sql独特+计数

Question

我要做的是为每个用户计算同一列的特定实例。

例：
vagt_type的“ timeloen”值可能是该用户的10倍，其中usr = 1 ，而“ prov”的值是该用户的2倍。 我需要在单独的列中计数，而我使用DISTINCT只能使每个usr一次。

下面是我到目前为止已经制定了，但是，计数的所有实例vagt_type ，由于某种原因没有受到cast从日期到日期：

$test = $wpdb->get_results("SELECT distinct m.usr, CONCAT(n1.meta_value, ' ', n2.meta_value) AS fl_name, count(m1.vagt_type) as timeloen
FROM $main_table as m
LEFT JOIN Lausten_usermeta n1 ON m.usr=n1.user_id and n1.meta_key = 'first_name'
LEFT JOIN Lausten_usermeta n2 ON m.usr=n2.user_id and n2.meta_key = 'last_name'
LEFT JOIN $main_table m1 ON m.usr=m1.usr and m1.vagt_type = 'timeloen'
WHERE m.vagtDato BETWEEN CAST('$start' AS DATE) AND CAST('$end' AS DATE)
", ARRAY_A);

编辑：我的表的示例：

 id | usr | vagtDato   |  vagt_type
 13 |   1 | 2015-09-05 |    kursus
 16 |   1 | 2015-09-01 |    kursus
 11 |   1 | 2015-09-03 |    trappetur
 10 |   1 | 2015-09-02 |    provision
 9  |   1 | 2015-09-01 |    timeloen    
 15 |   1 | 2015-09-04 |    sygedag
 17 |   1 | 2015-09-02 |    timeloen    
 18 |   29| 2015-09-18 |    timeloen    
 19 |   1 | 2015-10-01 |    timeloen    

其他表存在的仅仅是join到用户表及其在这种情况下并不重要，我只用它来CONCAT用户的全名。

预期成绩：

usr | timeloen | provision | sygedag
1   | 3        | 1         | 1
29  | 1        | 0         | 0

编辑：-希望这对其他人有帮助:)最终成为我的整体解决方案：

$test = $wpdb->get_results("SELECT a.usr, a.vagtDato, b.timeloen, c.provision, d.kursus, e.trappetur, f.sygedag
FROM $main_table a
LEFT JOIN (SELECT usr, count(vagt_type) as timeloen
FROM $main_table  WHERE vagt_type = 'timeloen'
AND vagtDato between DATE('$start') AND DATE('$end')
GROUP BY usr ) b on b.usr=a.usr

LEFT JOIN (SELECT usr, count(vagt_type) as provision
FROM $main_table  WHERE vagt_type = 'provision'
AND vagtDato between DATE('$start') AND DATE('$end')
GROUP BY usr ) c on c.usr=a.usr

LEFT JOIN (SELECT usr, count(vagt_type) as kursus
FROM $main_table  WHERE vagt_type = 'kursus'
AND vagtDato between DATE('$start') AND DATE('$end')
GROUP BY usr ) d on d.usr=a.usr

LEFT JOIN (SELECT usr, count(vagt_type) as trappetur
FROM $main_table  WHERE vagt_type = 'trappetur'
AND vagtDato between DATE('$start') AND DATE('$end')
GROUP BY usr ) e on e.usr=a.usr

LEFT JOIN (SELECT usr, count(vagt_type) as sygedag
FROM $main_table  WHERE vagt_type = 'sygedag'
AND vagtDato between DATE('$start') AND DATE('$end')
GROUP BY usr ) f on f.usr=a.usr

WHERE a.vagtDato between DATE('$start') AND DATE('$end')
GROUP BY a.usr
", ARRAY_A);

Answer 1

我相信您必须输入以下内容：

SELECT
    usr
    , ( SELECT( count( * ) FROM $TABLE where vagt_type = 'timeloen' ) AS timeloen
    , ( SELECT( count( * ) FROM $TABLE where vagt_type = 'provision ' ) AS provision 
    , ( SELECT( count( * ) FROM $TABLE where vagt_type = 'sygedag' ) AS sygedag
FROM ( $main_table )
    ( $LEFTJOINS )
GROUP BY usr --to get the distincts users

要么

SELECT
    usr
    , count( * )
FROM ( $main_table )
    ( $LEFTJOINS )
GROUP BY 1, 2 --to get the distincts users