[SOLVED] function?

[DarkPrince2005]

Ok I'm just trying to return a value depending on an if statement. Is there something wrong with my function or what?

 

function country_amount()
{
$a=run_query("SELECT userid,country FROM users_details where userid like '149'"); 
//return $a['country'];

if ($a['country'] == 'ZA') 
{
$netcashgross1 = "55";
//return $netcasrhgross1; 
global $netcashgross; 
}
else //if($a['country'] != 'ZA')
{
$netcashgross1 = "120";
//return $netcashgross1;
global $netcashgross; 
} 

return $netcashgross1;
}

 

return code

<?php

include "functions.php";
//include "config.php";
country_amount();
//$a=$netcashgross;
echo "$netcashgross1";

?>

 

it returns something but never what the if statement says.

[DarkPrince2005]

it continues to only return 120, doesn't matter what i change the userid to.

[BLaZuRE]

First of all, if you want a value returning function, you need to GIVE the function values using parameters.  These values go inside the () of the function.  For the actual function, use:

function nameFunction($parameter1,$parameter2) {...}

where $parameter1 and $parameter2 are the new names those values will become when you call the function and ... is just whatever code you want inside.  The return statement needs to be INSIDE the function to return stuff from inside the function (a return statement ends your function execution).  Anything outside the function will be executed from top-to-bottom, left-to-right, like always (functions are excluded from this method, they're only executed when called).  You can have any number of parameters (0 to infinite).

To call the function, use:

nameFunction("Here is some text",2005);

each parameter must be separated by a comma.

Also, you may want to do

echo $variable; instead of echo "$variable";

.  It's slightly more efficient and variables don't have to be in double quotes.

Try going to http://www.w3schools.com/PHP/php_functions.asp to better understand how to use functions.

[ignace]

Create all conversions 100-based so you can use them as percentages, use google:

 

100 usd to <currency>

 

Then divide currency through 100 (<currenty>/100) the currency conversion for usd to eur = 0.7

 

function country_amount($netcashgross) {
    global $cconv;/*currency conversions*/
    $r = run_query('SELECT userid,country FROM users_details WHERE userid =149');
    $country = strtoupper($r['country']);
    $conv = !empty($cconv[$country]) ? $cconv[$country] : 1;
    return $netcashgross * $conv;
}

print country_amount($netcashgross);

   

[DarkPrince2005]

How would you write the function? Cause I'm having trouble underrstanding what you mean.

 

function country_amount()
{
$a=run_query("SELECT userid,country FROM users_details where userid like '149'"); 
//return $a['country'];

if ($a['country'] == 'ZA') //or whatever South Africa is defined as in the database 
{
$netcashgross = "55";
//return $netcashgross; //if return value is required. Leave out if not //global $netcashgross; //if return value is required. Leave out if not 
}
else //if($a['country'] != 'ZA')
{
$netcashgross = "120";
//return $netcashgross; //if return value is required. Leave out if not //global $netcashgross; //if return value is required. Leave out if not 
} 

return $netcashgross; //if return value is required. Leave out if not 
}

[DarkPrince2005]

I dont want to convert the currency, the 55 and 120 are fixed vales

[BLaZuRE]

Have you tried using print_r to print your query and see its contents?  Maybe nothing is equal to ZA that is returned from you query?

[DarkPrince2005]

I unfortunately know for a fact that there is, the database only has 2 entries one equal to ZA and another not. For testing purposes.

[ignace]

I dont want to convert the currency, the 55 and 120 are fixed vales

 

You can't use fixed values unless you only have 1 product to sell or are all your products labeled 55 USD?

[TeNDoLLA]

Do a var_dump($a) inside the function and outside the if clause and paste the results here.