Jquery dialog tutorial - PHP Tutorials: Create a ‘Like’ Button – jQuery AJAX (Part 5/7)

10 great comment(s) for this post

1. 

   evoardent 2:38 am 8/02 of 2012

   guys make sure you have the very first line correct :

   function like_add(article_id) {

   I had the same issue and it happened to be that i skipped adding the article_id parameter to the function.

   Hope this helps

   And Sir alex – you are amazing. I have no words – your tuts are awesome. Keep up the good work – God bless!

   Reply
2. 

   Kid24you 3:03 am 8/02 of 2012

   @3RandomFails did you find the problem?

   Reply
3. 

   3RandomFails 3:27 am 8/02 of 2012

   like_add function doesn’t work, the onclick() can reach it but the function does nothing
   i have exactly copied you code down to every pixel.
   Help me

   Reply
4. 

   doldol1 3:27 am 8/02 of 2012

   @waitingfordoom Look @ post date =]

   Reply
5. 

   GTrismegisto 4:20 am 8/02 of 2012

   If I add all those functions in a class, how should I write the “$.post()” for the ajax Request?

   Reply
6. 

   killerbng 4:26 am 8/02 of 2012

   @waitingfordoom Ill have to check it out, im on linux and gedit just doesnt cut it for me after having Dreamweaver awesomeness for years :P

   Reply
7. 

   TheRealGeriBoss 5:09 am 8/02 of 2012

   No likes? Man… you deserve 10^30 views & likes O_o
   Keep up your good work, you are needed on this world.

   Reply
8. 

   waitingfordoom 5:10 am 8/02 of 2012

   I would like to recommend for the php guys out there netbeans ide. best free editor.

   Reply
9. 

   chancez6 5:16 am 8/02 of 2012

   I looked at this code, and laughed. This is so awesome. I swear I learn about 10 things every time I watch a video of yours :)

   Reply
10. 

    chancez6 5:32 am 8/02 of 2012

    Thank you so much for these videos!

    Reply

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