javascript function replace wont work , why?

Question

why javascript function replace wont work ?

 social_share: function(e){
 var is_video,
 social_name = $(e).attr('class'),
 share_url = document.URL,
 share_title = $('title').text(),
 share_media,
 share;
 switch(social_name) {
 case 'twitter':
 share = 'https://twitter.com/share?url={share_url}&text={share_title}';
 break;
 case 'facebook':
 share = 'https://www.facebook.com/dialog/feed?app_id=123050457758183&link={share_url}&picture={share_media}&name={share_title}';
 break;
 case 'google':
 share = 'https://plus.google.com/share?url={share_url}';
 break;
 case 'pinterest':
 share = 'https://pinterest.com/pin/create/bookmarklet/?media={share_media}&url={share_url}&is_video={is_video}&description={share_title}';
 break;
 case 'mailto':
 share = '...';
 break;
 }
 share.replace('{share_title}', share_title)
 .replace('{share_url}', encodeURI(share_url))
 .replace('{share_media}', encodeURI(share_media))
 .replace('{is_video}', is_video);
 console.log(share);
 },

and console.log function return string share without any replaces... it would by https://twitter.com/share?url={share_url}&text={share_title} if twitter , and same other's

• feesar: and how about function encodeURI , it's good for urls for social shares ?
• musefan: out of interest, how did you know about the replace function in the first place? Just wondering what source you have seen that doesn't show you the correct way to use it
• feesar: its was my mistake , i didn't see it , need some rest...

Answer 1

.replace will return the new value back, so you need to save it to the variable:

share = share.replace( ... )

Answer 2

Strings are immutable in javascript. So changing the string by any of the string function will return a new string as the output. In the above code change the syntax as :

share = share.replace('{share_title}', share_title);