如何在C#中的开始和结束字符之间获取字符串

时间:2015-11-19 12:43:52

标签: c#

如何获取字符串的子字符串,如file_ {AAA} _ {BBB} .xml?

我的代码:

string[] result = Regex.Split(str, "{(.*)}"); result.ToList().ForEach(x => MessageBox.Show(x));

结果:

AAA
BBB

4 个答案:

答案 0 :(得分:1)

你可以试试这个:

string str = "file_{AAA}_{BBB}.xml";
var regex = new Regex("(?<=\{)[^}]*(?=\})");
var matches = regex.Matches(str); 

答案 1 :(得分:0)

超级贫民区,我吮吸正则表达式而你应该使用Rahul的答案,但是:

string f = "file_{AAA}_{BBB}.xml";
string o = String.Empty;

while (f.Contains('{') && f.Contains('}'))
{
    int openIndex = f.IndexOf('{');
    int closeIndex = f.IndexOf('}');

    o += f.Substring(openIndex + 1, closeIndex - openIndex - 1) + " ";
    f = f.Remove(0, closeIndex + 1);
}

o.Trim();
Console.WriteLine(o);

将输出AAA BBB

答案 2 :(得分:0)

来自Rahul Tripathi的原文,刚刚删除{}

        string str = "file_{AAA}_{BBB}.xml";
        var regex = new Regex("{.*?}");
        var matches = regex.Matches(str);
        List<string> result = new List<string>();
        foreach (var item in matches)
        {
            result.Add(item.ToString().Replace("{", "").Replace("}", ""));
        }

答案 3 :(得分:-1)

您可以遵循:

string filename = string.Format(@&#34; {0}&#34;,Path.GetFileName(fileUpload1.PostedFile.FileName));
string filepath =&#34; f:\ ClientDocument \&#34; + Path.GetFileName(fileUpload1.PostedFile.FileName);             int fileLength =(fileUpload1.PostedFile.ContentLength)/ 1024;

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