Forum Discussion

bessie's avatar
bessie
New Contributor
9 years ago

how to get the zip file after running HTTP request.

Hi,   The http request is about exporting application to local path.   I run the HTTP request in postman and click "Send and Download" button, and can save the exported zip file into local path.  ...
nmrao's avatar
nmrao
Champion Level 3
Looks there are no attachments and you are getting the binary data as Response.

So, you may need to extract the response content and save it as zip file using groovy script.
bessie's avatar
bessie
New Contributor
9 years ago

Yes, do you know how to extract the response content and save it as zip file? Would you please share code with me? Thanks again.

    • bessie's avatar
      bessie
      New Contributor
      9 years ago

      Now after running changed scripts, .zip file is created. However, cannot open it, please see attached picture. Would you please see which script line I missed? Thanks.

       

      def responseBody=testRunner.testCase.getTestStepByName("ExportApplication").httpRequest.response.responseContent;
//InputStream ins = new ByteArrayInputStream(responseBody.getBytes("UTF-8"))
InputStream ins = new ByteArrayInputStream(responseBody.getBytes())
log.info responseBody
def outFile = new FileOutputStream(new File('C:/temp'+'/TestCreateDeleteApp_1.0.zip'))
if (ins) {
com.eviware.soapui.support.Tools.writeAll(outFile, ins )
}
ins.close()
outFile.close()



      • jurven007's avatar
        jurven007
        Occasional Contributor
        8 years ago

        Hi All, 

         

        I used the script from the previous answer and was able to save a zip. When I open the zip and try to open the xml inside I get a Data error 'file is broken'.

        What am I doing wrong?