IMG SRC inside PHP with more PHP inside how to - php

Question

I always have trouble mixing languages and I recently started with MYSQL. In this case I have this code:
<?php
$data = mysql_query("SELECT * FROM Badges")
or die(mysql_error());
while($info = mysql_fetch_array( $data ))
{
Print "http://www.google.com/s2/favicons?domain=".$info['Website'] . "";
}
?>
And I need it to print an image instead of the link it's printing.
http://www.google.com/s2/favicons?domain=".$info['Website'] . " being the image's url
How would that be written? Thanks a lot

Answer 1

print '<img src="http://www.google.com/s2/favicons?domain=' . $info['Website'] . '" alt="" />';
Some other tips...
mysql_* are old functions, PDO is much better to use now it is available.
or die() is old too - have you considered exceptions?
echo is more commonly used than print, and you should use the case that is stated in the manual, e.g. print instead of Print.
You should learn about separation of concerns, e.g. you should do your query and data management on a different layer where you pass the relevant data to your view which would consist solely of HTML and some PHP constructs used to generate it.

Answer 2

i usually find it easier to try and express what differs from case to case in an abstract manner. in your case it's the website (after ?domain=) that differs, all the rest is the same. so the url to the image could abstractly be expressed as http://www.google.com/s2/favicons?domain={website} where {website} is a place holder for future replacement.
replacement would then be performed using the function
$result = str_replace($what_to_replace, $what_to_replace_with, $original_string);
the advantage of this is that you're never mixing languages on one line, and this makes the code easier to develop :) just look at this quite easily read piece of code:
<?php
$img = '<img src="http://www.google.com/s2/favicons?domain={website}" />';
$data = mysql_query("SELECT * FROM Badges")
or die(mysql_error());
while($info = mysql_fetch_array( $data ))
{
$concrete_img = str_replace("{website}", $info['Website'], $img);
print $concrete_img;
}
?>