显示没有页面刷新的docx

I am curious to know that is it possible to show a docx file (which is available on server) inside a div without refreshing that page (using google doc viewer API or any other possible way).

Let me clear my requirement:-

HTML+JQUERY code:

  <div class="browse-flie col-md-7">
      <div class="alert alert-danger" style = "display:none;"></div>
    <input type="file" id="uploaddocfile">
    <button name="upload" value="Upload" class="btn EditBtn uplaod_file">Upload</button>
  </div>
  <div id = "myid" style = "height:500px;width:600px;"></div>
 <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
    <script type="text/javascript">
        $('.uplaod_file').on('click', function() {
          var file_data = $('#uploaddocfile').prop('files')[0];
          var ext = file_data.name.split('.').pop();
          if(ext == 'docx'){
            var form_data = new FormData();     
            form_data.append('file', file_data);                  
            $.ajax({
                    url: 'upload.php', // point to server-side PHP script 
                    dataType: 'text',  // what to expect back from the PHP script, if anything
                    cache: false,
                    contentType: false,
                    processData: false,
                    data: form_data,                         
                    type: 'post',
                    success: function(php_script_response){
                      var response = $.parseJSON(php_script_response);
                      if(response.status == 'error'){
                         $('.alert-danger').css({'display':'block'});
                         $('.alert-danger').html('file upload failed please try again');
                      }
                      if(response.status == 'success'){
                         $('#myid').html("http://docs.google.com/gview?url=http://loalhost:8888/Grade/uploads/IEP Form-1.docx&embedded=true");
                      }
                    }
             });
          }else{
            $('.alert-danger').css({'display':'block'});
            $('.alert-danger').html('file extension must be docx'); return false;
          }
        });
    </script>

PHP code:

<?php

// A list of permitted file extensions
$allowed = array('docx');

//echo "<pre/>";print_r($_FILES); die;

if(isset($_FILES['file']) && $_FILES['file']['error'] == 0){

    $extension = pathinfo($_FILES['file']['name'], PATHINFO_EXTENSION);

    if(!in_array(strtolower($extension), $allowed)){
        echo '{"status":"error"}';
        exit;
    }

    if(move_uploaded_file($_FILES['file']['tmp_name'], 'uploads/'.$_FILES['file']['name'])){
        echo '{"status":"success"}';
        exit;
    }
}

echo '{"status":"error"}';
exit;

Now:

I have simple file upload code through jquery and php (working perfectly fine).

Now what I want is, after upload file when it comes success as a response then I want to show a doc file using API to my <div id = "myid" style = "height:500px;width:600px;"></div> (without refresh)

So what I tried:

$('#myid').html("http://docs.google.com/gview?url=http://loalhost:8888/Grade/uploads/IEP Form-1.docx&embedded=true");

(link changed for security reason but docx is available on server and I am able to open it directly)

But obviously it's not going to work.

So how can I do that?

Normally you'd do :

if(response.status == 'success'){
     $('#myid').load("http://docs.google.com/gview?url=http://loalhost:8888/Grade/uploads/IEP Form-1.docx&embedded=true");
}

Instead of .html (which does something else).

However it's unlikely to work because the document you are accessing is not one that is managed by GoogleDocs (there may be cross-origin issues), so make sure you're allowing remote requests from Google to be made.

Instead you could try

if(response.status == 'success'){
    $('#myid').html("<iframe src=\"http://docs.google.com/gview?url=http://loalhost:8888/Grade/uploads/IEP Form-1.docx&embedded=true\"></iframe>");
}

In case Google allows embedding of gview in iframes (though they may not).