WHERE (Status = 'V' OR Status = 'D') AND WARD_ID = 1

SELECT BED_STRENGTH 
FROM Ward_Master 
WHERE Ward_ID = 1

declare @a int
declare @b int

SELECT @a = COUNT(WARD_id) FROM IP_Admission WHERE (Status='V' OR Status='D') AND WARD_ID=1

SELECT @b = BED_STRENGTH FROM Ward_Master WHERE Ward_ID=1

select @b - @a

查询的输出是

   2
   6

现在我需要答案4。感谢您的帮助。

您可以这样尝试：

SELECT COUNT(WARD_id) 
FROM IP_Admission 
WHERE (Status = 'V' OR Status = 'D') AND WARD_ID = 1

SELECT BED_STRENGTH 
FROM Ward_Master 
WHERE Ward_ID = 1

declare @a int
declare @b int

SELECT @a = COUNT(WARD_id) FROM IP_Admission WHERE (Status='V' OR Status='D') AND WARD_ID=1

SELECT @b = BED_STRENGTH FROM Ward_Master WHERE Ward_ID=1

select @b - @a

您可以这样尝试：

SELECT COUNT(WARD_id) 
FROM IP_Admission 
WHERE (Status = 'V' OR Status = 'D') AND WARD_ID = 1

SELECT BED_STRENGTH 
FROM Ward_Master 
WHERE Ward_ID = 1

declare @a int
declare @b int

SELECT @a = COUNT(WARD_id) FROM IP_Admission WHERE (Status='V' OR Status='D') AND WARD_ID=1

SELECT @b = BED_STRENGTH FROM Ward_Master WHERE Ward_ID=1

select @b - @a

---

所有病房ID都试试这个

SELECT M.Ward_ID,BED_STRENGTH -COUNT(WARD_id)  
FROM IP_Admission A  
     JOIN  Ward_Master M ON A.WARD_ID = M.WARD_ID 
WHERE (A.Status='V' OR A.Status='D') 
GROUP BY M.Ward_ID

---

所有病房ID都试试这个

SELECT M.Ward_ID,BED_STRENGTH -COUNT(WARD_id)  
FROM IP_Admission A  
     JOIN  Ward_Master M ON A.WARD_ID = M.WARD_ID 
WHERE (A.Status='V' OR A.Status='D') 
GROUP BY M.Ward_ID

---

由于您是从两个不同的表中提取这些值的，因此最好的方法是在ward id上连接该表

SELECT wm.BED_STRENGTH - COUNT(ip.WARD_id) 
FROM Ward_Master wm inner join IP_Admission ip
         On wm.ward_id = ip.ward_id
WHERE (ip.Status='V' OR ip.Status='D')
Group by wm.BED_STRENGTH

这应该是可行的，但可能并不取决于模式。此方法的优点是，它将找到所有病房ID的值

最简单的解决方案是：

SELECT (subquery1) - (subquery2)

---

这将只显示您指定的病房id的值。

因为您从两个不同的表中提取这些值，所以最好的方法是将病房id上的表连接起来

SELECT wm.BED_STRENGTH - COUNT(ip.WARD_id) 
FROM Ward_Master wm inner join IP_Admission ip
         On wm.ward_id = ip.ward_id
WHERE (ip.Status='V' OR ip.Status='D')
Group by wm.BED_STRENGTH

这应该是可行的，但可能并不取决于模式。此方法的优点是，它将找到所有病房ID的值

最简单的解决方案是：

SELECT (subquery1) - (subquery2)

---

这将只显示您指定的病房id的值。

您可以像这样使用内部联接。如果这不起作用，请告诉我

create table #Tmp1
(
    Ward_Id INT,
    Status Char(1)
)

create table #tmp2
(
    Ward_Id INT,
    BED_STRENGTH INT
)

INSERT INTO #Tmp1 VALUES (1,'V'),(1,'D')

INSERT INTO #tmp2 VALUES (1,6)

SELECT COUNT(WARD_id) 
FROM #Tmp1 
WHERE (Status='V' OR Status='D') AND WARD_ID=1

SELECT BED_STRENGTH FROM #Tmp2 WHERE Ward_ID=1

SELECT #tmp2.BED_STRENGTH -  COUNT(#tmp1.WARD_id) 
FROM #Tmp1 
INNER JOIN #tmp2 ON #tmp2.Ward_Id = #tmp1.Ward_Id 
WHERE (Status='V' OR Status='D') AND #tmp1.WARD_ID=1
GROUP BY #tmp2.BED_STRENGTH

DROP TABLE #Tmp1 

DROP TABLE #Tmp2 

---

可以像这样使用内部联接。如果这不起作用，请告诉我

create table #Tmp1
(
    Ward_Id INT,
    Status Char(1)
)

create table #tmp2
(
    Ward_Id INT,
    BED_STRENGTH INT
)

INSERT INTO #Tmp1 VALUES (1,'V'),(1,'D')

INSERT INTO #tmp2 VALUES (1,6)

SELECT COUNT(WARD_id) 
FROM #Tmp1 
WHERE (Status='V' OR Status='D') AND WARD_ID=1

SELECT BED_STRENGTH FROM #Tmp2 WHERE Ward_ID=1

SELECT #tmp2.BED_STRENGTH -  COUNT(#tmp1.WARD_id) 
FROM #Tmp1 
INNER JOIN #tmp2 ON #tmp2.Ward_Id = #tmp1.Ward_Id 
WHERE (Status='V' OR Status='D') AND #tmp1.WARD_ID=1
GROUP BY #tmp2.BED_STRENGTH

DROP TABLE #Tmp1 

DROP TABLE #Tmp2 

---

谢谢大家，我解决了我的问题

   SELECT M.Bed_Strength - COUNT(A.WARD_id)  as diff
  FROM IP_Admission A  
JOIN  Ward_Master M ON A.WARD_ID = M.WARD_ID 
 WHERE (A.Status='V' OR A.Status='D') and M.Ward_ID=1 group by M.Bed_Strength;

---

谢谢大家，我解决了我的问题

   SELECT M.Bed_Strength - COUNT(A.WARD_id)  as diff
  FROM IP_Admission A  
JOIN  Ward_Master M ON A.WARD_ID = M.WARD_ID 
 WHERE (A.Status='V' OR A.Status='D') and M.Ward_ID=1 group by M.Bed_Strength;

---

Msg 8120，级别16，状态1，第1行列“ward_master.Bed_Strength”在选择列表中无效，因为它未包含在聚合函数或GROUP BY子句中。如果您在查询中使用更多列，则需要按所有未使用聚合函数的列进行分组Msg 8120，级别16，状态1，第1行列“ward_master.Bed_Strength”在选择列表中无效，因为它既不包含在聚合函数中，也不包含在GROUP BY子句中。如果在查询中使用更多列，则需要按所有未使用聚合函数的列进行分组，这就是我发布的答案。我的答案也是相同的。为了更好地理解，我已经详细阐述了。这是我发布的答案。我的答案也是一样的。为了更好地理解，我进行了详细阐述。