的jQuery的onclick不發送數據

Question

爲什麼我會得到下面的下面的錯誤：

Uncaught TypeError: Object [object global] has no method 'data'

HTML：

$attributes = array('class' => 'main post', 'onsubmit' => 'validateModules(event)', 'dataid' => $moviesparx_website_id); echo form_open("admin/websites/{$website_id}/page/op/{$action}/{$id}", $attributes); ?> 

    <ul style="width: 25%; border-right: 1px solid #ccc; padding-right:10px;" class="postForm"> 
     <li> 
      <input style="width:100%;" placeholder="Page Title" id="post_title" name="post_title" onclick="urlCheck()" dataid="<?php echo website_id;?> type="text" value="<?php echo set_value('post_title', $post['post_title']); ?>" /> 
      <?php echo form_error('post_title'); ?> 
     </li> 

JS：

function urlCheck() 
{ 
     var id = self.data("id"); 

     $(".postForm").on('click', '#post_title', function(e){ 
     e.preventDefault(); 
     $.ajax({ 
      url  : '<?=base_url()?>/page/op', 
      data  : { post_title : $("#post_title").val(), 'website_id' : id }, 
      type  : 'POST', 
      success : function(resp){ 
       alert(resp); 
      }, 
      error : function(resp){ 
       console.log("Error in ajax request"); 
      } 
     }); 
    }); 
}; 

Answer 1

function urlCheck() { 
    var id = self.data("id"); 

    $(".postForm").on('click', '#post_title', function (e) { 
     e.preventDefault(); 
     $.ajax({ 
      url: '<?=base_url()?>/page/op', 
      data: { 
       post_title: $("#post_title").val(), 
        'website_id': id 
      }, 
      type: 'POST', 
      success: function (resp) { 
       alert(resp); 
      }, 
      error: function (resp) { 
       console.log("Error in ajax request"); 
      } 
     }); 
    }); 
} 

urlCheck(); <-- did you add this line ?