Question

我有桌子，

tbl_user

 id uniqid name
 1  123456 AAAA
 2  333333 BBBB

tbl_transaction

 id userid code value
 1  1      2202 500000
 2  1      2204 700000

以及如何得到像这样的结果

id uniqueid name code1 value1 code2 value2
1  123456   AAAA 2202  500000  2204  700000

在我的查询中我只知道使用加入我的代码

SELECT x.id, x.uniqueid, x.name, y.value
        FROM tbl_user x
        LEFT JOIN tbl_transaction y
            ON y.userid = x.id

Answer 1

一种方法，使用GROUP_CONCAT将可能的值分组为字符串，然后使用SUBSTRING_INDEX获取每个值（如果存在）。

SELECT x.id, 
        x.uniqueid, 
        x.name, 
        IF(COUNT(y.id) >= 1, SUBSTRING_INDEX(GROUP_CONCAT(y.code ORDER BY y.code), ',', 1), NULL) AS code1,
        IF(COUNT(y.id) >= 1, SUBSTRING_INDEX(GROUP_CONCAT(y.value ORDER BY y.code), ',', 1), NULL) AS value1,
        IF(COUNT(y.id) >= 2, SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(y.code ORDER BY y.code), ',', 2), ',', -1), NULL) AS code2,
        IF(COUNT(y.id) >= 2, SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(y.value ORDER BY y.code), ',', 2), ',', -1), NULL) AS value2,
        IF(COUNT(y.id) >= 3, SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(y.code ORDER BY y.code), ',', 3), ',', -1), NULL) AS code3,
        IF(COUNT(y.id) >= 3, SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(y.value ORDER BY y.code), ',', 3), ',', -1), NULL) AS value3,
        IF(COUNT(y.id) >= 4, SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(y.code ORDER BY y.code), ',', 4), ',', -1), NULL) AS code4,
        IF(COUNT(y.id) >= 4, SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(y.value ORDER BY y.code), ',', 4), ',', -1), NULL) AS value4,
        IF(COUNT(y.id) >= 5, SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(y.code ORDER BY y.code), ',', 5), ',', -1), NULL) AS code5,
        IF(COUNT(y.id) >= 5, SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(y.value ORDER BY y.code), ',', 5), ',', -1), NULL) AS value5
FROM tbl_user x
LEFT JOIN tbl_transaction y
ON y.userid = x.id
GROUP BY x.id, 
        x.uniqueid, 
        x.name

但是你需要编码来处理值的最大列数

Answer 2

我真的不知道它是否适合你，但只是试一试，

SELECT a.id,a.uniqueid,a.name,b.code,b.value,c.code,c.value FROM test.tbl_user as a,test.tbl_transaction as b,test.tbl_transaction as c where a.id=b.userid AND b.code!=c.code limit 1;

MySql查询加入两个表并将两行合并为一行

2 个答案: