回合一個數字並在oracle中保留尾隨零

Question

好吧，所以（說）我有一個數字，如7836.721。我想一個Oracle查詢，可以給我下面的結果：

7837.000 

數量應四捨五入到一個人的地方，小數點後的所有數字被轉換到零。

我用什麼：

select round(7836.721,0) from dual 

這給了我：

7837 

任何幫助，將不勝感激。

Answer 1

聽起來像是你需要TO_CHAR ...

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

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

這並假定你之後的任何數量的3位小數，而不是7.12要去7.00是。

---

ETA：

這似乎是一個很奇怪的規定，但在這裏你去：地方後

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