1

我有以下 SQL CLR C# UDF:

using System;
using System.Data;
using System.Data.SqlClient;
using System.Data.SqlTypes;
using Microsoft.SqlServer.Server;
using System.Collections;
using System.Text;

public partial class UserDefinedFunctions
{
    [Microsoft.SqlServer.Server.SqlFunction]
    public static SqlString clrFn_GetDigits(string theWord)
    {

        if (theWord == null) { theWord = ""; }
        string newWord = "";

        char[] KeepArray = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '\\', '/', '-', ' '};


        foreach (char thischar in theWord)
        {
            foreach (char keepchar in KeepArray)
            {
                if (keepchar == thischar)
                {     
                    newWord += thischar;
                }    
            }  
        }

        return (SqlString)(newWord.Trim());
    }
}

到目前为止,这很好用,除了以下地址:

141A, Some Street Avenue
4b, St Georges Street
16E Test Avenue

我希望我的函数返回 141A、4b 和 16E

有任何想法吗?

4

1 回答 1

2

我没有测试下面的代码,但是按照这些思路,需要进行一些错误检查以确保转换不会失败,但是这个解决方案可以为您提供所需的

    char[] KeepArray = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '\\', '/', '-', ' ' };


        foreach (char thischar in theWord) {

            if (KeepArray.Contains(thischar)) {

                newWord += thischar;
            }
            else if (Char.IsLetter(thischar) && newWord.Length > 0){

                try {
                    if (Char.IsDigit((Convert.ToChar(newWord.Substring(newWord.Length - 1, 1))))) {
                        newWord += thischar;

                    }

                }
                catch {

                }
            }

        }
于 2013-02-11T10:58:04.570 回答