Question

I have an ul with a li that cannot find a .PNG image associated with the list item ‘li’ in the code here:

$pngFilename= 'C:/xampp/htdocs/myProj/' . 'just_a.png';
echo 'pngFilename is "', $pngFilename, '" -- that was the .png image filename.';


// within a 'ul' is this li item that displays an image (the ul code is simplified 
// to only show the item that (needs to but doesn't!) display a .PNG image.)
echo '<ul>';
echo '<li>';
echo '<a href="http://localhost/myProj/just_an.htm">';
echo '<img src="', $pngFilename, '"';            
echo 'alt="http://localhost/myProj/the_other.png"';

// NOTE -- I left out the close /> to the img statement when I copied my code here but 
// it was in fact in my source code.
echo '/>'; 

echo '</a>';
echo '</li>';
echo '</ul>';

Before I wrote the above php code, I tested just the raw html and it was fine — my .PNG file called
C:/xampp/htdocs/myProj/just_a.png was displayed correctly.

But when I switched to php server-side generation of the html, the C:/xampp/htdocs/myProj/just_a.png image does not appear, only the ‘cant find it’ default small image appears that looks like a piece of paper torn horizontally.

Any ideas? The .png file exists and so does the directory and the html correctly displayed the image, but when I put the html into php ‘echo’ calls I must be screwing something up, just not sure what.

To make sure I have the correct path and filename you’ll see I echo it out at the top of the code.
The php variable $pngFilename is displayed when I do ‘View Source’ in the browser as:

pngFilename is "C:/xampp/htdocs/myProj/just_a.png"  -- that was the .png image filename.

The only other thing to mention is that the ‘alt’ link, the ‘alt=”http://localhost/myProj/the_other.png”‘ — this link (not the image) shows as blue underlined link text.

Why did this work in my html but breaks when I use the ‘echo’ in php? After all, the ‘echo’ simply sends the html to the client side — and that .png file is 100% definitely there and displays fine when I run the above html outside of php’s “echo” command.

Answer 1

This is because the path to the image changes depending on how your viewing the page. in a local context or from a server context; and since you are using an absolute path instead of a Relative path the system can’t adjust for the change in the location of the image. Unlike when using a PHP function that calls on the php file system functions that do use the internal file system. what your doing is having it send a text file to the browser which isn’t rendered as HTML code until after PHP has finished. because of that it has no access to the php file system to resolve the path to the image on the server. the way to fix it would be to use the path to the image Relative to the PHP script or use the web accessible path to the image

$pngFilename= 'C:/xampp/htdocs/myProj/' . 'just_a.png';

Should for example be

$pngFilename= 'http://localhost/myProj/' . 'just_a.png';

or if the image and the php file are in the same directory you could just do

$pngFilename= 'just_a.png';