Question

我正在开发一个网站，要求我显示用户输入每天平均数量的图表。我已经有一个SQL查询，它会将此信息返回给我：

SELECT sum(number)/count(number) as average, date FROM stats WHERE * GROUP BY date

这给了我正在寻找的结果，但结果以三位小数精度给出。我想围绕这个数字。当然，我可以在PHP或我的模板引擎中完成它，但我很好奇是否有办法在数据库中完成所有这些。

有没有办法将输出转换为整数（在MySQL中）？

Answer 1

SELECT 
  CAST(sum(number)/count(number) as UNSIGNED) as average, 
  date 
FROM stats 
WHERE * 
GROUP BY date

Answer 2

MySQL中CAST的有效类型如下

BINARY[(N)]
CHAR[(N)]
DATE
DATETIME
DECIMAL[(M[,D])]
SIGNED [INTEGER]
TIME
UNSIGNED [INTEGER]

所以你可以使用

SELECT CAST(sum(number)/count(number) AS UNSIGNED) as average...

或SIGNED如果SUM部分可以加起来为负数。

Answer 3

如何使用MySQL FORMAT函数？

mysql> SELECT FORMAT(12345.123456, 4);
+-------------------------+
| FORMAT(12345.123456, 4) |
+-------------------------+
| 12,345.1235             |
+-------------------------+
1 row in set (0.00 sec)

mysql> SELECT FORMAT(12345.123456, 0);
+-------------------------+
| FORMAT(12345.123456, 0) |
+-------------------------+
| 12,345                  |
+-------------------------+
1 row in set (0.00 sec)

Answer 4

使用DIV operator。

mysql> SELECT 5 DIV 2;
    -> 2

整数除法。与FLOOR（）类似，但BIGINT值是安全的。对于超出BIGINT范围的非整数操作数，可能会出现不正确的结果。

Answer 5

SELECT convert(int, sum(number)/count(number)) as average,
  date
FROM stats
WHERE * GROUP BY date

或

SELECT 
  CAST(sum(number)/count(number) as INT) as average, 
  date 
FROM stats 
WHERE * 
GROUP BY date

Answer 6

用户mysql函数round()
示例 round(23.33) 将在 msyql 中返回 23。

您的示例查询将更新为

SELECT round(sum(number)/count(number)) as average, date FROM stats WHERE * GROUP BY date

在SQL中选择/转换输出为整数

6 个答案: