Question

我有这种文字格式：

name:
last name:
birthday:
years old:
parent:
school:

我有下一个信息..

name:name1
last name:lastname1
birthday:13/03/1991
years old:20
parent:fatherx
school:university x

我如何获得：

name1
lastname1
13/03/1991
20
fatherx
university x

...对于不同的变量？不要忘记用户有时他们没有信息，例如他们有空

parent:

Answer 1

结肠上的

Split。例如，如果您将每条线存储在一个单独的字符串中，则可以执行以下操作：

string s = "name:angel rodrigo";
string name= s.Split(':')[1];  // Get everything after the colon

Answer 2

您可以使用以下代码创建键值对字典。

     List<string> fields = new List<string>
     {
        "name:",
        "last name:",
        "birthday:",
        "years old:",
        "parent:",
        "school:",
     };
     string rawData =
        @"name:angel rodrigo
last name:uc ku
birthday:13/03/1991
years old:20
parent:fernando uc puc
school:university x";

     var data = 
        fields.ToDictionary(
           field => field.TrimEnd (':'),
           field => Regex.Match(rawData, "(?<=" + Regex.Escape(field) + ").*"));
     foreach (var kvp in data)
     {
        Console.WriteLine(kvp.Key + "  =>  " + kvp.Value);
     }

产生这个结果：

name  =>  angel rodrigo
last name  =>  uc ku
birthday  =>  13/03/1991
years old  =>  20
parent  =>  fernando uc puc
school  =>  university x

你如何在另一个文本中获得子串？

2 个答案: