PHP Function using an Arg Value

Question

I have in my mind what I need to do to make the arg value available to all of the included functions that I am pulling into this script. As I am fairly new with the concept of functions, I am a bit lost as to why this doesn't work as I am thinking it would.

I am calling this script and passing an arguement at commandline. In this case I will use:

php mode_set.php test

I first tried setting a variable for the arguement value:

$mode = $argv[1];

and just referencing the value; however, I quickly learned that it wouldn't be available to other functions.

This led me to realize that in order for it to be available to other functions, I needed to create a function for it:

<?php
$mode = $argv[1];

//Returns the Mode Value
function show_mode()
{
    if($mode == "test")
    {
        return "test";
    }
    elseif($mode == "parse")
    {
        return "parse";
    }
    else
    {
        return "false";
    }

}

if(show_mode() == "test")
{
Do something useful...not working.
}
?>

What I am getting is it seems to be ignoring the arguement, and instead always passes false.

John Conde · Accepted Answer · 2014-03-17 13:16:07Z

3

There's no reason why that variable would not be available to other functions since it is in the global namespace. You probably are making the same error as you are making below.

It's not available to that function unless you pass it as a parameter or use the global keyword:

Method 1: pass as parameter (recommended)

$mode = $argv[1];

//Returns the Mode Value
function show_mode($mode)
{
    if($mode == "test")
    {
        return "test";
    }
    elseif($mode == "parse")
    {
        return "parse";
    }
    else
    {
        return "false";
    }

}

Method 2: use global keyword (not recommended)

$mode = $argv[1];

//Returns the Mode Value
function show_mode()
{
    global $mode;

    if($mode == "test")
    {
        return "test";
    }
    elseif($mode == "parse")
    {
        return "parse";
    }
    else
    {
        return "false";
    }

}

edited Mar 17, 2014 at 13:16

answered Mar 17, 2014 at 13:10

John Conde

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Comments

Amal · Accepted Answer · 2014-03-17 13:16:54Z

2

You can't access $mode because it is outside the scope of the function; that is, PHP can only see the variables within the function when it's inside it. For $mode to be accessible inside the function, you need to pass it as an argument: function($arg).

Your function can be simplified to just:

function show_mode($mode)
{
    if ($mode == "test" || $mode == "parse") {
        return $mode;
    }
    return FALSE;   
}

In simple English: if $mode is either test or parse, return $mode. Otherwise return FALSE.

Calling the function:

$mode = $argv[1];
echo show_mode($mode);

edited Mar 17, 2014 at 13:16

answered Mar 17, 2014 at 13:11

Amal

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Comments

Fabio · Accepted Answer · 2014-03-17 13:10:44Z

1

You need to pass your argument to your function

$mode = $argv[1];

//Returns the Mode Value
function show_mode($mode)
{

answered Mar 17, 2014 at 13:10

Fabio

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4 Comments

Daniel V. Over a year ago

This worked, thanks! However, it still doesn't appear to be available in an included function via another php file. What I mean by this is: If I include an additional file with additional functions, this function isn't available to it.

Fabio Over a year ago

I'm sorry i don't understand what you meant

Daniel V. Over a year ago

Sorry, let me try to explain it more clearly. The file that I reference above is my "main" file which houses all of my includes. In another file that I include, I do all of my main file parsing. So, two files: 1) Includes and Overall Directives (mode function is here) 2) The included file parser When I reference the function show_mode() in #2, it doesn't read a value. if(show_mode($mode) === "test") { Testing Directives } elseif(show_mode($mode) === "parse") { Parsing Directives } else { echo "Parsing Failed"; } This always echos "Parsing Failed"

Fabio Over a year ago

The value of $mode has to be available where you want to use the function

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