0
+----------------------+---------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+----------------------+---------------+------+-----+---------+-------+
| name | varchar(10) | YES | MUL | NULL | |
| slno | varchar(20) | YES | | NULL | |
| type | int(2) | YES | | NULL | |
| details | text | YES | | NULL | |
+----------------------+---------------+------+-----+---------+-------+
名称,slno并键入一起形成键。
的样本数据:
+---------+------+------+-------------------------------+
| name | slno | type | details |
+---------+------+------+-------------------------------+
| name1 | 11 | 1 | {"data":["feats1","feats2"] } |
| name1 | 11 | 2 | {"data":["feats1","feats2"] } |
| name1 | 12 | 1 | {"data":["feats5","feats6"] } |
| name1 | 12 | 2 | {"data":["feats5","feats6"] } |
| name2 | 11 | 1 | {"data":["feats3","feats4"] } |
| name2 | 11 | 2 | {"data":["feats3","feats4"] } |
| name2 | 12 | 1 | {"data":["feats7","feats8"] } |
| name2 | 12 | 2 | {"data":["feats10"] } |
+---------+------+------+-------------------------------+
所以基本上与NAME = '名称1' 中的每个条目,有一个与同slno但名称= '名2' 类似的条目。
我想要做的是对具有相同slno和类型但名称不同的行设置相同的细节,即上面的示例数据集应该如下所示。 name2行'的细节应该匹配name1行的细节,如果他们有相同的slno和类型。
+---------+------+------+-------------------------------+
| name | slno | type | details |
+---------+------+------+-------------------------------+
| name1 | 11 | 1 | {"data":["feats1","feats2"] } |
| name1 | 11 | 2 | {"data":["feats1","feats2"] } |
| name1 | 12 | 1 | {"data":["feats5","feats6"] } |
| name1 | 12 | 2 | {"data":["feats5","feats6"] } |
| name2 | 11 | 1 | {"data":["feats1","feats2"] } |
| name2 | 11 | 2 | {"data":["feats1","feats2"] } |
| name2 | 12 | 1 | {"data":["feats5","feats6"] } |
| name2 | 12 | 2 | {"data":["feats5","feats6"] } |
+---------+------+------+-------------------------------+
我试过了,但是不能拿出执行上述结果的命令。有人可以帮忙吗?
所以slno和细节应该形成一个单独的表? – Strawberry
不,我给出了样本数据(第二个表格)以及它应该如何转换到下面(最后一个表格) – user3248186
您如何选择给定'slno/type'对的参考值?我的意思是,你为什么选择'name1'的值而不是'name2'的值? –