RE: [PHP] Is this possible ?

[Dave Carrera]

Hi Again,

Fixed it.

Took my code inside the function that needed to return two vars / values for
use and made it a standard if else statement.

Amended other areas and all works as I wanted it.

I would still appreciate comments about what I was trying to do with the
function approach.

Thank you 

Dave C


-----Original Message-----
From: Toby Irmer [mailto:[EMAIL PROTECTED] 
Sent: 18 January 2004 10:29
To: Dave Carrera; [EMAIL PROTECTED]
Subject: Re: [PHP] Is this possible ?


One way:

<?
Function MyFunc()
{
    if(isset($_POST[var])){
        $sql = mysql_query("select * from table_name where
field=\"$_POST[var]\"");

        $num = mysql_num_rows($sql); // I want to use this result outside
this function.

        $returnsomething ="blah blah";
     }
    return array($returnsomething, $num);
}

list($text, $numrows) = MyFunc();
?>

hth

toby

----- Original Message -----
From: "Dave Carrera" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Sunday, January 18, 2004 11:22 AM
Subject: [PHP] Is this possible ?


Hi List,

I have a function that makes a call to mysql based on certain vars.

---example----

Function MyFunc(){
 if(isset($_POST[var])){
 $sql = mysql_query("select * from table_name where field=\"$_POST[var]\"
");  $returnsomething ="blah blah";  }  return $returnsomething; }

And that all works fine no probs here but.....

I want to use a result somewhere in my script that is not returned by
return. Let me show you...

---example----

Function MyFunc(){
 if(isset($_POST[var])){
 $sql = mysql_query("select * from table_name where field=\"$_POST[var]\"
");

 $num = mysql_num_rows($sql); // I want to use this result outside this
function.

$returnsomething ="blah blah";
 }
 return $returnsomething;
}

So $num contains a number that I want to use outside the function which is
not covered by return.

I know return stops a script and returns what I want it to return but how do
I send out of the function the var I want.

I have tried $GLOBAL[var]=$num; but that don’t work, but I thought I
would'nt anyway just tried it and yes I know I have to declare it inside my
new function using global $var; to use it.

So I ask is this achiveable or how can I do this.

Thank you in advance

Dave C


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Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.560 / Virus Database: 352 - Release Date: 08/01/2004
 

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Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.560 / Virus Database: 352 - Release Date: 08/01/2004
 

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