Problems with strings in PHP (not adding two strings together properly)

Question

if we say:

$termstringDomain1 = "something";
$url = "somethingelse";
$url = 'http://' . $termstringDomain1 . '/' . $url;

the result if gives me is: http:///somethingelse instead of http://something/somethingelse

so basically it is ignoring $termstringDomain1 Any idea why?

This runs fine, your problem exists elsewhere in some code that you haven't pasted. — Ben James
– Ben James, Commented Jan 28, 2010 at 20:09
but how would it be somewhere else, all what I am doing is that above then (echo $url) — user220755
– user220755, Commented Jan 28, 2010 at 20:11
is that really your $termstringDomain1? maybe your actual variable value is borked — davidosomething
– davidosomething, Commented Jan 28, 2010 at 20:11
when I echo $termstringDomain1, it prints (something) on the screen so it is working ... — user220755
– user220755, Commented Jan 28, 2010 at 20:12
@unknown: Your problem is elsewhere. I just copied/pasted your code and could not repro the problem. Why type of server are you on? Which version of PHP? — Sampson
– Sampson, Commented Jan 28, 2010 at 20:13

Sampson · Accepted Answer · 2010-01-28 20:11:42Z

2

I am unable to reproduce this issue:

$foo = "foo";
$bar = "bar";
$bar = "sayFoo: " . $foo . ", sayBar: " . $bar;
print $bar; // 'sayFoo: foo, sayBar: bar'

Your problem is likely elsewhere. If I copy/paste what you provided, I get the following:

http://something/somethingelse

Check your variable casing. $domain1 is not the same as $Domain1.

answered Jan 28, 2010 at 20:11

Sampson

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Comments

davethegr8 · Accepted Answer · 2010-01-28 20:13:07Z

1

Can you try putting echo infront of all the declarations, to try to track the problem down?

echo $termstringDomain1 = "something";
echo $url = "somethingelse";
echo $url = 'http://' . $termstringDomain1 . '/' . $url;

answered Jan 28, 2010 at 20:13

davethegr8

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