php - How can I post data through a form with ajax and get it working? -


this question has answer here:

i have piece of ajax script trying post form me. without ajax form post , send data. want ajax post not refresh page , posts data too. there multiple forms on 1 page.

my js script looks this:

        function post_form(action)         {             var token = $('.forms').attr('id');             var itemid = $('.forms').find('input.id').val();             var instaurl = 'https://api.instagram.com/v1/media/'+itemid+'/likes?access_token='+token+'';             console.log(token);             console.log(itemid);             console.log(instaurl);             var datastring = token;             $.ajax({                 type: "post",                 url: instaurl,                 data: datastring,                 crossdomain: true,                 datatype: 'jsonp',                 beforesend: function()                 {                    $("#loading").fadein("slow");                     if ( action == "like" )                     {                         $("#open"+comment_id).hide();                         $("#loading_like_or_unlike"+comment_id).html('<img src="loader.gif" align="absmiddle" alt="loading...">');                     }                     else if ( action == "unlike" )                     {                         $("#close"+comment_id).hide();                         $("#loading_like_or_unlike"+comment_id).html('<img src="loader.gif" align="absmiddle" alt="loading...">');                     }                     else {}                  },                 success: function(response)                 {                     if ( action == "like" )                     {                         $("#close"+comment_id).show();                     }                     else if ( action == "unlike" )                     {                         $("#open"+comment_id).show();                     }                     else {}                     $("#loading").fadeout("slow");                 }             });             event.preventdefault();         }  $(document).ready(function() {     $('button.like').each(function() {         $(this).on('click', function(){             post_form();         });     }); });

now in markup have form has id in hidden input value. form once posted looks id , uses case switcher ans unlike switch. uses instagram php library connect , data images able see:

try {      $instagram = new instagram\instagram;      $instagram->setaccesstoken($_session['instagram_access_token']);     $token = $_session['instagram_access_token'];     //$clientid = $_session['client_id'];      $current_user = $instagram->getcurrentuser();     $tag = $instagram->gettag('folkclothing');     $media = $tag->getmedia(isset($_get['max_tag_id']) ? array( 'max_tag_id' => $_get['max_tag_id'] ) : null);       $liked_media = $current_user->getlikedmedia();     /* echo 'https://api.instagram.com/v1/media/'. $item->getid() .'/likes?access_token='.$token.''; */      if ( isset( $_post['action'] ) ) {      echo '<br/>form submitted, inspect sent';                 print_r($_post);          $id = $_post['id'];                  switch( strtolower( $_post['action'] ) ) {             case 'like':                 $current_user->addlike( $id );                 break;             case 'unlike':                 $current_user->deletelike( $id );                 break;                 }        }  } catch ( exception $e ) {     // yes there error     $error = $e->getmessage();  }  // view rendering stuff   // display error if ( $error  != '' )  {     echo "<h2>error: ".$error."</h2>"; }    echo '<section id="images">';  foreach ( $media $item ) {      echo '<article class="instagram-image">';     // define form , set action post send data script     echo '<form id="'. $token .'" class="forms" action="'; echo url::current(); echo '" method="post">';          $id = $item->getid();          echo '<a title="' . $item->getcaption() .'" class="fancybox" href="' . $item->link . '"><img alt="' . $item->getcaption() .'" src="' . $item->images->standard_resolution->url . '" /></a>';         echo '<div class="formsubmit-feedback"></div>';         //echo '<img src="/public/img/377.gif" alt="loader"/>';         if ( $current_user->likes($item) ){             echo '<button class="ajax instabtn unlike icon-heart" type="submit" name="action" value="unlike"></button>';         } else {             echo '<button class="ajax instabtn icon-heart" type="submit" name="action" value="like"></button>';         }         echo '<input class="id" type="hidden" name="id" value="'; echo $id; echo '">';          echo '<p>'; echo $item->likes->count; echo '</p>';         //echo '<p>'.$item->getid().'</p>';         //echo '<p>by: <em>' . $item->user->username . '</em> </p>';         //echo '<p>date: ' . date('d m y h:i:s', $item->created_time) . '</p>';         //echo '<p>$item->comments->count . ' comment(s). ' . $item->likes->count . ' likes. ';      echo '</form>';     echo '</article>'; } echo '</section>';

the form works know sure need know how can post right place , switch likes/unlikes image.

does know way around @ all?

thanks

so database changes happen page doesn't reflect change properly?

i'm not sure if ajax success has action in scope...try echoing action in php script , using ajax response var control images.


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