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"
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
简单而简短:
// this means match a space or not a space (anything) up to 4 characters
var lines = Regex.Matches(str, @"[\s\S]{0,4}").Cast<Match>().Select(x => x.Value);
其他回答
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();
}
六年后o_O
仅仅因为
public static IEnumerable<string> Split(this string str, int chunkSize, bool remainingInFront)
{
var count = (int) Math.Ceiling(str.Length/(double) chunkSize);
Func<int, int> start = index => remainingInFront ? str.Length - (count - index)*chunkSize : index*chunkSize;
Func<int, int> end = index => Math.Min(str.Length - Math.Max(start(index), 0), Math.Min(start(index) + chunkSize - Math.Max(start(index), 0), chunkSize));
return Enumerable.Range(0, count).Select(i => str.Substring(Math.Max(start(i), 0),end(i)));
}
or
private static Func<bool, int, int, int, int, int> start = (remainingInFront, length, count, index, size) =>
remainingInFront ? length - (count - index) * size : index * size;
private static Func<bool, int, int, int, int, int, int> end = (remainingInFront, length, count, index, size, start) =>
Math.Min(length - Math.Max(start, 0), Math.Min(start + size - Math.Max(start, 0), size));
public static IEnumerable<string> Split(this string str, int chunkSize, bool remainingInFront)
{
var count = (int)Math.Ceiling(str.Length / (double)chunkSize);
return Enumerable.Range(0, count).Select(i => str.Substring(
Math.Max(start(remainingInFront, str.Length, count, i, chunkSize), 0),
end(remainingInFront, str.Length, count, i, chunkSize, start(remainingInFront, str.Length, count, i, chunkSize))
));
}
AFAIK所有的边缘情况都处理好了。
Console.WriteLine(string.Join(" ", "abc".Split(2, false))); // ab c
Console.WriteLine(string.Join(" ", "abc".Split(2, true))); // a bc
Console.WriteLine(string.Join(" ", "a".Split(2, true))); // a
Console.WriteLine(string.Join(" ", "a".Split(2, false))); // a
使用正则表达式和Linq:
List<string> groups = (from Match m in Regex.Matches(str, @"\d{4}")
select m.Value).ToList();
我觉得这样更有可读性,但这只是个人观点。它也可以是一行代码:)。
它基于@dove解决方案,但作为扩展方法实现。
好处:
扩展方法 涵盖角落案例 分割字符串与任何字符:数字,字母,其他符号
Code
public static class EnumerableEx
{
public static IEnumerable<string> SplitBy(this string str, int chunkLength)
{
if (String.IsNullOrEmpty(str)) throw new ArgumentException();
if (chunkLength < 1) throw new ArgumentException();
for (int i = 0; i < str.Length; i += chunkLength)
{
if (chunkLength + i > str.Length)
chunkLength = str.Length - i;
yield return str.Substring(i, chunkLength);
}
}
}
使用
var result = "bobjoecat".SplitBy(3); // bob, joe, cat
为简洁起见,删除了单元测试(请参阅以前的修订版)
它不是很漂亮,也不是很快,但它是有效的,它是一行程序,它是LINQy:
List<string> a = text.Select((c, i) => new { Char = c, Index = i }).GroupBy(o => o.Index / 4).Select(g => new String(g.Select(o => o.Char).ToArray())).ToList();