Oracle中怎么判斷一列數據是否為數字類型， 沒有isnumber()這樣的函數， to_number()的時候又會出錯， 想找到是那行數據導致出現to_number()出問題了， 怎么辦？

我簡單的google了以下， 看到：

簡單的SQL:

select aa.rowid, instr(translate(aa.a,

'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ',

'XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX'),'X') as IS_A

FROM aa

對于一列中的其他特殊字符就得自己添加到Z后面了，要理解上面SQL的含義先理解幾個Oracle函數了：

---------------

translate( string1, string_to_replace, replacement_string )

string1 is the string to replace a sequence of characters with another set of characters.

string_to_replace is the string that will be searched for in string1.

replacement_string - All characters in the string_to_replace will be replaced with the corresponding character in the replacement_string.

Applies To:

* Oracle 8i, Oracle 9i, Oracle 10g, Oracle 11g

For example:

translate('1tech23', '123', '456); ?? ?would return '4tech56'

translate('222tech, '2ec', '3it'); ?? ?would return '333tith'

---------

instr( string1, string2 [, start_position [, nth_appearance ] ] )

string1 is the string to search.

string2 is the substring to search for in string1.

start_position is the position in string1 where the search will start. This argument is optional. If omitted, it defaults to 1. The first position in the string is 1. If the start_position is negative, the function counts back start_position number of characters from the end of string1 and then searches towards the beginning of string1.

nth_appearance is the nth appearance of string2. This is optional. If omitted, it defaults to 1.

Note:

If string2 is not found in string1, then the instr Oracle function will return 0.

Applies To:

* Oracle 8i, Oracle 9i, Oracle 10g, Oracle 11g

For example:

instr('Tech on the net', 'e') ?? ?would return 2; the first occurrence of 'e'

instr('Tech on the net', 'e', 1, 1) ?? ?would return 2; the first occurrence of 'e'

instr('Tech on the net', 'e', 1, 2) ?? ?would return 11; the second occurrence of 'e'

instr('Tech on the net', 'e', 1, 3) ?? ?would return 14; the third occurrence of 'e'

instr('Tech on the net', 'e', -3, 2) ?? ?would return 2.

自己簡單的測試了以下：

select to_number('3 3') from dual;

也會出問題了， 就將空格添加到上面的SQL中就可以了。

select aa.rowid, instr(translate(aa.a,

'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ ',

'XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX'),'X') as IS_A

FROM aa

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