如何将$连接到特定字符串

Question

这是我的程序在输出中有$不会打印。 我将使用正确的syntex作为'$'+，但它不起作用

SELECT USERS.ID,CONCAT(USERS.FIRSTNAME,' ',USERS.LASTNAME)AS USERNAME,
     ('$'+ SUM(CPS_HISTORY.CHARGED_AMOUNT+CPS_HISTORY.TRANSACTION_FEE+CPS_HISTORY.SERVICE_CHARGE+CPS_HISTORY.COMBO_PAK_PRICE))AS REVENUE FROM USERS JOIN CPS_HISTORY ON CPS_HISTORY.SUBSCRIBER_ID = USERS.ID GROUP BY USERS.ID ORDER BY REVENUE DESC LIMIT 5;

+---------+---------------+---------+
| ID      | USERNAME      | REVENUE |
+---------+---------------+---------+
| 1803399 | Michael Rowan |     772 |
| 1697091 | NULL          |     676 |
| 1790000 | ree Green     |     626 |
| 1766654 | Jose M NUFIO  |     625 |
| 1731854 | Ashlee Durgin |     622 |
+---------+---------------+---------+
5 rows in set, 13305 warnings (0.15 sec)

concat函数的问题是：更改了数据。

+---------+-----------------+---------+
| ID      | USERNAME        | REVENUE |
+---------+-----------------+---------+
| 1753814 | Joseph Hearn    | $99     |
| 1806377 | Gideon Anderson | $99     |
| 1800992 | Camryn Revitte  | $99     |
| 1802344 | Tanner Chik     | $99     |
| 1594358 | NULL            | $99     |
+---------+-----------------+---------+
5 rows in set (0.00 sec)

Answer 1

使用concat来概括这些值：

SELECT USERS.ID,CONCAT(USERS.FIRSTNAME,' ',USERS.LASTNAME)AS USERNAME,
     concat('$', SUM(CPS_HISTORY.CHARGED_AMOUNT+CPS_HISTORY.TRANSACTION_FEE+CPS_HISTORY.SERVICE_CHARGE+CPS_HISTORY.COMBO_PAK_PRICE))AS REVENUE 
     FROM USERS JOIN CPS_HISTORY ON CPS_HISTORY.SUBSCRIBER_ID = USERS.ID GROUP BY USERS.ID ORDER BY REVENUE DESC LIMIT 5;

Answer 2

数据不会更改，只会更改类型。

您正在按REVENUE列排序，该列以前是数字，但是现在，在CONCAT() ，它是一个字符串。 而$99作为字符串要在$772之前（如果按DESC排序）。

有两种解决方案：

1. 在您的应用程序中添加$而不是在SQL中添加，或

2. 使用子查询，该子查询在内部查询中进行数据选择和排序，在外部查询中进行格式化。

   就像是

    SELECT ID, USERNAME, CONCAT('$', REVENUE) AS REVENUE FROM (SELECT USERS.ID,CONCAT(USERS.FIRSTNAME,' ',USERS.LASTNAME) AS USERNAME, SUM(CPS_HISTORY.CHARGED_AMOUNT+CPS_HISTORY.TRANSACTION_FEE+CPS_HISTORY.SERVICE_CHARGE+CPS_HISTORY.COMBO_PAK_PRICE)) AS REVENUE FROM USERS JOIN CPS_HISTORY ON CPS_HISTORY.SUBSCRIBER_ID = USERS.ID GROUP BY USERS.ID ORDER BY REVENUE DESC LIMIT 5) AS INNER 

   应该做。 （未经测试！）