Php SQL在MySQL中使用JOIN选择公共数据和剩余数据

我有三张桌子：wi_区、wi_集团和wi_培训。我需要根据地区统计小组和培训。为此，我使用了以下SQL

SQL1 查询统计每个地区的组数。同样地

SQL2 该查询统计每个地区的培训。现在我需要组合从SQL1和SQL2获得的所有结果，并以

dst_name || group_count || grp_created_date || training_count || trn_created_date

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问题是，每当我使用SQL1左连接SQL2时，它都会显示与SQL1对应的结果，其中无法获得SQL2的结果，反之亦然。请帮我在MySQL中解决这个问题

我想你可以加入过滤表，然后按地区名称分组。像这样：

SELECT dist.dst_name AS dst_name,
COUNT(grp.grp_id) AS group_count, MAX(grp.grp_created_date) AS grp_created_date,
COUNT(trn.trn_id) AS training_count, MAX(trn.trn_created_date) AS trn_created_date 
FROM wi_district AS dist
LEFT JOIN (
  SELECT dst_id, trn_id, trn_created_date
  FROM wi_training
  WHERE trn_deleted=0
    AND trn_beneficiary_type IN (-1,2,8,9,10)
) AS trn ON trn.dst_id=dist.dst_id
LEFT JOIN (
  SELECT grp_dst_id, grp_id, grp_created_date
  FROM wi_group
  WHERE grp_deleted=0
    AND grp_type IN (3)
) AS grp ON grp.grp_dst_id = dist.dst_id
GROUP BY dist.dst_name

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为什么使用像dst_name这样的名称？最好写下全名，这样几个月后你也会明白它的意思

无论如何，这个查询应该可以做到这一点

select d.dst_name as district
,      count(distinct g.grp_id) as group_count
,      max(grp_created_date) as group_created_date 
,      count(distinct trn_id) as training_count
,      max(trn_created_date) as trn_created_date 
from   wi_district d
left join wi_group g on  d.dst_id = g.grp_dst_id
                     and g.grp_deleted = 0 
                     and g.grp_type in (3)
left join wi_training t on  d.dst_id = t.dst_id
                        and t.trn_deleted = 0
                        and t.trn_beneficiary_type IN (-1,2,8,9,10)
group by d.dst_name

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您需要从分组表中驱动它。像这样

SELECT  D.dst_name,  COUNT(grp_id) AS group_count, MAX(grp_created_date) as grp_created_date, COUNT(trn_id) AS training_count, MAX(trn_created_date) as trn_created_date 
FROM    wi_district D
LEFT JOIN wi_group G ON D.dst_id = G.grp_dst_id AND G.grp_deleted=0 AND G.grp_type IN (3)
LEFT JOIN wi_training T ON  D.dst_id = T.dst_id AND T.trn_deleted=0 AND T.trn_beneficiary_type IN (-1,2,8,9,10)
GROUP BY wi_district.dst_name

如果您只需要其中一个表上存在的行，请添加一个子句：

WHERE NOT G.grp_dst_id IS NULL OR NOT D.dst_id IS NULL

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然后不要使用左连接。左联接是左表中的所有匹配记录，右表中的任何匹配记录。如果您需要两个表中的所有记录，则需要外部联接。不客气。只需确保计数中包含DISTINCT关键字，否则组计数将与每个地区的培训数相乘。如果是答案，则将其标记为答案：

SELECT  D.dst_name,  COUNT(grp_id) AS group_count, MAX(grp_created_date) as grp_created_date, COUNT(trn_id) AS training_count, MAX(trn_created_date) as trn_created_date 
FROM    wi_district D
LEFT JOIN wi_group G ON D.dst_id = G.grp_dst_id AND G.grp_deleted=0 AND G.grp_type IN (3)
LEFT JOIN wi_training T ON  D.dst_id = T.dst_id AND T.trn_deleted=0 AND T.trn_beneficiary_type IN (-1,2,8,9,10)
GROUP BY wi_district.dst_name

WHERE NOT G.grp_dst_id IS NULL OR NOT D.dst_id IS NULL