我有一个更新服务器,它有一个XML来告诉应用程序的updater函数所需的详细信息/文件等。此时,我必须抓取xml,保存它,然后将保存的xml加载到数据集中如下:
Private Function GetXML() as XMLDocument
Dim request As System.Net.HttpWebRequest = System.Net.HttpWebRequest.Create("http://somewebsite.com/current.xml")
Try
Dim response As System.Net.HttpWebResponse = request.GetResponse()
If response.StatusCode = System.Net.HttpStatusCode.OK Then
Dim stream As System.IO.Stream = response.GetResponseStream()
Dim reader As New System.IO.StreamReader(stream)
Dim contents As String = reader.ReadToEnd()
Dim document As New System.Xml.XmlDocument()
document.LoadXml(contents)
Return document
Catch ex As Exception
End Try
End Function
Private Sub DoUpdate()
Dim ServerXML as XMLDocument = GetServerXML()
ServerXML.Save("Server.xml")
Dim ServerDataSet as new DataSet
ServerDataSet.ReadXml("Server.xml")
End Sub
必须有一个更快的方法,而不保存本地文件,但我显然卡住了,并采取了无穷无尽的互联网知识。
一些简单的事情:
Dim stream As System.IO.Stream = response.GetResponseStream()
Dim reader As New System.IO.StreamReader(stream)
ServerDataset.ReadXML(reader)
去互联网做你的事! :)谢谢大家。
答案 0 :(得分:0)
事实证明:
bool arrCheck(char n[],char pos1,char pos2,char pos3,int size)
{
int first = 0,second = 0, third = 0;
for (int i=0;i<size;i++) {
if (n[i]==pos1) {
first = 1;
} else if (n[i]==pos2) {
second = 1;
} else if (n[i]==pos3) {
third = 1;
}
}
if( first+ second + third == 3)
return true;
else
return false;
}
事实上;工作。我有一个XML格式化问题,将其保存到文件然后重新打开它会起作用,但是通过读取器加载它不会。