I've implemented the getFeature function in my OL4 as in the example: https://openlayers.org/en/latest/examples/getfeatureinfo-tile.html

  map.on('singleclick', function(evt) {
    document.getElementById('info').innerHTML = '';
    var viewResolution = /** @type {number} */ (view.getResolution());
    var url = wmsSource.getGetFeatureInfoUrl(
        evt.coordinate, viewResolution, 'EPSG:3857',
        {'INFO_FORMAT': 'text/html'});
    if (url) {
      document.getElementById('info').innerHTML =
          '<iframe seamless src="' + url + '"></iframe>';

How can I handle the result in order to create a tooltip? I don't want to use the <iframe>

If I console.log the var URL, I have back the URL itself and not the XML page, how can I have back the XML page?

I also tried with var xmlDoc = parser.parseFromString(url,"text/xml"); but what I get back is not the XML.

  • 1
    You can use Overlay as an HTMLElement. and just check the comments on your past question. that's the answer how you do. – Chase Choi Nov 28 '17 at 13:17
  • I answer you in the other post, in this post we are talking about different stuff, if you click on the street line in this web traffwebdemo.co.uk/parking you have back data from mapserver, and it's manipulated to appear in the popup. – Sebastiano Nov 28 '17 at 15:04
  • the comment is for your first question about iframe. – Chase Choi Nov 29 '17 at 0:40
  • As I don't want the iframe and I don't know how to use the Overlay I found a way with XMLHttpRequest() you can see the solution below in my answer. It seems that OL4 doesn't create a real xml request, it gives you an URL and with iframe you can access that URL and import the content, but I want to manipulate the content before to show it on the screen. – Sebastiano Nov 29 '17 at 10:52

I found a way.

if (url) {

var xhttp = new XMLHttpRequest();
xhttp.open("GET", url, true);
xhttp.onreadystatechange = function (aEvt) {
  if (xhttp.readyState == 4) {
    if (xhttp.status == 200)  {
      document.getElementById('popup-content').innerHTML = xhttp.responseText;

} // end if url

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