xsd格式编号,小数点后5位

时间:2018-05-03 08:08:25

标签: oracle xsd-validation

我有xsd代码,它将LINE_AMOUNT值打印为14,952.59,现在我想将其显示为14,952.59000(小数点后5位)。

如何实现这一目标?

谢谢

2 个答案:

答案 0 :(得分:1)

SQL Fiddle

查询1 要么(如果要将当前NLS值用于十进制和千位字符):

SELECT TO_CHAR(
         14952.59,
         'FM9G999G999G999G990D00000'
       )
FROM   DUAL

<强> Results

| TO_CHAR(14952.59,'FM9G999G999G999G990D00000') |
|-----------------------------------------------|
|                                  14,952.59000 |

查询2 或:

SELECT TO_CHAR(
         14952.59,
         'FM9,999,999,999,990.00000'
       )
FROM   DUAL

<强> Results

| TO_CHAR(14952.59,'FM9,999,999,999,990.00000') |
|-----------------------------------------------|
|                                  14,952.59000 |

更新SQL Fiddle

查询3

SELECT TO_CHAR(
         TO_NUMBER( 
           '13,214,952.59',
           'FM9G999G999G999G990D99999'
         ),
         'FM9G999G999G999G990D00000'
       ) AS formatted_value
FROM   DUAL

<强> Results

|  FORMATTED_VALUE |
|------------------|
| 13,214,952.59000 |

查询4

SELECT TO_CHAR(
         TO_NUMBER( 
           '13,214,952.59',
           'FM9,999,999,999,990.99999'
         ),
         'FM9,999,999,999,990.00000'
       ) AS formatted_value
FROM   DUAL

<强> Results

|  FORMATTED_VALUE |
|------------------|
| 13,214,952.59000 |

答案 1 :(得分:0)

试试这个,我觉得它对你有用。

select trim(to_char(14952.59,9999999999.99999)) from dual


OUTPUT
14952.59000