回合一个数字并在oracle中保留尾随零

Question

好吧，所以（说）我有一个数字，如7836.721。我想一个Oracle查询，可以给我下面的结果：

7837.000 

数量应四舍五入到一个人的地方，小数点后的所有数字被转换到零。

我用什么：

select round(7836.721,0) from dual 

这给了我：

7837 

任何帮助，将不胜感激。

Answer 1

听起来像是你需要TO_CHAR ...

SELECT to_char(TRUNC(7836.721), '9999990.000') numbr 
FROM dual; 

NUMBR 
------------ 
    7837.000 

这并假定你之后的任何数量的3位小数，而不是7.12要去7.00是。

---

ETA：

这似乎是一个很奇怪的规定，但在这里你去：地方后

WITH sample_data AS (SELECT 123 numbr FROM dual UNION ALL 
        SELECT 123.12 numbr FROM dual UNION ALL 
        SELECT 123.1234 numbr FROM dual UNION ALL 
        SELECT 123.12345 numbr FROM dual UNION ALL 
        SELECT 0.12345 numbr FROM dual UNION ALL 
        SELECT 1.23456789e50 numbr FROM dual UNION ALL 
        SELECT NULL numbr FROM dual UNION ALL 
        SELECT 1.23e-50 numbr FROM dual) 
SELECT numbr, 
     ROUND(numbr) || CASE WHEN numbr != TRUNC(numbr) THEN 
           RPAD('.', LENGTH(numbr -TRUNC(numbr)), '0') END new_numbr 
FROM sample_data; 

    NUMBR NEW_NUMBR 
---------- -------------------------------------------------------------------------------- 
     123 123 
    123.12 123.00 
    123.1234 123.0000 
123.12345 123.00000 
    0.12345 0.00000 
1.23456789 1.23456789E+50 

    1.23E-50 0.000000000000000000000000000000000000000