MySql嵌套选择,变量在PHP中不起作用
我有一个嵌套的mysql查询,每6行给我一个表列(poutput_edat)的平均值MySql嵌套选择,变量在PHP中不起作用,php,mysql,mysqli,Php,Mysql,Mysqli,我有一个嵌套的mysql查询,每6行给我一个表列(poutput_edat)的平均值 set @rownum := 0; set @sum := 0; SELECT the_avg FROM ( select poutput_edat, @rownum := (@rownum + 1) as rownum, @sum := IF(@rownum mod 6 = 1,0 + poutput_edat,@sum + poutput_edat
set @rownum := 0;
set @sum := 0;
SELECT the_avg FROM (
select poutput_edat, @rownum := (@rownum + 1) as rownum,
@sum := IF(@rownum mod 6 = 1,0 + poutput_edat,@sum + poutput_edat)
as running_sum, IF(@rownum mod 6 = 0,@sum / 6,NULL) as the_avg
from raw_edat /*WHERE date_edat = CURDATE()*/ ORDER BY invTime_edat desc)
as average
我在php mysqli中使用此查询时遇到问题,它可以完美地直接查询数据库。这是php mysql的一个限制吗
Jimmy能否尝试在独立的
FROM
子句中声明变量,例如:
SELECT the_avg FROM (
select poutput_edat, @rownum := (@rownum + 1) as rownum,
@sum := IF(@rownum mod 6 = 1,0 + poutput_edat,@sum + poutput_edat)
as running_sum, IF(@rownum mod 6 = 0,@sum / 6,NULL) as the_avg
FROM raw_edat /*WHERE date_edat = CURDATE()*/, (SELECT @rownum := 0, @sum := 0) a
ORDER BY invTime_edat desc)
as average
它会返回什么错误消息?没有错误,只是一个空结果。谢谢你,Darshan,现在这对我来说运行得很好。我想这是我的新手失误。吉米