Oracle函数以无序方式比较字符串 [英] Oracle function to compare strings in a not ordered way

问题描述

我需要一个函数来比较两个字符串而不考虑 oracle 中的顺序.即asd"和sad"应该被认为是平等的.有没有类似的功能?或者我需要编写自己的函数?

I need a function to make a comparison between two strings withouth considering the order in oracle. i.e. "asd" and "sad" should be considered as equal. Are there similar functions? Or I need to write my own function?

推荐答案

这可以通过一个简单的 java 函数按字母顺序对字符串的字符进行排序来完成:

This can be done with a simple java function to sort the characters of a string alphabetically:

CREATE AND COMPILE JAVA SOURCE NAMED SORTSTRING AS
public class SortString {
  public static String sort( final String value )
  {
    final char[] chars = value.toCharArray();
    java.util.Arrays.sort( chars );
    return new String( chars );
  }
};
/

然后您可以创建一个 PL/SQL 函数来调用:

Which you can then create a PL/SQL function to invoke:

CREATE FUNCTION SORTSTRING( in_value IN VARCHAR2 ) RETURN VARCHAR2
AS LANGUAGE JAVA NAME 'SortString.sort( java.lang.String ) return java.lang.String';
/

然后你可以对排序后的字符串做一个简单的比较:

Then you can do a simple comparison on the sorted strings:

SELECT CASE
       WHEN SORTSTRING( 'ads' ) = SORTSTRING( 'das' )
       THEN 'Equal'
       ELSE 'Not Equal'
       END
FROM   DUAL;

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