Echoing through PHP an HTML image path which is dynamic - php

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.

Answer 2

Your $lineofstring variable is in a string using single quotes. Single quotes tell PHP to use the string literally so your variable is not being recognized as a variable. Change it to double quotes or use concatenation to accomplish your goals.
$image = "<img src=\"img/$lineofstring\">";
Or:
$image = '"<img src="img'.$lineofstring.'">';

Answer 3

$profilepicture="images/profiles/".$myid.".gif";
$noprofilepicture="images/profiles/no_avatar.gif";
echo "<IMG SRC='";
if (file_exists($profilepicture)) {
echo "".$profilepicture."";
} else {
echo "".$noprofilepicture."";
}
echo "'>";