请教一个简单SQL:select * from eload_goods where goods_sn like '(select left(goods_sn,9) as goods_sn from eload_goods where goods_id = 126300 limit 1)%'
这个SQL有啥错误?搜不到结果.select * from eload_goods where goods_sn = (select left(goods_sn,9) as goods_sn from eload_goods where goods_id = 126300 limit 1)
这样有搜索是有结果的.注:子查询是正确的.子查询返回结果为:IE0384701
goods_sn为varchar数据类型
这个SQL有啥错误?搜不到结果.select * from eload_goods where goods_sn = (select left(goods_sn,9) as goods_sn from eload_goods where goods_id = 126300 limit 1)
这样有搜索是有结果的.注:子查询是正确的.子查询返回结果为:IE0384701
goods_sn为varchar数据类型
但我绝不相信你的表中有 (select left(goods_sn,9) as goods_sn from eload_goods where goods_id = 126300 limit 1)... 这样的字符串
版主,您错了.
(select left(goods_sn,9) as goods_sn from eload_goods where goods_id = 126300 limit 1)是绝对正确的.也有这样的字符串.问题己解决,附上答案:
SELECT * FROM eload_goods WHERE goods_sn LIKE (SELECT CONCAT(LEFT(goods_sn,9),'%') AS goods_sn FROM eload_goods WHERE goods_id = 126300 LIMIT 1)
select * from eload_goods where goods_sn like '(select left(goods_sn,9) as goods_sn from eload_goods where goods_id = 126300 limit 1)%'
不知字符串吗
如果goods_id是主键的话,limit 1完全没有必要,而且不被支持
我理解你的意思,'select...' 是字符串,所以我是请教解决的写法嘛。