PHP passing variables into a function

Question

This is really basic PHP. Can someone tell me why this does not work and what I need to do to make it work.

<?php
$test_var=12;
proc_scrn($test_var);

proc_scrn($local_pid)
{
echo "tp12",$local_pid ;
}
?>

+1, try adding function before proc_scrn. And I am not sure if the echo-part will work. Try changing it to echo "tp12".$local_pid; — OptimusCrime
– OptimusCrime, Commented Nov 2, 2011 at 11:59
@Nanocom Functions do not need to be declared before being called — Phil
– Phil, Commented Nov 2, 2011 at 12:02
@OptimusCrime That echo statement is fine and is actually faster than concatenation — Phil
– Phil, Commented Nov 2, 2011 at 12:02

Nick · Accepted Answer · 2011-11-02 11:59:22Z

5

Well, you haven't actually created a function there. This would work:

<?php
$test_var=12;
proc_scrn($test_var);

function proc_scrn($local_pid='')
{
echo "tp12: ".$local_pid;
}
?>

answered Nov 2, 2011 at 11:59

Nick

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Comments

Vamsi Krishna B · Accepted Answer · 2011-11-02 12:00:33Z

2

function proc_scrn($local_pid)
{
// something
}

answered Nov 2, 2011 at 12:00

Vamsi Krishna B

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a perfect answer with no offtopic chatter.

Wasim Karani · Accepted Answer · 2012-07-30 13:51:19Z

0

Pretty simple

<?php 
$var=1;
function proc_scrn($var1){
    echo "tp12: ".$var1;
}
proc_scrn($var);
?>

Wasim Karani

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answered Nov 2, 2011 at 12:17

user962293