UrlFetchApp.fetch in google app scripts delete url parameter

Question

I'm building a simple spreadsheet that monitors a series of url for their social sharing data using the sharedcount.com API.

For some reason the UrlFetchApp.fetch() eliminate the url I pass as a parameter of the endpoint to interrogate.

This is my code:

var sheet = SpreadsheetApp.getActiveSheet(),
          urls = sheet.getRange('A:A').getValues().slice(1),
          apikey = "XXXXXXXXXXXXXXXXX",
          apiEndPoint = "https://free.sharedcount.com/"
          data = [];

      urls.forEach(function(url, idx) {

        apiurl = apiEndPoint + '?url=' + encodeURIComponent(url) + '&apikey=' + apikey

        Logger.log(apiurl)

        sharedCountData = UrlFetchApp.fetch(apiurl)
        data[idx] = sharedCountData
      })

The logger function returns the correctly built url (eg: https://free.sharedcount.com/?url=http%3A%2F%2Fwww.teamvaxitalia.it%2F&apikey=XXXXXXXXXXXXXX), but running my code I receive a 401 error when it reach fetch(), in which the passed endpoint is this: https://free.sharedcount.com/?url=&apikey=XXXXXXXXXXXXXX and the error is "Not a valid URL". The url parameter is missing!

Any idea about what's going on?

Thanks

SOLVED: For some reason with UrlFetchApp.fetch it was needed to use https://free.sharedcount.com/url?url= as endpoint, not just https://free.sharedcount.com/?url=, even if in browser both work.

Bakaburg · Accepted Answer · 2016-04-24 09:42:31Z

1

SOLVED: For some reason with UrlFetchApp.fetch it was needed to use https://free.sharedcount.com/url?url= as endpoint, not just https://free.sharedcount.com/?url=, even if in browser both work.

But suggestions on why this was happening are welcome.

answered Apr 24, 2016 at 9:42

Bakaburg

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