Go

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. Go

    success := false

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

    http := chilkat.NewHttp()
    req := chilkat.NewHttpRequest()

    // ...
    // 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 := chilkat.NewHttpResponse()
    success = http.HttpSReq("www.somedomain.com",443,true,req,resp)
    if success == false {
        fmt.Println(http.LastErrorText())
        http.DisposeHttp()
        req.DisposeHttpRequest()
        resp.DisposeHttpResponse()
        return
    }

    // Get the response body (which is expected to be binary)
    respBody := chilkat.NewBinData()
    resp.GetBodyBd(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 := chilkat.NewMime()
    success = mime.LoadMimeBd(respBody)
    if success == false {
        fmt.Println(mime.LastErrorText())
        http.DisposeHttp()
        req.DisposeHttpRequest()
        resp.DisposeHttpResponse()
        respBody.DisposeBinData()
        mime.DisposeMime()
        return
    }

    numParts := mime.NumParts()
    if numParts < 2 {
        fmt.Println("Expected multipart MIME with at least 2 sub-parts.")
        http.DisposeHttp()
        req.DisposeHttpRequest()
        resp.DisposeHttpResponse()
        respBody.DisposeBinData()
        mime.DisposeMime()
        return
    }

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

    part0 := chilkat.NewMime()
    mime.PartAt(0,part0)

    // Should be OK because we checked NumParts above..
    xmlStr := part0.GetBodyDecoded()
    fmt.Println(*xmlStr)
    fmt.Println("----")

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

    part1 := chilkat.NewMime()
    mime.PartAt(1,part1)

    success = part1.SaveBody("qa_output/attachedZip.zip")

    // Alternatively, we could extract the binary data to a BinData and use elsewhere..
    zipData := chilkat.NewBinData()
    success = part1.GetBodyBd(zipData)
    success = zipData.WriteFile("qa_output/attachedZip_again.zip")

    fmt.Println("OK.")

    http.DisposeHttp()
    req.DisposeHttpRequest()
    resp.DisposeHttpResponse()
    respBody.DisposeBinData()
    mime.DisposeMime()
    part0.DisposeMime()
    part1.DisposeMime()
    zipData.DisposeBinData()