如何在C＃中根据文件名格式过滤文件

Question

我需要使用某种文件名格式从FTP服务器下载文件。服务器中的文件名可能如下所示：

1. O形0098-00009801_08052014054258.xml
2. 订单ID 3359_08062014121815.xml
3. O形0302-00022043_07312014085513.xml
4. O形LABELS_94920235569876.XML
5. JO ANN_08062014170911.xml
6. O形0233-00026378_07312014155245.xml
7. LABELS_08052014174907.xml

如果名称格式为O-XXXX-XXXXXXXX_XXXXXXXXXXXXXXXX，我只想下载文件

我使用的是以下代码，但效果不是很好。仅选择上述文件名格式的文件最有效的方法是什么？

using (Stream responseStream = response.GetResponseStream())
{
    using (StreamReader reader = new StreamReader(responseStream))
    {
        while (!reader.EndOfStream)
        {
            fileName = reader.ReadLine();
            if (fileName.ToUpper().IndexOf("O-") > -1)
                files.Add(fileName);
        }
    }
}

结果是只选择这些文件：

1. O形0098-00009801_08052014054258.xml
2. O形0233-00026378_07312014155245.xml
3. O形0302-00022043_07312014085513.xml

Answer 1

假设您已将文件列表存储在List<string>中，则提取很简单

List<string> files = new List<string>()
{
    "O-0098-00009801_08052014054258.xml",
    "ORDER ID 3359_08062014121815.xml",
    "O-0302-00022043_07312014085513.xml",
    "O-LABELS_94920235569876.XML",
    "JO ANN_08062014170911.xml",
    "O-0233-00026378_07312014155245.xml",
    "LABELS_08052014174907.xml"
};

Regex r = new Regex(@"O-\d{4}-\d{8}_\d{14}");
var result = files.Where(x => r.IsMatch(x)).ToList();

Answer 2

你可能只是这样做：

List<string> orders = new List<string>()
{
    "O-0098-00009801_08052014054258.xml",
    "ORDER ID 3359_08062014121815.xml",
    "O-0302-00022043_07312014085513.xml",
    "O-LABELS_94920235569876.XML",
    "JO ANN_08062014170911.xml",
    "O-0233-00026378_07312014155245.xml",
    "LABELS_08052014174907.xml"
};

List<string> valid = new List<string>()
{
    "O-0098-00009801_08052014054258.xml",
    "O-0302-00022043_07312014085513.xml",
    "O-0233-00026378_07312014155245.xml"
};

List<string> order = orders.Where(x => valid.Contains(x)).ToList();

如果你有List有效/总数，你可以用上述方式完成。否则你可以使用史蒂夫回答的正则表达式。