用PHP悬停显示图像

Question

I have a small problem with my PHP code and It would be very nice if someone could help me. I want to display an image when hovering over a link. This is the link with the PHP code that I have now:

<a href="<?php the_permalink(); ?>"><?php if ( has_post_thumbnail() ) {the_post_thumbnail();} else if ( has_post_video() ) {the_post_video_image();}?></a>

This code shows a image, but I want to execute this code when hovering over the link with the image:

<?php echo print_image_function(); ?>

The code also shows a image that belongs to a category. I don't want the initial image to disappear I simply want to show the second image on top off the first image when hovering over the first image.

I don't know if it is helpful but I use Wordpress and I am not a PHP expert. I even don't know if this is going to work. Thats why I am asking if somebody can help me with this.

Thanks in advance

THANKS EVERYONE

I want to thank everybody that took the time to read my post and helped me by giving their solution. I didnt exspect so many answers in such a fast time. After spending a few hours trying it to get it to work with PHP, CSS and Javacript, I stumbled upon the following question on this website: Solution

It was exactly where I was looking for and with a few modifications to fit my needs, I got it to work. Sometimes things can be so easy while you are looking for the hard way. So for everyone that has the same problem: You can use one of the solutions that where given by the awesome people here or take a look at the link above.

Thanks again! :)

Answer 1

You can do this with CSS (if you so please and this fits with your overall architecture) - here is an example using the :hover condition and :after pseudo element.
html

<a href="#" class="foo"><img src="http://www.gravatar.com/avatar/e5b801f3e9b405c4feb5a4461aff73c2?s=32&d=identicon&r=PG" /></a>​

css

.foo {
    position: relative;
}
.foo:hover:after {
    content: ' ';
    background-image: url(http://www.gravatar.com/avatar/ca536e1d909e8d58cba0fdb55be0c6c5?s=32&d=identicon&r=PG);
    position: absolute;
    top: 10px;
    left: 10px;
    height: 32px;
    width: 32px;   
}​

http://jsfiddle.net/rlemon/3kWhf/ demo here

Edit:

Always when using new or experimental CSS features reference a compatibility chart http://caniuse.com/ to ensure you are still in your supported browsers. For example, :after is only supported starting IE8.

Answer 2

You cannot randomly execute server side code on the client side.

Try using javascript AJAX requests instead.

Answer 3

PHP is a server-side language; you can't get PHP to execute after the page has loaded (because PHP completely finishes parsing before the page loads). If you want hover events, you need JS.

Answer 4

Firstly you don't need the elseif statement. An else will serve the same purpose, unless you intend to have blank tags where neither a thumbnail or a video image are present.

<a href="<?php the_permalink(); ?>">
    <?php 
        if ( has_post_thumbnail() ) 
        {
            the_post_thumbnail();
        } 
        else
        {
            the_post_video_image();
        }
    ?>
</a>

You can't explicitly use PHP for client side functionality. You will need to use javascript or jquery to supply the on hover capability.

Jquery example:

$(function() {

    $('a').hover(function() {

        // Code to execute whenever the <a> is hovered over

        // Example: Whenever THIS <a> tag is hovered over, display
        //          the hidden image that has a class of .rollover

        $(this + ' .rollover').fadeIn(300);

    }, function() {

        // Execute this code when the mouse exits the hover area
        // Example (inline with above example)

        $(this + ' .rollover').fadeOut(300);

    });

});

To have an image placed on top of another image you would need to make sure your CSS uses absolute positioning for the images with the image that is to overlay the other on hover is given a z-index value higher than the image to sit underneath it.

Hope this helps.

Answer 5

您需要一些JavaScript和/或CSS来实现这一点，因为PHP位于服务器端，而不是客户端浏览器。