我想知道在.NET中是否有一个用来解析字符串位的内置方式。
例如,我有以下字符串:
"bsarbirthd0692"由以下部分组成,以后将与数据交叉引用:
Indexes Purpose
0-3 (name)
4-9 (description)
10-13 (date mm-yy)我希望有这样的本土化的东西:
string name, desc, date;
string.ParseFormat("{0:4}{1:5}{2:4}", "bsarbirthd0692", out name, out desc, out date);在.NET或流行的库中,是否有一种本地的方法来做到这一点?
发布于 2015-04-15 15:15:58
由于已知格式,且不应更改子字符串,因此应适用于您
string data = "bsarbirthd0692";
string name, desc, date;
name = data.Substring(0, 4);
desc = data.Substring(4, 6);
date = data.SubString(10);编辑
还有一些扩展方法,您可以创建这些方法来执行任何您想做的事情。这显然比以前的建议复杂得多。
public static class StringExtension
{
/// <summary>
/// Returns a string array of the original string broken apart by the parameters
/// </summary>
/// <param name="str">The original string</param>
/// <param name="obj">Integer array of how long each broken piece will be</param>
/// <returns>A string array of the original string broken apart</returns>
public static string[] ParseFormat(this string str, params int[] obj)
{
int startIndex = 0;
string[] pieces = new string[obj.Length];
for (int i = 0; i < obj.Length; i++)
{
if (startIndex + obj[i] < str.Length)
{
pieces[i] = str.Substring(startIndex, obj[i]);
startIndex += obj[i];
}
else if (startIndex + obj[i] >= str.Length && startIndex < str.Length)
{
// Parse the remaining characters of the string
pieces[i] = str.Substring(startIndex);
startIndex += str.Length + startIndex;
}
// Remaining indexes, in pieces if they're are any, will be null
}
return pieces;
}
}用法1:
string d = "bsarbirthd0692";
string[] pieces = d.ParseFormat(4,6,4);结果:

用法2:
string d = "bsarbirthd0692";
string[] pieces = d.ParseFormat(4,6,4,1,2,3);结果:

发布于 2015-04-15 15:20:56
您可以使用Regexp进行此操作。
string str= "bsarbirthd0692";
var regex = "(?<name>.{4})(?<desc>.{6})(?<date>.{4})";
MatchCollection matches = Regex.Matches(str, regex);
foreach(Match m in matches){
Console.WriteLine(m.Groups["name"].ToString());
Console.WriteLine(m.Groups["desc"].ToString());
Console.WriteLine(m.Groups["date"].ToString());
}发布于 2015-04-15 15:17:42
没有类似的东西,但是写一些东西来实现:
IEnumerable<string> inputString.BreakIntoLengths(4, 6, 4)签名:
public IEnumerable<string> BreakIntoLengths(this string input, params int[] lengths);很简单:
public IEnumerable<string> BreakIntoLengths(this string input, params int[] lengths) {
var pos = 0;
foreach (var len in lengths) {
yield return input.Substring(pos, len);
pos += len;
}
}(实际的实现有一些错误检查。)
注意:我已经删除了格式字符串,如接口:它似乎没有提供任何价值。一旦返回集合,就很容易按索引分配条目。
https://stackoverflow.com/questions/29653991
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