我正在尝试获取已发送到我的应用程序的XML文件的正文数据。
但当我复制内容并将其放在新文件中时,内容是不正确的。
原始XML:
<?xml version="1.0" encoding="utf-8"?>
<eM_NM_001 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<MESSAGE>
....
</MESSAGE>
</eM_NM_001>
粘贴正文后的新文件:
-----------------------8d618ce02a1a097
Content-Disposition: form-data; name="file"; filename="14_20180912162639.xml"
Content-Type: application/octet-stream
<?xml version="1.0" encoding="utf-8"?>
<eM_NM_001 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<MESSAGE>
...
</MESSAGE>
</eM_NM_001>
-----------------------8d618ce02a1a097--
我只希望第二个文件也包含
xml
而不是标题数据…(至少我认为是标题数据)
我到底怎么解决这个问题?
到目前为止我得到的是:
//POST api/<controller>
[HttpPost]
public string Post()
{
var result = "";
if (Request.Content != null)
{
string map = @"D:\bestelApp\bestelApp\Content\bestellingen\ontvangen";
string count = Directory.GetFiles(map, "*.xml").Count().ToString();
string extention = ".xml";
string fileName = "bestelling" + (count != "0" ? count : "") + extention;
string path = map + "//" + fileName;
using (Stream output = File.OpenWrite(path))
{
using (Stream input = HttpContext.Current.Request.GetBufferedInputStream())
{
input.CopyTo(output);
}
}
result = new bestellingenController().ConvertXmlToObj(fileName, path);
}
return result;
}
到目前为止我所做的:
[HttpPost]
public void Post(HttpRequestMessage request) {
var doc = new XmlDocument();
doc.Load(request.Content.ReadAsStreamAsync().Result);
doc.Save(@"D:\bestelApp\bestelApp\Content\bestellingen\ontvangen");
}
var context = new HttpContextWrapper(HttpContext.Current);
context.Request.ContentType = null;
HttpRequestBase request = context.Request; //This was an attempt to set everything to null except the xml.
//Get the data from the HTTP stream
var body = new StreamReader(HttpContext.Current.Request.InputStream).ReadToEnd();
var xmlDocument = new XmlDocument();
xmlDocument.LoadXml(body);
xmlDocument.Save(@"D:\bestelApp\bestelApp\Content\bestellingen\ontvangen\" + fileName);
I had already asked this question but poorly..