How to call external url in jquery

Question

I am trying to put comments on Facebook wall using jquery   But my ajax call not alowing external url    can anyone explain how can we use external url with jquery    below is my code    var fbUrl  https   graph facebook com 16453004404 481759124404 comments access token my token      ajax               url  fbURL       data   message   commentdata      type   POST       success  function  resp            alert resp              error  function e           alert  Error    e                 its giving xmlhtttprequest error

User · Answer

it is Cross-site scripting problem  Common modern browsers doesn t  allow to send request to another url

User · Answer

I think the only way is by using internel PHP code like MANOJ and Fernando suggest    curl post get in php file on your server --  call this php file with ajax  The PHP file let say  fb php     commentdata   GET  commentdata     fbUrl  https   graph facebook com 16453004404 481759124404 comments access token my token   curl setopt  ch  CURLOPT URL  fbUrl   curl setopt  ch  CURLOPT POST  1      POST data here curl setopt  ch  CURLOPT POSTFIELDS           message    commentdata       receive server response     curl setopt  ch  CURLOPT RETURNTRANSFER  true     server output   curl exec   ch   echo  server output  curl close   ch     Than use AJAX GET to   fb php commentmeta your comment goes here   from your server    Or do this with simple HTML and JavaScript from externel server   Message   lt input type  text  id  message  gt   lt input type  submit  onclick  PostMessage    gt   lt script gt  function PostMessage     var comment   document getElementById  message   value      window location assign  http   yourdomain tld fb php commentmeta   comment     lt  script gt

User · Answer

Hi url should be calling a function which in return will give response    ajax   url  function to call url                  try using calling API facebook method

User · Answer

Follow the below simple steps you will able to get the result   Step 1- Create one internal function getDetailFromExternal in your back end  step 2- In that function call the external url by using cUrl like below function   function getDetailFromExternal  p1  p2              url    http   request url with parameters            ch   curl init            curl setopt array  ch  array              CURLOPT URL   gt   url              CURLOPT RETURNTRANSFER   gt  true                                  curl setopt  ch  CURLOPT SSL VERIFYHOST  0           curl setopt  ch  CURLOPT SSL VERIFYPEER  0            output   curl exec  ch           curl close  ch           echo  output          exit          Step 3- Call that internal function from your front end by using javascript jquery Ajax

User · Answer

google the javascript same origin policy  in a nutshell  the url you are trying to use must have the same root and protocol   so  http   yoursite com cannot access https   yoursite com or  http   anothersite com  is you absolutely MUST bypass this protection  which is at the browser level  as galimy pointed out   consider the ProxyPass module for your favorite web server

User · Answer

All of these answers are wrong   Like I said in my comment  the reason you re getting that error because the URL fails the  Same origin policy   but you can still us the AJAX function to hit another domain  see Nick Cravers answer on this similar question      You need to trigger JSONP behavior   with   getJSON   by adding  amp callback     on the querystring  like this     getJSON  http   en wikipedia org w api php action query amp prop revisions amp rvprop content amp titles   title   amp format json amp callback     function data        doSomethingWith data               You can test it here       Without using JSONP you re hitting the   same-origin policy which is blocking   the XmlHttpRequest from getting any   data back    With this in mind  the follow code should work   var fbURL  https   graph facebook com 16453004404 481759124404 comments access token my token      ajax       url  fbURL   amp callback         data   message   commentdata      type   POST       success  function  resp            alert resp              error  function e            alert  Error    e

User · Answer

JQuery and PHP  In PHP file  contenido php     lt  php  mURL     GET  url     echo file get contents  mURL     gt    In html    lt script type  text javascript  src  js jquery jquery min js  gt  lt  script gt   lt script type  text javascript  gt      function getContent pUrl  pDivDestino           var mDivDestino         pDivDestino              ajax               type    GET               url    contenido php               dataType    html               data                    url   pUrl                            success   function data                                                                  mDivDestino html data                                       lt  script gt    lt a href     onclick  javascript getContent  http   www google com     contenido    gt Get Google lt  a gt   lt div id  contenido  gt  lt  div gt

[jquery] How to call external url in jquery?

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