Help with adding text after a variable please

[Chappers]

Hi,

Just a small thing - I've got this code:
[code]<?php
$url = "http://example.com/";
echo "$url images/example.jpg";
?>[/code]
Now that'll echo: http://example.com/ images/example.jpg

But, of course, I don't want a space between the [b]example.com/[/b] and the [b]images/example.jpg[/b]. If I just put the echo as: $urlimages/example.jpg, it doesn't work of course. What's the proper way of doing this?

Many thanks,
James

[shocker-z]

<?php
$url = "http://example.com/";
echo $url.'images/example.jpg';
?>


That will do the job using . to seperate the variable and string.


Regards
Liam

[Chappers]

Thanks a lot for that quick reply, much appreciated.

All the best,
James

[shocker-z]

no problem :)

im out to get tipsey now ;) Mrs is out working :-D

[Chappers]

D'oh! Still can't get it working, because my code is like this:
[code]<?php
$url = "http://example.com/";
?>

<table align="center" width="100%" cellspacing="0" cellpadding="0" border="0">
<tr>
<td align="center">

<?php
echo "<link rel='stylesheet' href='$url css/styles.css' type='text/css'>";
?>[/code]

Where I don't want the current space between [b]$url[/b] and [b]css/styles.css[/b]

Any ideas please?


Enjoy your night out ;-)

[wildteen88]

You keep putting a space after your variable name in the echo statement. So you are always going to get the space. What you'll want to do is to come out of the echo by adding a double quote before your variable ($url) and then a period (.) now add another period at the end of your variable name and another double quote. So your echo statement should be like this:
[code=php:0]echo "<link rel='stylesheet' href='" . $url . "css/styles.css' type='text/css'>";[/code]


Or you can just ad curly braces around your variable:
[code=php:0]echo "<link rel='stylesheet' href='{$url}css/styles.css' type='text/css'>";[/code]

[Chappers]

Both methods work perfectly, many thanks. The space was added to show where I wanted the two elements to join, but to make things clear I had to leave a space otherwise the variable would be read as $urlcss...

Best regards,
James