Question

我有两张表如下

T1：

id  code    field1  field2
1   1000    a1111   a2222
2   2000    b1111   b2222
3   1000    a3333   a4444
4   2000    b3333   b4444
5   2000    b5555   b6666
6   3000    c1111   c2222
7   3000    c3333   c4444
8   3000    c5555   c6666

T2：

t2id    t1_code     var1
1   1000           xxxx
2   2000           yyyyy
3   3000           zzz
4   3000           mmm

我希望结果表为：

code    field1  field2      t1_code     var1
1000    a3333   a4444       1000            xxxx
1000    a1111   a2222           null            null
2000    b3333   b4444       2000            yyyyy
2000    b5555   b6666           null            null
2000    b1111   b2222       null            null
3000    c1111   c2222       3000            mmm
3000    c3333   c4444       3000            zzz
3000    c3333   c4444       null            null

我试过了：

SELECT t1.code, t1.field1, t1.field2, t2.t1_code, t2.var1
FROM t2, t1
WHERE t1_code = code
ORDER BY code

没有给我答案。

请帮忙......

Answer 1

好的，你希望连接转到t1中的“第一个”匹配行（其中“first”表示最低id）。

这几乎是一种方法：

SELECT t1.code, t1.field1, t1.field2, t2.t1_code, t2.var1
FROM t1 left outer join
     (select t1.code, min(id) as minid
      from t1
      group by t1.code
     ) t1min
     on t1.id = t1min.minid left outer join
     t2
     on t2.t1_code = t1min.code;

这样做是将原始表连接到另一个表以查找每个代码的最小id。通过加入id，t1min.code只会在最小值上显示值。最后的连接只会为这些代码带来代码。

编辑：

实际问题有点复杂。我认为最好的方法是简单地枚举每个表中的值，并使用两个键进行连接：

select t1.code, t1.field1, t1.field2, t2.t1_code, t2.var1
from (select t1.*,
             @rn1 := if(@code = code, @rn1 + 1, 1) as rn, @code := code
      from t1 cross join
           (select @rn1 := 0, @code := '') const
      order by code, id
     ) t1 left outer join
     (select t2.*,
             @rn2 := if(@code = t1_code, @rn2 + 1, 1) as rn, @code := t1_code
      from t2 cross join
           (select @rn2 := 0, @code := '') const
      order by t1_code, t2id
     ) t2
     on t1.code = t2.t1_code and t1.rn = t2.rn
order by t1.id;

在mysql中使用单列连接多个列

1 个答案: