Search code examples
echoonmouseoveronmouseout

PHP echo and onmouseover

I have this HTML code and it works perfect:

<a href="www.link.com"><img onmouseover="this.src='http://www.example.com.img1.jpg';" onmouseout="this.src='http://www.example.com.img2.jpg';" src="http://www.example.com.img2.jpg" alt="ALT Text" class="img-responsive"/></br></a>

I want to add this code in PHP and replace the link and image urls with this variables: $link, $image_src1 and $image_src2.

I tried this code:

<a href=".$link."><img onmouseover="this.src='.$image_src1.';" onmouseout="this.src='.$image_src2.';" src=".$image_src2." alt="ALT Text" class="img-responsive"/></br></a>

I think I can not manage how to use quotes, because of this, I am getting an error.

Thank you in advance!

Solution

  • Try this:

    $link = 'http://www.google.com';
$image_src1 = "https://image.flaticon.com/icons/svg/148/148766.svg";
$image_src2 = "https://image.flaticon.com/icons/svg/149/149147.svg";
echo '<a href="'.$link.'"><img onmouseover="this.src='.$image_src1.';" onmouseout="this.src='.$image_src2.';" src="'.$image_src2.'" alt="ALT Text" class="img-responsive"/></br></a>';

    But you really need to look in your code logic, for mouseover/out. You can achieve same logic using CSS :hover property in simple and elegant way.