Passing URL variable with & from JS to PHP result in "&" omission

Question

It is a bit difficult to find the proper title for this question for me, so maybe this example will clarify my issue.

I am making an ajax request to pass some variables from a JS to a PHP. One of these variables is a URL with some options, namely

https://www.wondermap.it/cgi-bin/qgis_mapserv.fcgi?map=/home/ubuntu/qgis/projects/Demo_sci_WMS/demo_sci.qgs&SERVICE=WFS&VERSION=1.1.0&REQUEST=GetFeature&TYPENAME=impianti_risalita&

The PHP code is ignoring any options after the first & symbol, considering only this part

https://www.wondermap.it/cgi-bin/qgis_mapserv.fcgi?map=/home/ubuntu/qgis/projects/Demo_sci_WMS/demo_sci.qgs

The AJAX request to the PHP I am making at the moment looks like

https://localhost/shire/php/export_wfs.php?wfs_url=https://www.wondermap.it/cgi-bin/qgis_mapserv.fcgi?map=/home/ubuntu/qgis/projects/Demo_sci_WMS/demo_sci.qgs&SERVICE=WFS&VERSION=1.1.0&REQUEST=GetFeature&TYPENAME=impianti_risalita&format=ESRI%20Shapefile

being wfs_url and format the two parameters the PHP is supposed to process.

I think i am supposed to avoid placing the & symbols in the wfs_url parameter, but I have no idea what should i do instead. Any help would be appreciated.

EDIT

Here is the AJAX call:

var xhr;
if (window.XMLHttpRequest) xhr = new XMLHttpRequest(); // all browsers
else xhr = new ActiveXObject("Microsoft.XMLHTTP"); // for IE

// url is https://www.wondermap.it/cgi-bin/qgis_mapserv.fcgi?map=/home/ubuntu/qgis/projects/Demo_sci_WMS/demo_sci.qgs&SERVICE=WFS&VERSION=1.1.0&REQUEST=GetFeature&TYPENAME=impianti_risalita&
var php_url = window.location.protocol + "//" + window.location.hostname + '/shire/php/export_wfs.php?wfs_url=' + url + 'format=' + format_list[0];
xhr.open('GET', php_url, false);
xhr.onreadystatechange = function () {
    if (xhr.readyState===4 && xhr.status===200) {
        alert('Downloading...');
    }
}
xhr.send();

return false;
});

@ChrisG That could be a possibility, will work on it and come back with my finding. — umbe1987, Dec 30 '18 at 23:41
@umbe1987 Looks like you're not [URL Encoding](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/encodeURI) the request. — General_Twyckenham, Dec 30 '18 at 23:43
@General_Twyckenham I tried both placing `encodeURI()` in front of the url varialbe (`encodeURI(url)`) and the entire php_url string (`encodeURI(window.location.protocol + "//" + window.location.hostname + '/shire/php/export_wfs.php?wfs_url=' + url + 'format=' + format_list[0])`) in separate attempts, but I didn't succeed. Is that what you suggested? — umbe1987, Dec 30 '18 at 23:53
@umbe1987 the encodeURI function "encodes special characters, except: , / ? : @ & = + $ # ", so doesn't help. Try base64_encoding, as I show you in my answer. I've used it many times — Ass3mbler, Dec 30 '18 at 23:54
@skyboyer no, it's a call to a Web Feature Server (WFS). If you try it (of course, wondermap.it not localhost) you'll see that it's responding. — umbe1987, Dec 30 '18 at 23:55

score 2 · Accepted Answer · answered Dec 30 '18 at 23:46

2

Here's how to send it as POST request:

var php_url = '/shire/php/export_wfs.php';
var formData = new FormData();
formData.append('wfs_url', url);
formData.append('format', format_list[0]);
xhr.open('POST', php_url);
xhr.onreadystatechange = function () {
    if (xhr.readyState===4 && xhr.status===200) {
        alert('Server reply: ' + xhr.responseText);
    }
}
xhr.send(formData);

answered Dec 30 '18 at 23:46

That did it. Thank you for providing the code, I didn't know about `new formData()`. – umbe1987 Dec 31 '18 at 00:20
@umbe1987 See here: https://stackoverflow.com/questions/9713058/send-post-data-using-xmlhttprequest – Dec 31 '18 at 00:21

score 1 · Answer 2 · answered Dec 30 '18 at 23:51

1

try including this function (base64_encode):

var php_url = window.location.protocol + "//" + window.location.hostname + '/shire/php/export_wfs.php?wfs_url=' + base64_encode(url) + 'format=' + base64_encode(format_list[0]);

and on the server side:

$wfs_url = base64_decode($_GET['wfs_url']);
$format = base64_decode($_GET['format']);

answered Dec 30 '18 at 23:51

Ass3mbler

3,855
2
20
18

Thanks for your answer. I decided to accept the other though for two reasons: first of all, trying your suggestions i received `base64_encode is not defined javascript`; second, I preferred the other method. I am sure I would have succeeded if I would have overcame the base64 problem, so thanks again for your precious effort. – umbe1987 Dec 31 '18 at 00:19
@umbe1987 thank you for your kind explanation. The error you are getting is probably due to the inclusion of the function I've linked. I understand your choice, and I'm glad you've solved the issue. Good work and regards! – Ass3mbler Dec 31 '18 at 00:48

Passing URL variable with & from JS to PHP result in "&" omission

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