Postgresql查詢獲得一年內每月有多個表的計數

Question

根據reference link我習慣於在一年內使用單個表進行計數。但在這裏，我需要根據用戶類型和財產關係計算兩張關係表。

的樣本數據

居民表

resident_id | resident_user_id | resident_estate_id | created_at   | 
31   | 75    |  1   | 2016-12-07 11:22:23 | 
32   | 76    |  16   | 2016-12-07 11:22:23 | 
37   | 81    |  16   | 2016-12-07 11:22:23 | 
38   | 84    |  17   | 2016-12-07 11:22:23 | 

用戶表

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在這裏我們需要根據條件和一年內列出的數據。以下示例查詢用於獲取數據。

示例1：如果將用戶登錄類型3更改爲2，則意味着12月份的計數爲1，但需要計入零計數。

SELECT to_char(i, 'YYYY') as year_data, to_char(i, 'MM') as month_data, to_char(i, 'Month') as month_string, count(resident_id) as ios_total_count 
FROM generate_series(now() - INTERVAL '1 year', now(), '1 month') as i 
left join residents on (to_char(i, 'YYYY') = to_char(created_at, 'YYYY') and to_char(i, 'MM') = to_char(created_at, 'MM') and resident_estate_id = 17) 
left join users on (users.id=residents.resident_user_id and users.user_login_type = 3) 
GROUP BY 1,2,3 order by year_data desc, month_data desc 
limit 12 

輸出

"2016";"12";"December ";1 
"2016";"11";"November ";0 
"2016";"10";"October ";0 
"2016";"09";"September";0 
"2016";"08";"August ";0 
"2016";"07";"July  ";0 
"2016";"06";"June  ";0 
"2016";"05";"May  ";0 
"2016";"04";"April ";0 
"2016";"03";"March ";0 
"2016";"02";"February ";0 
"2016";"01";"January ";0 

例2：如果我用在獲得12月當月用計數也加入語句後的狀態。但在年內沒有獲得其他月份。

SELECT to_char(i, 'YYYY') as year_data, to_char(i, 'MM') as month_data, to_char(i, 'Month') as month_string, count(resident_id) as ios_total_count 
FROM generate_series(now() - INTERVAL '1 year', now(), '1 month') as i 
left join residents on (to_char(i, 'YYYY') = to_char(created_at, 'YYYY') and to_char(i, 'MM') = to_char(created_at, 'MM')) 
left join users on (users.id=residents.resident_user_id) 
where residents.resident_estate_id = 17 and users.user_login_type = 3 
GROUP BY 1,2,3 order by year_data desc, month_data desc 
limit 12 

期望輸出

"2016";"12";"December ";0 
"2016";"11";"November ";0 
"2016";"10";"October ";0 
"2016";"09";"September";0 
"2016";"08";"August ";0 
"2016";"07";"July  ";0 
"2016";"06";"June  ";0 
"2016";"05";"May  ";0 
"2016";"04";"April ";0 
"2016";"03";"March ";0 
"2016";"02";"February ";0 
"2016";"01";"January ";0 

Answer 1

您篩選其他行與where residents.resident_estate_id = 17 and users.user_login_type = 3 我假設你想，而不是水木清華這樣的：

SELECT 
    DISTINCT 
    to_char(i, 'YYYY') as year_data 
    , to_char(i, 'MM') as month_data 
    , to_char(i, 'Month') as month_string 
    , count(resident_id) 
    filter (where residents.resident_estate_id = 17 and users.user_login_type = 3) 
    over (partition by date_trunc('month',i)) as ios_total_count 
FROM generate_series(now() - INTERVAL '1 year', now(), '1 month') as i 
left join residents on (to_char(i, 'YYYY') = to_char(created_at, 'YYYY') and to_char(i, 'MM') = to_char(created_at, 'MM')) 
left join users on (users.id=residents.resident_user_id) 
where residents.resident_estate_id = 17 and users.user_login_type = 3 
order by year_data desc, month_data desc 
limit 12