Ok, what I would do is either pass in an array of key/value pairs OR pass in
2 variables, the key name and the value.
$keyValuePairs["name"]="Company1";
$keyValuePairs["address"]="123 Main Street";
$keyValuePairs["city"]="Anytown";
$newURL = testPassVar("http://www.calevans.com",$keyValuePairs)
function testPassVar($url, $keyValuePairs) {
/*
* There here in the function you unwind the array and add each to the URL
in the form of
* $thisKey=$keyValuePairs[$thiskey]
*
*/
return $url
} // function testPassVar($url, $keyValuePairs)
A simpler method is to pass in a single key and value and let the function
add them and return you the newly munged URL.
function testPassVar($url, $key, $value) {
//won't work if it's the first pair. need to test for that!
return $url."&".$key."=".$value;
}
$newURL = testPassVar("http://www.calevans.com?test=one","name","Cal");
Cal
http://www.calevans.com
-----Original Message-----
From: Abe [mailto:[EMAIL PROTECTED]]
Sent: Thursday, January 11, 2001 3:51 PM
To: Cal Evans; [EMAIL PROTECTED]
Subject: Re: [PHP] Function -> Sending URL's
Hey Cal -
this makes sense but the thing is that different parts of the site will call
such a function and they will all want different variables attached to the
end -
I could do it some other way - but I wanted to know if there was some way of
doing it by sending the whole URL(incl. the variables) and letting them be
dealt with by the function.
Thanks,
Abe
----- Original Message -----
From: "Cal Evans" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, January 11, 2001 9:40 PM
Subject: RE: [PHP] Function -> Sending URL's
> There's a couple points here:
>
> 1: echo isn't what you want to do. You want to RETURN the value from you
> function. Try this:
> <?
> function testPassVar($url) {
> $name = "TEST";
> echo "<a href=\"$url\">Click</a>";
> return "<a href=\"$url?name=$name\">Click</a>";
> }
>
> $url = 'asdfasdf.php3';
> $newURL = testPassVar($url);
>
> echo($newURL);
>
> echo "This is a $name";
> ?>
>
> 2: in the last line, you try to echo $name. $name is defined within the
> function testPassVar() and goes out of scope when that function is done
At
> this point in your script, $name does not exist.
>
> Hope this helps,
> Cal
>
> http://www.calevans.com
>
> -----Original Message-----
> From: Abe [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, January 11, 2001 3:35 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP] Function -> Sending URL's
>
>
> Hey there,
> this is a strange one - I want to send a URL to a function that includes
> varibles. Those variables should be taken from within the function - as
in
> the example below the link I want is:
>
> asdfasdf.php3?name=TEST , but that is not what I get as you can see.
>
> Does anybody know a way around this - The example is simpler than what I
am
> actually doing and the value of $company must come from the variable.
>
> Thanks,
> Abe
>
>
>
> <?
> function testPassVar($url) {
> $name = "TEST";
> echo "<a href=\"$url\">Click</a>";
> }
>
> $url = 'asdfasdf.php3?name=$company';
> testPassVar($url);
> echo "This is a $name";
> ?>
>
>
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>
>
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