Question

I am trying to echo a dynamic image path in PHP. I have this code that works

 <?php
 //this code works
 $image = '<img src="img/newlogo.jpg">';
 echo($image);
?>
//gives me an image

And this code that doesn’t

 <?php
 //this code doesn't
 $lineofstring='newlogo.jpg';

 $image = '<img src="img/$lineofstring">';
 echo($image);
?>

$lineofsting is actually going to be an image path which is stored in a mysql database where one row is filled with, for example: pictureabc.jpg, second row is picturexyz.jpg etc.

I am trying to pass on the searched imagepath name onto $lineofstring, which is then echo’d but no picture comes out. Where am i going wrong?

Answer 1

To interpolate variables in PHP, you need to use double quote marks ", e.g.

$image = "<img src=\"img/$lineofstring\">";

In this case, you need to escape the inner ".

You can also concatenate strings with .:

$image = '<img src="img/' . $lineofstring . '">';

if it’s easier.