Question

我有一个如下所示的列，我想要的输出如下例所示。 Col1是一种数值数据类型。

Col1                OutputCol1
1234    round to    2000
2300000 round to    3000000
456789.23 round to  500000

始终是第一位数字+ 1。我可以使用带有负值的圆函数，但如果第二个数字小于5，则它会向下舍入。

Answer 1

好吧。。。你可以在第一个数字上加1，然后用零填充：

select rpad(cast(substr(col1, 1, 1) as int) + 1,
            log(10, col1),
            '0'
           )

Answer 2

使用字符串操作。取第一个数字，添加一个，添加尾随零。如果数字后跟零（1000或10.00），则不添加1。

select col1, 
  case when nvl(to_number(substr(to_char(col1), 2)),0) = 0 then
    to_number(rpad(substr(to_char(trunc(col1)), 1, 1), length(to_char(trunc(col1))), '0'))
  else
    to_number(rpad(to_char(to_number(substr(to_char(trunc(col1)), 1, 1)) + 1), length(to_char(trunc(col1))), '0'))
  end as x,
  to_number(substr(to_char(trunc(col1)), 2))
from

Answer 3

这里有几种方法 - 一种是数学方法，另一种是用科学记数法操纵col1的字符串输出：

WITH sample_data AS (SELECT 1234 col1 FROM dual UNION ALL
                     SELECT 2300000 col1 FROM dual UNION ALL
                     SELECT 456789.23 col1 FROM dual UNION ALL
                     SELECT -183 col1 FROM dual UNION ALL
                     SELECT 1000 col1 FROM dual UNION ALL
                     SELECT 0.392 col1 FROM dual)
SELECT col1,
       ceil(col1 / order_of_magnitude_of_col1) * order_of_magnitude_of_col1 output1,
       CEIL(to_number(SUBSTR(sn, 1, 4))) * POWER(10, to_number(SUBSTR(sn, 6))) output2
FROM   (SELECT col1,
               power(10, floor(LOG(10, abs(col1)))) order_of_magnitude_of_col1,
               to_char(col1, 'fms0.0EEEE') sn
        FROM   sample_data);

      COL1    OUTPUT1    OUTPUT2
---------- ---------- ----------
      1234       2000       2000
   2300000    3000000    3000000
 456789.23     500000     500000
      -183       -100       -100
      1000       1000       1000
     0.392        0.4        0.4

在oracle中舍入到第一个数字+1

3 个答案: