Using a php function inside a php echo - php

Question

I'm trying to setup my WordPress website so that if a post / page has a featured image assigned, this image will be used as the page banner. If however the page doesn't have a featured image, it must onload select a random image out of six available options. I've tried to used this if statement below:
<div id="slider">
<div class="theslide">
<?php
if ( has_post_thumbnail() ) {
the_post_thumbnail();
}
else {
echo '<img src="/wp-content/uploads/link-ship-chandlers-banner-' . $random = rand(1,6); '.jpg">';
}
?>
</div>
</div>
It works, but the random number function is not closing preperly, so the code ends up looking like this:
<div id="slider">
<div class="theslide">
<img src="/wp-content/uploads/link-ship-chandlers-banner-6 </div>
</div>
</div>
Instead of like this:
<div id="slider">
<div class="theslide">
<img src="/wp-content/uploads/link-ship-chandlers-banner-6.jpg">
</div>
</div>
I'm assuming my syntax is wrong for using php inside of the echo, but everything I try either has the same problem or cause a php error.
Any help would be appreciated.
Thanks in advance
Willem

Answer 1

If you don't need to know which header was chosen, just do this:
echo '<img src="/wp-content/uploads/link-ship-chandlers-banner-' . rand(1,6) . '.jpg">';
Using more than one ; on a line is a no-go (you are essentially telling the php-interpreter that '.jpg">'; is an independant command - which will do nothing)
Assigning a variable (ie. $whatever = 'something) inside the echo statement won't be a problem in this case - though it really doesn't do any good either. What it does is create a new variable called $random that you could use afterwards to find out which header was used - but results will be unpredictable if used in the echo statement (ie. in your case the variable would contain [random number].jpg, not just the random number), instead assign the random number to $random first like this:
$random = rand(1,6);
echo '<img src="/wp-content/uploads/link-ship-chandlers-banner-' . $random . '.jpg">';
echo "We are using header {$random}, which was chosen at random.";
Note that the above example also shows an alternate way to include the variable in a string - using double-quotes you can simply write the variable directly inside the string itself. When doing so I recommend always using {} to wrap the variable though not needed in this case - this allows referencing more complex variables (such as array elements or object properties), and it makes the whole thing more readable too.
Other (IMO less readable, possibly more error-prone) solutions include:
$random = rand(1,6);
echo "<img src=\"/wp-content/uploads/link-ship-chandlers-banner-$random.jpg\">";
echo sprintf('<img src="/wp-content/uploads/link-ship-chandlers-banner-%d.jpg">', $random);
printf('<img src="/wp-content/uploads/link-ship-chandlers-banner-%d.jpg">', $random);

Answer 2

Actually this can work, but you end the line with ;.
So removing the ; and adding a .
echo '<img src="/wp-content/uploads/link-ship-chandlers-banner-' . $random = rand(1,6) . '.jpg">';
do this alternatively you don't need to set this variable
echo '<img src="/wp-content/uploads/link-ship-chandlers-banner-' . rand(1,6) . '.jpg">';