Question

例如，我有2个这样的表

data have;
input name $ status $;
datalines;
A    a
B    b
C    c
;;;;
run;

第二桌：

data addon;
input name $ status $;
datalines;
A    a
C    f
D    d
E    e
F    f
B    z
;;;;
run;

如何获得如下结果：

B    b
C    c
C    f
D    d
E    e
F    f
B    z

A-a行与2个表相同，因此被删除。我正在尝试使用左连接，但结果不对。请提前帮助和感谢。我真的很感激。

Answer 1

试试这个：

  SELECT COALESCE(table1.input, table2.input) AS input
         , COALESCE(table1.status, table2.status) AS status
    FROM table1
         FULL OUTER JOIN table2 ON table1.input = table2.input
         AND table1.status = table2.status
   WHERE (table1.input IS NULL OR table2.input IS NULL)
ORDER BY 1

输出

INPUT STATUS
----- ------
B     b      
B     z      
C     f      
C     c      
D     d      
E     e      
F     f

Answer 2

没有时间对此进行测试，但这是大致正确的。由于MySQL不支持除了。

，因此无法在SQLFiddle中工作

select * from (
  select * from have union select * from addon)
except
( select * from have, addon 
   where have.status=addon.status and have.name=addon.name)

Answer 3

另一种方式

data have;
 input name $ status $;
 datalines;
A    a
B    b
C    c
;;;;
run;

data addon;
 input name $ status $;
datalines;
A    a
C    f
D    d
E    e
F    f
B    z
;;;;
run;

Data Together;
 Set have addon;
 /* If the data sets were already sorted */
 /* By Name Status;                      */
/*  Then skip the Proc Sort              */
Run;

Proc sort data=together;
 by name status;
Run;

Data final;
 Set Together;
 by name status;
 if first.status and last.status;
Run;

Answer 4

SELECT t1.name,t1.status
FROM 
  (
    SELECT name,status
    FROM have
    UNION ALL
    SELECT name,status
    FROM addon
  ) as t1
JOIN have t2 ON t1.name!=t2.name AND t1.status!=t2.status
JOIN addon t3 ON t2.name=t3.name AND t2.status=t3.status

我为你创建了SQL fiddle。

Answer 5

data have;
input name $ status $;
datalines;
A    a
B    b
C    c
;;;;
run;

第二桌：

data addon;
input name $ status $;
datalines;
A    a
C    f
D    d
E    e
F    f
B    z
;;;;
run;

如何获得如下结果：

B    b
C    c
C    f
D    d
E    e
F    f
B    z

简单使用合并声明。在此步骤之前使用键对数据集进行排序

DATA RESULT;
KEEP H.NAME A.STATUS;
MERGE HAVE(IN = H) ADDON (IN = A);
BY NAME;
RUN;

数据步/ SQL连接/合并/联合2数据集/表并删除相同的行/观察

5 个答案: