谷歌将从外部 php 文件中的 jquery .load() 抓取链接

Question

This is my jQuery

$(document).ready(function() {
  $('#name').load('file.php?query=<?php echo urlencode($query); ?>', function() {
    $('#loading').hide();
  });
});

after the initial html loads it then loads the content from file.php into the div with the id=name. this enalbes me to show a loading image while the slow moving content loads. its slow because it uses a few different json apis to get its content. now, file.php has a bunch of different links on it. will google follow those links to other pages. or will google only follow the links on the initial loading of the webpages html?

i ask this because the dynamically loaded content loaded with jquery doesn show up in the webpages source code when i look at it with my browser.

Answer 1

No Google will not see that content. Google does not run any client side javascript, so the content above will never be loaded.

• Use Google Webmaster tools fetch as Googlebot to see what Google sees
• http://support.google.com/webmasters/bin/answer.py?hl=en&answer=158587
• Visit Google page to find out how to make your ajax applications crawlable (see Cymen comment above).
• https://developers.google.com/webmasters/ajax-crawling/

Answer 2

I believe that if you add an empty anchor to your page, it will be found by Google, but of course, won't show up to your viewers.

In my case, I do a PHP scandir() to dynamically create a list of files:

echo "<li rel=$curr_name>".$curr_name."</li>\n";

The viewer sees the list and can click on each item. The jQuery click() function then gets added to each <li> and uses the rel 's value to do its thing (replacing a div with the content of the page).

Now, if we add an empty anchor to the mix:

echo "<li rel=$curr_name>".$curr_name."<a href=\".$curr_path."\"></a></li>\n";

the user still sees the same list, and the jQuery will still do its thing. BUT, there will also be an anchor to each of the files that Google will find and follow.

I couldn't find anything that says a search engine would penalize you for this. And if you're worried, you could always wrap it around an image.

Answer 3

@jhanifen is right

However, your jquery does not look right, it should not have php in it but instead should be javascript and the backend file will take care of the rest, maybe something like this...

$(document).ready(function() {
  query = 'SELECT * FROM `table`';
  $('#name').load('file.php?query=' + escape(query), function() {
    $('#loading').hide();
  });
});