返回COUNT每一行，而不是在POSTGRES

一筆或單行

SELECT (select count(active) AS true 
     from sooper_entry 
     where active = 't' 
     and entry_id_ref = 28) AS true, 
     (select count(active) AS false 
     from sooper_entry 
     where active = 'f' 
     and entry_id_ref = 28) AS false;

所以我可以得到所有的真與假的計數，但我需要一個真正的假count在關聯數組中返回。

期望的結果：

true | false | uId 
------+-------+----- 
    16 |  0 | 1 
    10 |  2 | 3 
    13 | 10 | 4 
    19 |  8 | 5 
    12 |  3 | 8 
    21 |  0 | 12 
(6 rows)

來源

2011-07-25 roberthuttinger

SELECT SUM(active::BOOLEAN::INT) AS active, 
     SUM((NOT active::BOOLEAN)::INT) AS inactive, 
     entry_id_ref 
FROM sooper_entry 
GROUP BY 
     entry_id_ref

來源

2011-07-25 15:21:04 Quassnoi

完美和簡單！非常感謝，你做搖滾！（顯然我還不能給任何人+1，有人請爲我做！）我現在可以+1！ – roberthuttinger

@robert：請將其中一個答案（您認爲最有幫助的）標記爲已接受。要做到這一點，請點擊答案的選票數下面的複選標記，以便它變成綠色。 – Quassnoi

從我這裏稍微加了一點：'SELECT SUM（active :: BOOLEAN :: INT）as active，SUM（（NOT active :: BOOLEAN）:: INT）as inactive，entry_id_ref FROM sooper_entry GROUP BY entry_id_ref' – roberthuttinger

SELECT 
    sum(case when active = 't' then 1 else 0 end) AS true, 
    sum(case when active = 'f' then 1 else 0 end) AS false, 
    entry_id_ref 
FROM sooper_entry 
GROUP BY entry_id_ref

來源

2011-07-25 15:19:21 marc

這將返回等於計數。 – Quassnoi

謝謝！每行返回14 | 3（示例） – roberthuttinger

@roberthuttinger：固定的。 – marc

返回COUNT每一行，而不是在POSTGRES

回答

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