假设我有一个字符串:

string str = "1111222233334444"; 

我如何把这个字符串分成一定大小的块?

例如,将它分解为4的大小将返回字符串:

"1111"
"2222"
"3333"
"4444"

当前回答

根据其他帖子的答案,以及一些使用示例:

public static string FormatSortCode(string sortCode)
{
    return ChunkString(sortCode, 2, "-");
}
public static string FormatIBAN(string iban)
{
    return ChunkString(iban, 4, "  ");
}

private static string ChunkString(string str, int chunkSize, string separator)
{
    var b = new StringBuilder();
    var stringLength = str.Length;
    for (var i = 0; i < stringLength; i += chunkSize)
    {
        if (i + chunkSize > stringLength) chunkSize = stringLength - i;
        b.Append(str.Substring(i, chunkSize));
        if (i+chunkSize != stringLength)
            b.Append(separator);
    }
    return b.ToString();
}

其他回答

这样写一行代码怎么样?

List<string> result = new List<string>(Regex.Split(target, @"(?<=\G.{4})", RegexOptions.Singleline));

对于这个正则表达式,最后一个块是否小于4个字符并不重要,因为它只查看它后面的字符。

我知道这不是最有效的解决方案,但我不得不把它扔出去。

如有必要分割几个不同的长度: 例如,日期和时间的指定格式为stringstrangeStr = "07092016090532";07092016090532(日期:07.09.2016时间:09:05:32)

public static IEnumerable<string> SplitBy(this string str, int[] chunkLength)
    {
        if (String.IsNullOrEmpty(str)) throw new ArgumentException();
        int i = 0;
        for (int j = 0; j < chunkLength.Length; j++)
        {
            if (chunkLength[j] < 1) throw new ArgumentException();
            if (chunkLength[j] + i > str.Length)
            {
                chunkLength[j] = str.Length - i;
            }
            yield return str.Substring(i, chunkLength[j]);
            i += chunkLength[j];
        }
    }

使用:

string[] dt = strangeStr.SplitBy(new int[] { 2, 2, 4, 2, 2, 2, 2 }).ToArray();
List<string> SplitString(int chunk, string input)
{
    List<string> list = new List<string>();
    int cycles = input.Length / chunk;

    if (input.Length % chunk != 0)
        cycles++;

    for (int i = 0; i < cycles; i++)
    {
        try
        {
            list.Add(input.Substring(i * chunk, chunk));
        }
        catch
        {
            list.Add(input.Substring(i * chunk));
        }
    }
    return list;
}

这应该比使用LINQ或这里使用的其他方法更快更有效。

public static IEnumerable<string> Splice(this string s, int spliceLength)
{
    if (s == null)
        throw new ArgumentNullException("s");
    if (spliceLength < 1)
        throw new ArgumentOutOfRangeException("spliceLength");

    if (s.Length == 0)
        yield break;
    var start = 0;
    for (var end = spliceLength; end < s.Length; end += spliceLength)
    {
        yield return s.Substring(start, spliceLength);
        start = end;
    }
    yield return s.Substring(start);
}
class StringHelper
{
    static void Main(string[] args)
    {
        string str = "Hi my name is vikas bansal and my email id is bansal.vks@gmail.com";
        int offSet = 10;

        List<string> chunks = chunkMyStr(str, offSet);

        Console.Read();
    }

    static List<string> chunkMyStr(string str, int offSet)
    {


        List<string> resultChunks = new List<string>();

        for (int i = 0; i < str.Length; i += offSet)
        {
            string temp = str.Substring(i, (str.Length - i) > offSet ? offSet : (str.Length - i));
            Console.WriteLine(temp);
            resultChunks.Add(temp);


        }

        return resultChunks;
    }
}