Question

我有一个表，要求我使用两个行共享的唯一值将某些行配对。

例如在下表中;

+--------+----------+-----------+-----------+----------------+-------------+
| id     | type     |  member   |  code     | description    | matching    |
+--------+----------+-----------+-----------+----------------+-------------+
| 1000   |transfer  |  552123   | SC120314  |  From Gold     |             |
| 1001   |transfer  |  552123   | SC120314  |  To Platinum   |             |
| 1002   |transfer  |  833612   | SC120314  |  From silver   |             |
| 1003   |transfer  |  833612   | SC120314  |  To basic      |             |
| 1004   |transfer  |  457114   | SC150314  |  From Platinum |             |
| 1005   |transfer  |  457114   | SC150314  |  To silver     |             |
| 1006   |transfer  |  933276   | SC180314  |  From Gold     |             |
| 1007   |transfer  |  933276   | SC180314  |  From To basic |             |
+--------+----------+-----------+-----------+----------------+-------------+

基本上我需要的查询/例程是找到＆＃39;成员＆＃39;中值的行。每行匹配的列。然后看看＆＃39;代码中的值是否为＆＃39;找到的相同行的列也匹配。

如果两行的两列都匹配，则为匹配的＆＃39;分配一个值。两行的列。对于两行，此值应该相同，并且仅对它们唯一。

唯一代码绝对可以是任何东西，只要它对匹配行是唯一的。是否有任何查询/例程可以执行此操作？

Answer 1

我不确定我是否正确理解了这个问题，但是如果您想挑选并更新code和member列匹配的行并将matching设置为某些每个相关行的唯一值，我相信这会起作用：

UPDATE <table> A 
  INNER JOIN (SELECT * FROM <table>) B ON 
    B.member = A.member && B.code = A.code && A.id <> B.id
SET A.matching = (A.id + B.id);

匹配值将设置为两行的id列的总和。注意如果有两行以上的行可以匹配，则以这种方式更新matching字段将无效。

对您的示例表运行上述查询将产生：

+------+----------+--------+----------+---------------+----------+
| id   | type     | member | code     | description   | matching |
+------+----------+--------+----------+---------------+----------+
| 1000 | transfer | 552123 | SC120314 | From Gold     | 2001     |
| 1001 | transfer | 552123 | SC120314 | To Platinum   | 2001     |
| 1002 | transfer | 833612 | SC120314 | From Silver   | 2005     |
| 1003 | transfer | 833612 | SC120314 | To basic      | 2005     |
| 1004 | transfer | 457114 | SC150314 | From Platinum | 2009     |
| 1005 | transfer | 457114 | SC150314 | To silver     | 2009     |
| 1006 | transfer | 933276 | SC180314 | From Gold     | 2013     |
| 1007 | transfer | 933276 | SC180314 | From To basic | 2013     |
+------+----------+--------+----------+---------------+----------+

Answer 2

我可以给你一个简单的查询，告诉你什么可以做你需要的。

tst 是表格的名称。

SELECT *, COUNT( t2.id ) as matching FROM tst t LEFT JOIN tst t2 ON t2.member = t.member GROUP BY t.id

比较不同的行并显示结果

2 个答案: