mysql - 一起显示来自两个单独查询的结果

Question

我必须一起显示来自两个不同查询的数据。

查询一：

select count(DISTINCT(nickname)) users_got_a_card from capture_captive where `number_successed`>0;

查询 2：

select count(DISTINCT(`nickname`)) users_shown_captive from `capture_captive`;

我想同时显示这两个查询的数据。

Answer 1

根据您的需要，您可以在
每列中执行以下操作：

select ( select count(DISTINCT(nickname)) users_got_a_card from capture_captive where 'number_successed'>0) as column1,
 (select count(DISTINCT(`nickname`)) users_shown_captive from 'capture_captive')as column2

或每行：

select count(DISTINCT(nickname)) users_got_a_card from capture_captive where number_successed>0 
UNION ALL  
select count(DISTINCT(nickname)) users_shown_captive from capture_captive

Answer 2

如果您希望它们作为单行的两列，您可以尝试

SELECT 
(
    SELECT COUNT(DISTINCT(nickname))  
      FROM capture_captive 
      WHERE `number_successed`> 0
) users_got_a_card,
(
    SELECT COUNT(DISTINCT(`nickname`))  
      FROM `capture_captive`
) users_shown_captive

或者

SELECT users_got_a_card, users_shown_capt
  FROM
(
    SELECT COUNT(DISTINCT(nickname)) users_got_a_card 
      FROM capture_captive 
      WHERE `number_successed`> 0
) a CROSS JOIN
(
    SELECT COUNT(DISTINCT(`nickname`)) users_shown_capt
      FROM `capture_captive`
) b

或者

SELECT COUNT(DISTINCT(IF(`number_successed` > 0, `nickname`, NULL))) users_got_a_card,
       COUNT(DISTINCT(`nickname`)) users_shown_capt
  FROM `capture_captive`

这是SQLFiddle演示

Answer 3

select count(DISTINCT(case when `number_successed`>0 
           then `nickname` end)) users_got_a_card, 
       count(DISTINCT(nickname)) users_shown_captive
from capture_captive 

Answer 4

这很简单。

首先，从第一个查询通常的方式获取结果并将其存储在一个名为 的变量中$var1，
接下来，从第二个查询通常的方式获取结果并将其存储在一个名为 的变量$var2中，
最后，创建一个代码以将它们一起显示：

echo $var1," ",$var2;

就如此容易。