如果我有一个类似于这样的MySQL表:例如2014000是产品笔,有四种不同价格的abcd类型。和2014001是另一个产品电话,还有三种不同价格的efg。现在我想让一个产品的名称和价值一行。
id code name value
1 2014000 A 10
2 2014000 B 9
3 2014000 C 11
4 2014000 D 12
5 2014001 E 100
6 2014001 F 110
7 2014001 G 120
尊重结果:
code name1 value1 name2 value2 name3 value3 name4 value4
----------- --------- --------- --------- --------- ----------- --------- --------- ---------
2014000 A 10 B 9 C 11 D 12
2014001 E 100 F 110 G 120 null null
CREATE TABLE T(ID INT,代码INT,NAME CHAR(1),VALUE VARCHAR(10));
插入T VALUES(1,2014000,'A','10'),(2,2014000,'B','9'),(3,2014000,'C','11'),( 4,2014000,'D','12'),(5,2014001,'E','100'),(6,2014001,'F','110'),(7,2014001,'G', '120');
-----后面只有两行,如果有四行以上,怎么办呢?
SELECT T1.code, T2.name AS name1, T2.value AS value1, T3.name AS name2, T3.value AS value2 FROM( SELECT code,MIN(ID) AS ID1,CASE COUNT(code) WHEN 1 THEN NULL ELSE MAX(ID) END AS ID2 FROM T GROUP BY code ) T1 LEFT JOIN T T2 ON T1.ID1 = T2.ID LEFT JOIN T T3 ON T1.ID2 = T3.ID
答案 0 :(得分:0)
SELECT t.code
, t1.name name1
, t1.value value1
, t2.name name2
, t2.value value2
, t3.name name3
, t3.value value3
, t4.name name4
, t4.value value4
FROM ( SELECT code
, group_concat(ID) ids
FROM T
GROUP BY code
) t
LEFT JOIN T t1
ON find_in_set(t1.ID,t.ids)=1
LEFT JOIN T t2
ON find_in_set(t2.ID,t.ids)=2
LEFT JOIN T t3
ON find_in_set(t3.ID,t.ids)=3
LEFT JOIN T t4
ON find_in_set(t4.ID,t.ids)=4