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  • select loginname,substr(lettertype,0,4) value,count(1) from slj_info where lettertype is not null and createtime between '${kssj} 00:00:00' and '${jssj} 23:59:59' and loginname in (select loginname from ${SWITCH(gh,'1','dh','2','wl','3','jb','4','db')} order by loginname) group by loginname,substr(lettertype,0,4) order by loginname 执行时报缺失右括号,什么鬼?
  • select * from 表名; 表名 :dh,wl,jb,db,四个表, 我的想法是:设置一个参数,${gh} , 等于1 dh表, 等于2 wl表, 等于3 jb表, 等于4 db表, 要怎么写sql语句?
  • 104774怎么把它去掉?
  • select id,createtime, f_keytoname(lettersrc,'00020001') lettersrc, f_keytoname(lettertype,'00020002') lettertype,letternote,dbloginname from slj_info where createtime........ 其中f_keytoname是什么意思?
  • 时间字段在sql语句中只能查询出有记录的数,如何让他按照顺序补全? h0-h23这个字段库中没有,怎么拿到?或者其他的方法获得显示值是h0-h23,真实值为0-23的一个序列? 没有值的情况下怎么让它为零? select substr(t,1,7) day,substr(t,12,2) time ,count(*) num from (select id,fromtime t from stable where fromtime>='2018-04-09 00:00:00' and fromtime <='2018-05-14 23:59:59') group by substr(t,1,7),substr(t,12,2) order by substr(t,1,7),substr(t,12,2) 结果图2 怎么拿解决这三个问题?
  • 第一条select substr(lettertype,0,4),count(1) from slj_info where lettersrc=2002 and createtime>='${stime}' and createtime<='${etime}' group by substr(lettertype,0,4) 第二条select substr(lettertype,0,4),count(1) from slj_info where lettersrc=2001 and createtime>='${stime}' and createtime<='${etime}' group by substr(lettertype,0,4) 第三条select substr(lettertype,0,4),count(1) from slj_info where lettersrc=2003 and createtime>='${stime}' and createtime<='${etime}' group by substr(lettertype,0,4) 怎么写成一句话?
  • 数据库语句查询, 比如字段有id, name, riqi, sql: select * from table where riqi>='2017-05-08'and riqi<= '2017-05-08' and name='6738' order by riqi; 其中name不是必须要的条件,这个语句在帆软里怎么写?

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