Question

$ i table：id，item_name

$ t table：id，member_name，item_name

$ u table：id，member_name，membership_status

我想

列出$ i中的所有item_names，即使$ t表中没有任何项目名称
加上$ t
然后将计数结果限制为$ u。中的membership_status ='active'。

这是我到目前为止所回答的问题的第一部分：

$find_actives = mysql_query ("select $i.item_name,  
                  count($t.item_name) from $i left  
                  join $t on $i.item_name = $t.item_name group by $t.item_name")

但是当我将其更改为下面的查询时，它不再列出$ i中的所有项目。

$find_actives = mysql_query ("select $i.item_name,  
                  count($t.item_name) from $i left  
                  join $t on $i.item_name = $t.item_name 
                  where $t.member_name IN (select distinct member_name from $u
                  where membership_status = 'active' group by member_name)
                  group by $t.item_name")

我希望结果如下：

item_name1 - count = 5

item_name2 - count = 0 //这是缺失的行

item_name3 - count = 2

我怎样才能做到这一点？

Answer 1

如果我正确地阅读你的问题，我认为这就是你要找的：

SELECT
    $i.item_name,
    COUNT(temp.item_name) AS item_count
FROM $i
LEFT JOIN (
    SELECT $t.item_name AS item_name
    FROM $t
    INNER JOIN $u
      ON $t.member_name = $u.member_name
    WHERE $u.membership_status = 'active'
) AS temp
  ON $i.item_name = temp.item_name
GROUP BY $i.item_name

左加入哪里

1 个答案: