SELECT * from (
SELECT bwp.pattern_id as patternId ,tp.id as tpId,p.`name` as pname,sum( IF(pi.type = 1, 1, 0) ) as whiteApp
,rt.name as teName,
tp.b_time as bTime,tp.e_time as eTime
FROM ro_template_pattern as tp
INNER JOIN ro_pattern as p on p.id=tp.pattern_id
INNER JOIN ro_template AS rt ON tp.template_id = rt.id
INNER JOIN ro_black_white_pattern as bwp on bwp.pattern_id =p.id
INNER JOIN ro_public_information as pi on pi.id =bwp.public_information_id
where 1=1
<if test="templateId !=null and templateId !='' ">
and tp.template_id=#{templateId}
</if>
<if test="weekDay !=null">
and tp.week_day=#{weekDay}
</if>
<if test="customerId != '' and customerId != null">
AND p.customer_id = #{customerId,jdbcType=VARCHAR}
</if>
<if test="state != '' and state !=null">
AND rt.state = #{state,jdbcType=VARCHAR}
</if>
<if test="type != '' and type!=null">
AND p.type = #{type,jdbcType=VARCHAR}
</if>
<if test="inventoryCoe != '' and inventoryCoe != null">
AND rt.inventory_code = #{inventoryCoe,jdbcType=VARCHAR}
</if>
and tp.is_delete=0
and p.is_delete=0
and bwp.is_delete=0
GROUP BY bwp.pattern_id ,tp.id,rt.name ORDER BY tp.create_time asc
) as b
LEFT JOIN (SELECT bwr.pattern_id as patternId, sum( IF(bwr.type = 2, 1, 0) ) as whiteWeb FROM ro_black_white_rule as bwr GROUP BY bwr.pattern_id ) as c
on b.patternId =c.patternId
mysql把两个表中统计一起展示
最新推荐文章于 2024-03-19 18:59:02 发布
1万+

被折叠的 条评论
为什么被折叠?



