Dear friends,
I have got the Java code from the vendor to post a XML message with a JPG file to the webservice. Here you can find the Java code:
I have tried to tranlate this but without success, anybody know how to post the stuff like the code above?
This is the part of my code:
Im struggling with the objHttp.send (myDom.XML) part and dont know how to add the XML and the file to the send commando like in the java code. Could somebody help me with this code? Three golden stars for the final solution -
I have got the Java code from the vendor to post a XML message with a JPG file to the webservice. Here you can find the Java code:
Code:
import java.io.File;
import javax.ws.rs.core.MediaType;
import org.apache.commons.httpclient.HttpClient;
import org.apache.commons.httpclient.URI;
import org.apache.commons.httpclient.methods.PutMethod;
import org.apache.commons.httpclient.methods.multipart.FilePart;
import org.apache.commons.httpclient.methods.multipart.MultipartRequestEntity;
import org.apache.commons.httpclient.methods.multipart.Part;
import org.apache.commons.httpclient.methods.multipart.StringPart;
public class MultipartRecord
{
public static void main(String[] args)
{
try
{
File f = new File("c:\\image.jpg");
HttpClient client = new HttpClient();
PostMethod postMethod = new
PostMethod("[URL unfurl="true"]http://localhost/networking/rest/record/HOMEPAGE/");[/URL]
postMethod.setRequestHeader("Cookie", "JSESSIONID=A741270662A86F796DA16646F0708C43");
FilePart filePart = new FilePart("image_field", f);
filePart.setContentType(MediaType.APPLICATION_OCTET_STREAM);
StringPart sp = new StringPart("__xml_data__",
"<platform><record><field1>Some text field</field1>"
+ "<image_field>image.jpg</image_field></record></platform>");
sp.setContentType(MediaType.APPLICATION_XML);
final Part[] parts = { sp, filePart };
postMethod.setRequestEntity(new MultipartRequestEntity(parts, postMethod.getParams()));
int executeMethod = client.executeMethod(postMethod);
}
catch(Exception e)
{
System.out.println("Exception");
}
}
}
I have tried to tranlate this but without success, anybody know how to post the stuff like the code above?
This is the part of my code:
Code:
Public Sub main()
Dim query As String
Dim strUrl As String
Dim strUserName As String
Dim strPassword As String
Dim strUpload As String
Dim strFile
Dim myDom As MSXML2.DOMDocument
Dim myDomResult As MSXML2.DOMDocument
Dim objNodeList As IXMLDOMNodeList
Dim str As String
Set myDom = CreateObject("MSXML2.DOMDocument")
Set myDomResult = CreateObject("MSXML2.DOMDocument")
'Load entire Document before moving on
myDom.async = False
strWebAddress = "localhost/networking/rest/record/HOMEPAGE"
strUrl = "[URL unfurl="true"]https://"[/URL] & strWebAddress
Set objHttp = CreateObject("MSXML2.XMLHTTP.6.0")
Dim myFSO, f
Set myFSO = CreateObject("Scripting.FileSystemObject")
Set f = myFSO.getFile("C:\image.jpg")
strUpload = "<platform><record><field1>Example</field1>" & _
"<image_field>" & f & "</image_field></record></platform>"
myDom.loadXML (strUpload)
objHttp.Open "POST", strUrl, False
objHttp.setRequestHeader "Content-Type", "application/xml"
'This is the difficult part??
objHttp.send (myDom.XML)
'objHttp.send "__xml_data__" & (myDom.XML) & "image_field" & f
result2 = objHttp.responseText
Range("a3") = (result2)
End Sub
Im struggling with the objHttp.send (myDom.XML) part and dont know how to add the XML and the file to the send commando like in the java code. Could somebody help me with this code? Three golden stars for the final solution -