我想截取result 字段中“ /" 前12位作为一个字段,该咋写
select substr(result,instr(result,'/')-12,12) from sa_job_queue j,sa_user u where j.crateuserguid=u.guid
楼主你好, 可以参考 select left(result,charindex('/',result)-1) as result_01 from sa_job_seque
left(result,12) as re_result