Problem with variable in url

Oni

Hey,

I'm trying to make a small piece of code that will set the url of links according to a number in the url. So, add what's in the url to the end of a link.

My code:

If ($ref > '0' ) {
include('http://site.com/top.php?ref=$ref'); 
}

And inside the file the statement is including are a few "$ref" values added after the urls.

The code seems only to work on one page. The only difference between the working page and the nonworking ones is that it uses the $ref variable again in the main page later on...

Sorry if my writings kind of confusing...I'm a bit confused myself! 😉

Thanks!

rincewind456

Confused? me too!

If you mean that the link ends up looking like

http://site.com/top.php?ref=$ref

then it's because your using the wrong type of quotes to enclose the include path, you need to change the code to

If ($ref > '0' ) {
include("http://site.com/top.php?ref=$ref");
}

Blulagoon

Its worth checking the html page source for where you use the $ref variable on the page you use it twice, you should then be able to see how the browser is interpreting the variable.

FWIW I usually write it like this:

include("http://site.com/top.php?ref=".$ref);

Blu

bradgrafelman

Yeah... PHP doesn't parse single-quoted strings for variables; only double-quoted.

It's a good practice to just escape out of the string and concatenate the variable and remaining pieces (if any) of the string together, as Blulagoon has shown. The reason is that people easily forget that it is NOT proper (and probably erraneous) to reference indeces of arrays inside double-quoted strings UNLESS you surround them by braces ( {} ).

n_wattam

i thought when using the include or include_once fucntion you cant pass varibles across in that manner?

Oni

You can only pass it the variables that way if you include the file as a url. (Atleast, as far as I know)

Thanks for the help all! Works great 😉