PHP syntax help

Hi

I don't know why the thumb1.jpg and big1.jpg are not showing up.
Here's the code:

$imgNo =1;
<?php echo '<li><img src="images/thumb"'.$imgNo.'".jpg" alt="White" title="RapperW" border="0" onclick="swap(\'largeid\',\'images/big$imgNo.jpg\');"/></li>'; ?>

Any clues?

 

i think you need to use " instead of ' for your string above, data inside ' doesn't get parsed, so it'd be ignoring the backslashed ' in your string because of it, using " will fixed that.

 

What do you mean ?

 

Dunno about big1.jpg but the first one, you have it outputting something like thumb"1.jpg" which is wrong of course.

$imgNo =1;
<?php echo '<li><img src="images/thumb'.$imgNo.'.jpg" alt="White" title="RapperW" border="0"

Should fix it.
That of course assumes you have $imgNo1 inside the <?php tags.

 

Try this:

<?php

     echo "<li><img src=\"images/thumb/$imgNo.jpg\" alt=\"White\" title=\"RapperW\" border=\"0\" onclick=\"swap('largeid','images/big$imgNo.jpg');\"><li>";

?>

Code (markup):

That should do it. Make sure that second var gets parsed though (in the onclick). You might have to change the "" '' in there I think.

 

Just curious, Whats the difference between single quote and double quote? I meant they both seems to work.

 

double quotes, php will parse double quotes, but won't parse single quotes.

$var = "bob";
print "$var";
[/bob]
you'll get "bob"
[code]
$var = "bob";
print '$var';
[/bob]
will print out $var

Code (markup):

 

<img src="images/thumb"'.$imgNo.'".jpg" ...etc.

You are essentially printing <img src="images/thumb"1".jpg .....

 

<?php
$imgNo =1;
echo "<li><img src=\"images/thumb$imgNo.jpg\" alt=\"White\" title=\"RapperW\" border=\"0\" onclick=\"swap(\'largeid\',\'images/big$imgNo.jpg\');\"/></li>"; ?>



I think this will help you just copy and paste it.......... :

 

Just concatenate it:

echo '<li><img src="images/thumb"' . $imgNo . '".jpg" alt="White" title="RapperW" border="0" onclick="swap(\'largeid\',\'images/big' . $imgNo . '.jpg\');"/></li>';

PHP:

 

Example:

$var = 'Hello';

echo "$var world!";
echo '$var world!';

/* Theoretical fastest way */
echo $var , ' world!';

?>

Run to see the difference. Parses things such as \n, \r, \t and variables etc if it's IN double quotes, otherwise not.

Dan