sql distinct + count

Question

我想要做的是爲每個用戶統計同一列的特定實例。

例如：
vagt_type可能具有的「timeloen」 10倍的值，其中usr = 1和「省」 2次該用戶的值。我需要單獨列中的計數，並且我使用DISTINCT來僅讓每個usr得到一次。

下面是我到目前爲止已經制定了，但是，纔是最重要的vagt_type所有實例，由於某種原因沒有受到cast從日期到日期：

$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); 

編輯： 例如我的表：

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  

其他表存在的僅僅是join用戶表及其在這種情況下並不重要，我只用它到CONCAT用戶的全名。

預期結果：

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

編輯： - 希望這可以幫助別人:) 什麼結束了我的整體解決方案：

$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

我會相信你必須把東西如：

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 

或

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