有没有办法更好地解析字符串?

时间:2015-04-15 15:11:21

标签: c# .net string parsing

我想知道在.NET中是否有内置的解析字符串位的方法。

例如,我有以下字符串:

"bsarbirthd0692"

由以下将与数据交叉引用的部分组成:

Indexes   Purpose
0-3       (name)
4-9       (description)
10-13     (date mm-yy)

我希望有一些像土生土长的东西:

string name, desc, date;
string.ParseFormat("{0:4}{1:5}{2:4}", "bsarbirthd0692", out name, out desc, out date);

在.NET或流行的库中是否有以本机方式执行此操作?

5 个答案:

答案 0 :(得分:50)

由于格式已知,因此不应更改Substring应该适合您

string data = "bsarbirthd0692";
string name, desc, date;
name = data.Substring(0, 4);
desc = data.Substring(4, 6);
date = data.SubString(10);

修改

您还可以创建扩展方法来执行您想要的任何操作。这显然比以前的建议更复杂

public static class StringExtension
{
    /// <summary>
    /// Returns a string array of the original string broken apart by the parameters
    /// </summary>
    /// <param name="str">The original string</param>
    /// <param name="obj">Integer array of how long each broken piece will be</param>
    /// <returns>A string array of the original string broken apart</returns>
    public static string[] ParseFormat(this string str, params int[] obj)
    {
        int startIndex = 0;
        string[] pieces = new string[obj.Length];
        for (int i = 0; i < obj.Length; i++)
        {
            if (startIndex + obj[i] < str.Length)
            {
                pieces[i] = str.Substring(startIndex, obj[i]);
                startIndex += obj[i];
            }
            else if (startIndex + obj[i] >= str.Length && startIndex < str.Length)
            {
                // Parse the remaining characters of the string
                pieces[i] = str.Substring(startIndex);
                startIndex += str.Length + startIndex;
            }

            // Remaining indexes, in pieces if they're are any, will be null
        }

        return pieces;
    }
}

用法1:

string d = "bsarbirthd0692";
string[] pieces = d.ParseFormat(4,6,4);

结果:

enter image description here

用法2:

string d = "bsarbirthd0692";
string[] pieces = d.ParseFormat(4,6,4,1,2,3);

结果:

enter image description here

答案 1 :(得分:16)

您可以将Regexp用于此

string str= "bsarbirthd0692";
var regex = "(?<name>.{4})(?<desc>.{6})(?<date>.{4})";
MatchCollection matches = Regex.Matches(str, regex);
foreach(Match m in matches){
    Console.WriteLine(m.Groups["name"].ToString());
    Console.WriteLine(m.Groups["desc"].ToString());
    Console.WriteLine(m.Groups["date"].ToString());
}

答案 2 :(得分:11)

没有类似的东西,但是要写一些东西来实现:

IEnumerable<string> inputString.BreakIntoLengths(4, 6, 4)
带签名的

public IEnumerable<string> BreakIntoLengths(this string input, params int[] lengths);

非常简单:

public IEnumerable<string> BreakIntoLengths(this string input, params int[] lengths) {

  var pos = 0;
  foreach (var len in lengths) {
    yield return input.Substring(pos, len);
    pos += len;
  }
}

(实际的实现有一些错误检查。)

NB。我已经删除了类似接口的格式字符串:它似乎没有提供任何价值。返回集合后,很容易按索引分配条目。

答案 3 :(得分:2)

更好的是什么?像这样的东西?

var listStrings = new List<string>();
var tempIndex = 0;
var indexList = new List<int>{4, 6, 4}
foreach(var length in indexList)
{
    listStrings.Add(string.Substring(tempIndex , length);
    tempIndex += length 
}

答案 4 :(得分:2)

不,没有内置方式。我会使用像Substring这样的字符串方法:

string name, desc, date;
if (input.Length >= 4)
{
    name = input.Substring(0, 4);
    if (input.Length >= 10)
    {
        desc = input.Substring(4, 6);
        if (input.Length >= 14)
        {
            date = input.Substring(10, 4);
        }
    }
}