dqg2269 2014-04-04 15:01
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AJAX向PHP发送一个变量来处理问题

I have a script that is sending data to a PHP file when the user clicks on an , but it's not working properly..

Here's my jQuery:

jQuery( document ).ready(function( $ ) {
    $('.rve_button a').click(function(){
        console.log("You clicked it");
        var del_id = $(this).attr('id');
        $.ajax(
          type: "POST",
              url: "/wp-content/themes/vantage-child/parts/deleteticker.php",
              data: "id="+del_id,
              success: function(msg){
                 $(this).closest('tr').remove();
          });
         });
        });
});

The HTML & PHP:

if ($vars['watch']) { 
     echo "<td class=\"rve_button\">
             <a href=\"javascript:void(0);\" id=\"".$stocks[$searchresults]['ticker']."\" class=\"delete tooltip\" title=\"Remove from this watch list\">
                <i class=\"icon-remove\"></i>
             </a>
           </td>
"; 
}

The error being returned is this:

Uncaught SyntaxError: Unexpected token :

What is wrong with my ajax call? Why would it be throwing this error? If it matters, this is in Wordpress.

  • 写回答

1条回答 默认 最新

  • duanbi5906 2014-04-04 15:07
    关注

    You need an opening { for the first parameter of $.ajax(). And then you need to add a closing } for the function that is the value for 'success'.

    So...

    $.ajax({
        type: "POST",
        url: "/wp-content/themes/vantage-child/parts/deleteticker.php",
        data: "id="+del_id,
        success: function(msg) {
            $(this).closest('tr').remove();
        }
    });
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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