That's the thing see - I don't want to send the value -
but I guess I could do that with a case statement so if I want to append the
'address' then i could send the word address to the function and then have a
case statement check if the var is address then if it is, to append the
actual address value onto the it.
Thanks eveyone,
Abe
----- Original Message -----
From: "Cal Evans" <[EMAIL PROTECTED]>
To: "Abe" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Thursday, January 11, 2001 10:07 PM
Subject: RE: [PHP] Function -> Sending URL's
>
> 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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>
>
>
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