如何用c#中的一个空格替换字符串中的多个空格?
例子:
1 2 3 4 5
是:
1 2 3 4 5
如何用c#中的一个空格替换字符串中的多个空格?
例子:
1 2 3 4 5
是:
1 2 3 4 5
当前回答
我认为Matt的答案是最好的,但我不认为它是完全正确的。如果你想替换换行符,你必须使用:
myString = Regex.Replace(myString, @"\s+", " ", RegexOptions.Multiline);
其他回答
Regex即使执行简单的任务也会相当慢。这将创建一个可用于任何字符串的扩展方法。
public static class StringExtension
{
public static String ReduceWhitespace(this String value)
{
var newString = new StringBuilder();
bool previousIsWhitespace = false;
for (int i = 0; i < value.Length; i++)
{
if (Char.IsWhiteSpace(value[i]))
{
if (previousIsWhitespace)
{
continue;
}
previousIsWhitespace = true;
}
else
{
previousIsWhitespace = false;
}
newString.Append(value[i]);
}
return newString.ToString();
}
}
它将被这样使用:
string testValue = "This contains too much whitespace."
testValue = testValue.ReduceWhitespace();
// testValue = "This contains too much whitespace."
string sentence = "This is a sentence with multiple spaces";
RegexOptions options = RegexOptions.None;
Regex regex = new Regex("[ ]{2,}", options);
sentence = regex.Replace(sentence, " ");
不使用正则表达式:
while (myString.IndexOf(" ", StringComparison.CurrentCulture) != -1)
{
myString = myString.Replace(" ", " ");
}
可以在短弦上使用,但在有很多空格的长弦上表现不佳。
下面的代码将所有多个空格删除为一个空格
public string RemoveMultipleSpacesToSingle(string str)
{
string text = str;
do
{
//text = text.Replace(" ", " ");
text = Regex.Replace(text, @"\s+", " ");
} while (text.Contains(" "));
return text;
}
混合StringBuilder和Enumerable.Aggregate()作为字符串的扩展方法:
using System;
using System.Linq;
using System.Text;
public static class StringExtension
{
public static string CondenseSpaces(this string s)
{
return s.Aggregate(new StringBuilder(), (acc, c) =>
{
if (c != ' ' || acc.Length == 0 || acc[acc.Length - 1] != ' ')
acc.Append(c);
return acc;
}).ToString();
}
public static void Main()
{
const string input = " (five leading spaces) (five internal spaces) (five trailing spaces) ";
Console.WriteLine(" Input: \"{0}\"", input);
Console.WriteLine("Output: \"{0}\"", StringExtension.CondenseSpaces(input));
}
}
执行这个程序产生以下输出:
Input: " (five leading spaces) (five internal spaces) (five trailing spaces) "
Output: " (five leading spaces) (five internal spaces) (five trailing spaces) "