選擇計數（*）不會給出正確的計數

我想要計算以下所有三個結果的總和，但我只能計算一個結果。選擇計數（*）不會給出正確的計數

$sql = "(SELECT COUNT(*) FROM table1 WHERE somecondition) 
UNION 
(SELECT COUNT(*) FROM table2 WHERE somecondition) 
UNION 
(SELECT COUNT(*) FROM table3 WHERE somecondition)";

任何人都可以help.thanks

來源

2012-04-27 sanjay

您的查詢將返回三人行，用三個獨立的罪名。

你想要的是

select sum(c) from (
     (SELECT COUNT(*) AS c FROM table1 WHERE somecondition) 
     UNION 
     (SELECT COUNT(*) FROM table2 WHERE somecondition) 
     UNION 
     (SELECT COUNT(*) FROM table3 WHERE somecondition) 
) all_three

來源

2012-04-27 05:12:28 Thilo

+1，不要忘記派生表的別名。 – 2012-04-27 05:15:08

@PaulBellora：謝謝。像這樣？ – Thilo 2012-04-27 05:16:48

完美！ http://sqlfiddle.com/#!2/82ed0/8 – 2012-04-27 05:18:39

-1

您的查詢將返回3 rows-爲每個表。

使用此查詢，而不是：

$sql = "SELECT SUM(C) FROM(
(SELECT COUNT(*) as C FROM table1 WHERE somecondition) 
UNION 
(SELECT COUNT(*) as C FROM table2 WHERE somecondition) 
UNION 
(SELECT COUNT(*) as C FROM table3 WHERE somecondition) 
)";

（您需要調整，這取決於你所使用的數據庫查詢）

來源

2012-04-27 05:13:58

我正在使用mysql，我是一個新手，學習mysql/php。我上面嘗試過，但仍然沒有工作，我錯過了什麼？ – sanjay 2012-04-27 05:25:18

你可以嘗試這樣的事情....

SELECT sum(
(SELECT count(*) from table1) 
+ (SELECT count(*) from table2) 
+ (SELECT count(*) from table3) 
+ (SELECT count(*) from table4))as total from dual

來源

2012-04-27 05:15:36 Addicted

此查詢正在運行並正在運行 – Addicted 2012-04-27 05:30:35

如果您正在尋找作爲不同領域的每張桌子的計數...

SELECT (SELECT count(*) FROM table_1) As first_count, 
     (SELECT count(*) FROM table_2) As second_count, 
     (SELECT count(*) FROM table_3) As third_count

如果要總結所有3代表的所有計數然後

SELECT SUM ((SELECT count(*) FROM table_1) + 
     (SELECT count(*) FROM table_2) + 
     (SELECT count(*) FROM table_3)) AS total_count

來源

2012-04-27 13:39:35

選擇計數（*）不會給出正確的計數

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