Do this:
function MyFunc($num=NULL)
{
if(isset($_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;
}
$numrows=0;
$res_arr=MyFunc(&$numrows);
Notice the &, that'll make the value of $numrows changed... I did not test
this at all, but it should work.. If you only do MyFunc(); without any
parameters, it'll just return the result of the SQL statement..
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// DvDmanDT
MSN: dvdmandt¤hotmail.com
Mail: dvdmandt¤telia.com
"Dave Carrera" <[EMAIL PROTECTED]> skrev i meddelandet
news:[EMAIL PROTECTED]
Hi List,
I have a function that makes a call to mysql based on certain vars.
---example----
Function MyFunc(){
if(isset($_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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