How to construct variable external link

[zambala]

Hello,

I'm trying to create a variable external link or redirect, but I'm complete beginner with php...

1.In one instance I have a script for registration (address, desciption, features, etc) with Post--> I am trying to add another field for URL --> I have added this extra line and extra line in MySql database.... "property_link"

2. In another instance I have a script with Constructed link which currently go to "Booking Form" (depending on the item(property) --> but I need to replace with external link from registration Form for each item....
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- But I don't understand How to replace/construct it?!
Now it looks like
//$url=SITEPAGE_URL."&task=dobooking&selectedProperty=$property_uid}";// --> and everyone tells me I should join the new link something like $url=$_POST['url']; - but when I try, it doesn't work :( How it should be joined/redirected

[cffrost2]

Not sure what you're asking. Are you having them type a complete link and need to display it, replacing the current $url var?

[zambala]

Yes. Trying to replace current.

- current link use "selectedProperty=$property_uid" - and I need to keep it adding + url from registration form ($_POST);
- and for next Property would be again "selectedProperty=$property_uid" + $_POST
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[zambala]

in other words:

User is searching for properties;
Each unique property have a unique link depending on property number ($property_uid)

[cffrost2]

Ok. So this is the link you need to show, but need the url var added added to the end. Right?

Now it looks like 
//$url=SITEPAGE_URL."&task=dobooking&selectedProperty=$property_uid}";// --> and everyone tells me I should join the new link something like $url=$_POST['url']; - but when I try, it doesn't work :( How it should be joined/redirected 

Try this

$url=SITEPAGE_URL."&task=dobooking&selectedProperty=".$property_uid."&url=".$_POST['url];}"

Is that what you were after?