[英]sql distinct + count
我要做的是為每個用戶計算同一列的特定實例。
例:
vagt_type
的“ timeloen”值可能是該用戶的10倍,其中usr = 1
,而“ prov”的值是該用戶的2倍。 我需要在單獨的列中計數,而我使用DISTINCT
只能使每個usr一次。
下面是我到目前為止已經制定了,但是,計數的所有實例vagt_type
,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: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
其他表存在的僅僅是join
到用戶表及其在這種情況下並不重要,我只用它來CONCAT
用戶的全名。
預期成績:
usr | timeloen | provision | sygedag
1 | 3 | 1 | 1
29 | 1 | 0 | 0
編輯:-希望這對其他人有幫助:)最終成為我的整體解決方案:
$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);
我相信您必須輸入以下內容:
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
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