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
public static List<string> DevideByStringLength(string text, int chunkSize)
{
double a = (double)text.Length / chunkSize;
var numberOfChunks = Math.Ceiling(a);
Console.WriteLine($"{text.Length} | {numberOfChunks}");
List<string> chunkList = new List<string>();
for (int i = 0; i < numberOfChunks; i++)
{
string subString = string.Empty;
if (i == (numberOfChunks - 1))
{
subString = text.Substring(chunkSize * i, text.Length - chunkSize * i);
chunkList.Add(subString);
continue;
}
subString = text.Substring(chunkSize * i, chunkSize);
chunkList.Add(subString);
}
return chunkList;
}
其他回答
为什么不是循环?这里有一些东西可以很好地做到这一点:
string str = "111122223333444455";
int chunkSize = 4;
int stringLength = str.Length;
for (int i = 0; i < stringLength ; i += chunkSize)
{
if (i + chunkSize > stringLength) chunkSize = stringLength - i;
Console.WriteLine(str.Substring(i, chunkSize));
}
Console.ReadLine();
我不知道你会如何处理字符串不是因子4的情况,但不是说你的想法是不可能的,只是想知道它的动机,如果一个简单的for循环做得很好?显然,上面的内容可以被清除,甚至可以作为扩展方法加入。
或者正如评论中提到的,你知道它是/4
str = "1111222233334444";
for (int i = 0; i < stringLength; i += chunkSize)
{Console.WriteLine(str.Substring(i, chunkSize));}
public static List<string> SplitByMaxLength(this string str)
{
List<string> splitString = new List<string>();
for (int index = 0; index < str.Length; index += MaxLength)
{
splitString.Add(str.Substring(index, Math.Min(MaxLength, str.Length - index)));
}
return splitString;
}
List<string> chunks = new List<string>();
var longString = new string('s', 3000);
var chunkLength = 1273;
var ratio = (double)longString.Length / chunkLength;
var countOfChunks = Convert.ToByte(Math.Round(ratio, MidpointRounding.ToPositiveInfinity));
for (byte i = 0; i < countOfChunks; i++)
{
var remainingLength = longString.Length - chunkLength * i;
if (chunkLength > remainingLength)
chunks.Add(longString.Substring(i * chunkLength, remainingLength));
else
chunks.Add(longString.Substring(i * chunkLength, chunkLength));
}
使用IX库中的Buffer扩展
static IEnumerable<string> Split( this string str, int chunkSize )
{
return str.Buffer(chunkSize).Select(l => String.Concat(l));
}
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);
}