Delphi DLL

Parse Multipart Binary Http Response

See more HTTP Examples

This example demonstrates how to parse an HTTP response that is multipart and contains a binary file, such as a .zip or .pdf. Delphi DLL

uses
    Winapi.Windows, Winapi.Messages, System.SysUtils, System.Variants, System.Classes, Vcl.Graphics,
    Vcl.Controls, Vcl.Forms, Vcl.Dialogs, Vcl.StdCtrls, Http, HttpRequest, BinData, HttpResponse, Mime;

...

procedure TForm1.Button1Click(Sender: TObject);
var
success: Boolean;
http: HCkHttp;
req: HCkHttpRequest;
resp: HCkHttpResponse;
respBody: HCkBinData;
mime: HCkMime;
numParts: Integer;
part0: HCkMime;
xmlStr: PWideChar;
part1: HCkMime;
zipData: HCkBinData;

begin
success := False;

// This example requires the Chilkat API to have been previously unlocked.
// See Global Unlock Sample for sample code.

http := CkHttp_Create();
req := CkHttpRequest_Create();

// ...
// Insert code here to construct some kind of HTTP request.
// this example is to show how to parse a particular kind of response.
// ...
// ...

// Send the request (whatever it may be in your case) to get the HTTP response object.
resp := CkHttpResponse_Create();
success := CkHttp_HttpSReq(http,'www.somedomain.com',443,True,req,resp);
if (success = False) then
  begin
    Memo1.Lines.Add(CkHttp__lastErrorText(http));
    Exit;
  end;

// Get the response body (which is expected to be binary)
respBody := CkBinData_Create();
CkHttpResponse_GetBodyBd(resp,respBody);

// For this example, the response body contains something like this:

// ------=_Part_21302_2029949381.1547401515443
// Content-Type: application/xop+xml; charset=UTF-8; type="text/xml"
// Content-Transfer-Encoding: 8bit
// Content-ID: <rootpart@ws.jboss.org>
// 
// <env:Envelope xmlns:env='http://schemas.xmlsoap.org/soap/envelope/'><env:Header></env:Header><env:Body>...</env:Body></env:Envelope>
// ------=_Part_21302_2029949381.1547401515443
// Content-Type: application/octet-stream
// Content-Transfer-Encoding: binary
// Content-Id: <fileArchivio-7d302908-4d64-43d3-bf4e-79ce806d43b3@ws.jboss.org>
// 
// BINARY_CONTENT_HERE...
// 
// ------=_Part_21302_2029949381.1547401515443--
// 

// Load it into a Chilkat MIME object.
mime := CkMime_Create();
success := CkMime_LoadMimeBd(mime,respBody);
if (success = False) then
  begin
    Memo1.Lines.Add(CkMime__lastErrorText(mime));
    Exit;
  end;

numParts := CkMime_getNumParts(mime);
if (numParts < 2) then
  begin
    Memo1.Lines.Add('Expected multipart MIME with at least 2 sub-parts.');
    Exit;
  end;

// Get the 1st sub-part, which is the XML.

part0 := CkMime_Create();
CkMime_PartAt(mime,0,part0);

// Should be OK because we checked NumParts above..
xmlStr := CkMime__getBodyDecoded(part0);
Memo1.Lines.Add(xmlStr);
Memo1.Lines.Add('----');

// Save the 2nd part to a file.  (It is a .zip file in our test case..)

part1 := CkMime_Create();
CkMime_PartAt(mime,1,part1);

success := CkMime_SaveBody(part1,'qa_output/attachedZip.zip');

// Alternatively, we could extract the binary data to a BinData and use elsewhere..
zipData := CkBinData_Create();
success := CkMime_GetBodyBd(part1,zipData);
success := CkBinData_WriteFile(zipData,'qa_output/attachedZip_again.zip');

Memo1.Lines.Add('OK.');

CkHttp_Dispose(http);
CkHttpRequest_Dispose(req);
CkHttpResponse_Dispose(resp);
CkBinData_Dispose(respBody);
CkMime_Dispose(mime);
CkMime_Dispose(part0);
CkMime_Dispose(part1);
CkBinData_Dispose(zipData);

end;