在單個查詢中具有多個選定結果的多個select語句

Question

我想將多個選擇鏈接到一個查詢中，我已經查看了另一個線程，這在很大程度上有所幫助，但是這隻能解決單個選擇選項。

SELECT (
    SELECT COUNT(*) 
    FROM user_table 
    ) AS tot_user, 
    (
    SELECT COUNT(*) 
    FROM cat_table 
    ) AS tot_cat, 
    (
    SELECT COUNT(*) 
    FROM course_table 
    ) AS tot_course 

但我的問題是，我試圖從每個選擇像這樣獲得多個選項。

SELECT (
SELECT c.name AS Company, 
CONCAT(UPPER(u.forename),' ', u.surname) AS Username, 
count(c.case) AS 'No. of Private Cases Generated', 
) AS 'No. of Private Cases Generated', 
(
SELECT c.name AS Company, 
CONCAT(UPPER(u.forename),' ', u.surname) AS Username, 
count(c.case) AS 'No. of Web Cases Generated' 
) AS 'No. of Web Cases Generated' 

我不斷收到錯誤「＃1241 - 操作數應包含1列（S）」

從我讀這是因爲我想下一個選擇鏈接3個不同的值，這是我不想做。

最後這段代碼就是問題所在。

) AS 'No. of Web Cases Generated' 

我已經嘗試過這個解決方案，但顯然我錯過了專業知識來讓它正確。

) "" AS Company, 
    "" AS Username, 
    "" AS 'No. of Private Cases Generated', 
    AS 'No. of Web Cases Generated' 

這不起作用，因爲它不知道什麼值分配給什麼選擇器。

任何幫助將不勝感激。

Answer 1

是的，這是正確的行爲，因爲你實際上是試圖合併在一列，這是不可能

(SELECT c.name AS Company, 
CONCAT(UPPER(u.forename),' ', u.surname) AS Username, 
count(c.case) AS 'No. of Private Cases Generated', 
) AS 'No. of Private Cases Generated', 

想多列值你所尋找的是一個UNION很像

SELECT c.name AS Company, 
CONCAT(UPPER(u.forename),' ', u.surname) AS Username, 
count(c.case) AS 'No. of Private Cases Generated' 
FROM some_table --first part of query 

UNION   -- combine the data 

SELECT c.name AS Company, 
CONCAT(UPPER(u.forename),' ', u.surname) AS Username, 
count(c.case) AS 'No. of Web Cases Generated' 
FROM some_other_table; -- second part of query 

根據您的評論，執行JOIN然後

select xx.Company, xx.Username,xx.'No. of Private Cases Generated', 
xxx.Company1, xxx.Username1, xxx.'No. of Web Cases Generated' 
FROM (
SELECT c.name AS Company, 
CONCAT(UPPER(u.forename),' ', u.surname) AS Username, 
count(c.case) AS 'No. of Private Cases Generated', 
FROM some_table ) xx 

JOIN (
SELECT c.name AS Company, 
CONCAT(UPPER(u.forename),' ', u.surname) AS Username, 
count(c.case) AS 'No. of Web Cases Generated' 
FROM some_other_table) xxx 
ON xx.Company = xxx.Company;