我有一个通过.NET System.Net.WebRequest操作的http POST,如下所示:
...
XXXUtilities.Log.WriteLog(string.Format("XXXHTTPPost PostToUri has uri={0}, body={1}", uri, messageBodyAsString));
System.Net.WebRequest req = System.Net.WebRequest.Create(uri);
req.ContentType = "application/x-www-form-urlencoded";
req.Method = "POST";
byte[] bytes = System.Text.Encoding.ASCII.GetBytes(messageBodyAsString);
req.ContentLength = bytes.Length;
System.IO.Stream os = req.GetRequestStream();
os.Write(bytes, 0, bytes.Length);
os.Close();
try
{
using (System.Net.WebResponse resp = req.GetResponse())
{
if (resp == null) return null;
System.IO.StreamReader sr =
new System.IO.StreamReader(resp.GetResponseStream());
string rs = sr.ReadToEnd().Trim();
sr.Close();
resp.Close();
XXXUtilities.Log.WriteLog(string.Format("XXXHTTPPost PostToUri has string response = {0}", rs));
MongoDB.Bson.BsonDocument doc2 = new BsonDocument();
doc2.Add("Response", rs);
return doc2;
}
}
catch (System.Net.WebException e)
{...
这一切在大多数时候都很好。但是,查看这个创建的日志文件我发现了一些奇怪的东西。可疑日志条目如下所示:
18:59:17.0608 HPSHTTPPost PostToUri has uri=https://salesforce.ringlead.com/cgi-bin/2848/3/dedup.pl, body=LastName=Doe&FirstName=Jon
18:59:17.5608 HPSHTTPPost PostToUri has string response = Success
18:59:18.0295 HPSHTTPPost PostToUri has string response = Success
似乎Http Response被收到两次。这在技术上是否可行?即,Http POST是否可以一个接一个地接收两个响应?如果是这样,我的代码是否可能被调用两次,从而导致观察到的日志文件条目?非常感谢。
编辑: 为了回应日志代码可能被破坏的注释,这里是日志代码:
public class Log
{
public static void WriteLog(string commandText)
{
string clientDBName = "test";
string username = "test";
try
{
string filePath = "c:\\Data\\XXXLogs\\" + clientDBName + "logs\\";
string filename = System.DateTime.Now.ToString("yyyyMMdd_") + username + ".log";
DirectoryInfo dir = new DirectoryInfo(filePath);
if (!dir.Exists)
{
dir.Create();
}
System.IO.FileStream stream = new System.IO.FileStream(
filePath + filename
, System.IO.FileMode.Append); // Will create if not already exists
StreamWriter writer = new StreamWriter(stream);
writer.WriteLine(); // Writes a line terminator, thus separating entries by 1 blank line
writer.WriteLine(System.DateTime.Now.ToString("HH:mm:ss.ffff") + " " + commandText);
writer.Flush();
stream.Close();
}
catch { }
}
}