Whatsapp URL scheme does not allow query parametrs in iOS

Question

I am not able to share any link which has query parameters like second one.

whatsapp://send?abid=2&text=http://www.google.com WORKS FINE

whatsapp://send?abid=2&text=http://www.google.com/?search=123 DOES NOT WORK

Have you tried encoding the text http://www.w3schools.com/tags/ref_urlencode.asp — Rohit Rehan, Apr 24 '15 at 05:46
yup.. I am sending the whole string to NSURL after encoding. — nr5, Apr 24 '15 at 06:01

Esha · Answer 1 · 2015-04-24T06:33:06.340

0

You cannot use the ? for sending any parameter after the first one.

To pass the value of the search as 123 try using &search=123 rather than ?search=123.

If the http://www.google.com/?search=123 is a whole string then you have to encode it as that contains "?" , "/" and ":".

Your text value would be like this:

whatsapp://send?abid=2&text=http%3A%2F%2Fwww.google.com%2F%3Fsearch%3D123

which is the encoded string of the above. Maybe this could solve your problem.

edited Apr 24 '15 at 06:33

answered Apr 24 '15 at 06:05

Esha

1

Was this successfully? Because when I try adding query parameters, my string automatically cuts off at '&' of '?'.. – JarnoV May 03 '16 at 21:05

1 Answers1