Ead*_*del 21
您可以使用以下代码放置表单按钮并为其创建单击处理程序:
private void buttonGetFile_Click(object sender, EventArgs e)
{
OpenFileDialog dialog = new OpenFileDialog();
dialog.Filter = "Text files | *.txt"; // file types, that will be allowed to upload
dialog.Multiselect = false; // allow/deny user to upload more than one file at a time
if (dialog.ShowDialog() == DialogResult.OK) // if user clicked OK
{
String path = dialog.FileName; // get name of file
using (StreamReader reader = new StreamReader(new FileStream(path, FileMode.Open), new UTF8Encoding())) // do anything you want, e.g. read it
{
// ...
}
}
}
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jon*_*ham -2
请参阅本教程了解原始 HTTP POST:
http://msdn.microsoft.com/en-us/library/debx8sh9.aspx
参考.NET的WebClient类:
http://msdn.microsoft.com/en-us/library/system.net.webclient(v=vs.80).aspx
一个简单的 HTTP POST 可以这样完成:
string Upload_File_Content = ...;
string Url = ...;
using (var Http_Client = new WebClient()) {
var Post_Data = new NameValueCollection();
Post_Data["upload_file"] = Upload_File_Content;
var Response = Http_Client.UploadValues(Url,"POST",Post_Data);
}
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