veg*_*rby 22
如果您希望分隔符为"自己的分割",则可以使用Regex.Split,例如:
string input = "plum-pear";
string pattern = "(-)";
string[] substrings = Regex.Split(input, pattern); // Split on hyphens
foreach (string match in substrings)
{
Console.WriteLine("'{0}'", match);
}
// The method writes the following to the console:
// 'plum'
// '-'
// 'pear'
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此版本不使用LINQ或Regex,因此可能相对有效。我认为它可能比Regex更容易使用,因为您不必担心转义特殊的分隔符。它返回的an IList<string>比总是转换为数组更有效。这是一种扩展方法,非常方便。您可以将定界符作为数组或多个参数传入。
/// <summary>
/// Splits the given string into a list of substrings, while outputting the splitting
/// delimiters (each in its own string) as well. It's just like String.Split() except
/// the delimiters are preserved. No empty strings are output.</summary>
/// <param name="s">String to parse. Can be null or empty.</param>
/// <param name="delimiters">The delimiting characters. Can be an empty array.</param>
/// <returns></returns>
public static IList<string> SplitAndKeepDelimiters(this string s, params char[] delimiters)
{
var parts = new List<string>();
if (!string.IsNullOrEmpty(s))
{
int iFirst = 0;
do
{
int iLast = s.IndexOfAny(delimiters, iFirst);
if (iLast >= 0)
{
if (iLast > iFirst)
parts.Add(s.Substring(iFirst, iLast - iFirst)); //part before the delimiter
parts.Add(new string(s[iLast], 1));//the delimiter
iFirst = iLast + 1;
continue;
}
//No delimiters were found, but at least one character remains. Add the rest and stop.
parts.Add(s.Substring(iFirst, s.Length - iFirst));
break;
} while (iFirst < s.Length);
}
return parts;
}
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一些单元测试:
text = "[a link|http://www.google.com]";
result = text.SplitAndKeepDelimiters('[', '|', ']');
Assert.IsTrue(result.Count == 5);
Assert.AreEqual(result[0], "[");
Assert.AreEqual(result[1], "a link");
Assert.AreEqual(result[2], "|");
Assert.AreEqual(result[3], "http://www.google.com");
Assert.AreEqual(result[4], "]");
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这个问题有很多答案!我敲了一个被各种字符串拆分的字符串(原始答案仅适用于字符,即长度为 1)。这还没有经过全面测试。
public static IEnumerable<string> SplitAndKeep(string s, params string[] delims)
{
var rows = new List<string>() { s };
foreach (string delim in delims)//delimiter counter
{
for (int i = 0; i < rows.Count; i++)//row counter
{
int index = rows[i].IndexOf(delim);
if (index > -1
&& rows[i].Length > index + 1)
{
string leftPart = rows[i].Substring(0, index + delim.Length);
string rightPart = rows[i].Substring(index + delim.Length);
rows[i] = leftPart;
rows.Insert(i + 1, rightPart);
}
}
}
return rows;
}
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