我有这个方法从字符串URI中获取文件名。我该怎么做才能让它更健壮呢?
private string GetFileName(string hrefLink)
{
string[] parts = hrefLink.Split('/');
string fileName = "";
if (parts.Length > 0)
fileName = parts[parts.Length - 1];
else
fileName = hrefLink;
return fileName;
}
这是我的样本,你可以使用:
public static string GetFileNameValidChar(string fileName)
{
foreach (var item in System.IO.Path.GetInvalidFileNameChars())
{
fileName = fileName.Replace(item.ToString(), "");
}
return fileName;
}
public static string GetFileNameFromUrl(string url)
{
string fileName = "";
if (Uri.TryCreate(url, UriKind.Absolute, out Uri uri))
{
fileName = GetFileNameValidChar(Path.GetFileName(uri.AbsolutePath));
}
string ext = "";
if (!string.IsNullOrEmpty(fileName))
{
ext = Path.GetExtension(fileName);
if (string.IsNullOrEmpty(ext))
ext = ".html";
else
ext = "";
return GetFileNameValidChar(fileName + ext);
}
fileName = Path.GetFileName(url);
if (string.IsNullOrEmpty(fileName))
{
fileName = "noName";
}
ext = Path.GetExtension(fileName);
if (string.IsNullOrEmpty(ext))
ext = ".html";
else
ext = "";
fileName = fileName + ext;
if (!fileName.StartsWith("?"))
fileName = fileName.Split('?').FirstOrDefault();
fileName = fileName.Split('&').LastOrDefault().Split('=').LastOrDefault();
return GetFileNameValidChar(fileName);
}
用法:
var fileName = GetFileNameFromUrl("http://cdn.p30download.com/?b=p30dl-software&f=Mozilla.Firefox.v58.0.x86_p30download.com.zip");
大多数其他答案要么不完整,要么不处理路径之后的东西(查询字符串/哈希)。
readonly static Uri SomeBaseUri = new Uri("http://canbeanything");
static string GetFileNameFromUrl(string url)
{
Uri uri;
if (!Uri.TryCreate(url, UriKind.Absolute, out uri))
uri = new Uri(SomeBaseUri, url);
return Path.GetFileName(uri.LocalPath);
}
测试结果:
GetFileNameFromUrl(""); // ""
GetFileNameFromUrl("test"); // "test"
GetFileNameFromUrl("test.xml"); // "test.xml"
GetFileNameFromUrl("/test.xml"); // "test.xml"
GetFileNameFromUrl("/test.xml?q=1"); // "test.xml"
GetFileNameFromUrl("/test.xml?q=1&x=3"); // "test.xml"
GetFileNameFromUrl("test.xml?q=1&x=3"); // "test.xml"
GetFileNameFromUrl("http://www.a.com/test.xml?q=1&x=3"); // "test.xml"
GetFileNameFromUrl("http://www.a.com/test.xml?q=1&x=3#aidjsf"); // "test.xml"
GetFileNameFromUrl("http://www.a.com/a/b/c/d"); // "d"
GetFileNameFromUrl("http://www.a.com/a/b/c/d/e/"); // ""