My PHP Function isn't working

I'm 开发者_Python百科having trouble with the following code. What it should do is echo cats.php followed by example.php but it's not echoing the example.php. Any ideas why this might be happening?

$bookLocations = array(
    'example.php',
    'cats.php',
    'dogs.php',
    'fires.php',
    'monkeys.php',
    'birds.php',
);

echo $bookLocations[1];

function findfile($filenumber)
{
echo $bookLocations["$filenumber"];
}

findfile(0);

Try changing,

echo $bookLocations["$filenumber"];

to:

echo $bookLocations[$filenumber];

Edit* To expand on Thomas's correct answer, instead of using global variables, you could change your method to:

function findfile($filenumber, $bookLocations)
{
    echo $bookLocations[$filenumber];
}

i believe you may also need to declare the global variable in your function.

global $bookLocations;

Ok, there are two issues.

Variable Scope

Your function doesn't know the array $bookLocations, you need to pass it to your function like so:

function findfile($filenumber, $bookLocations)

Array key

You don't want to wrap your array key in quotes:

wrong: $bookLocations["$filenumber"];
right: $bookLocations[$filenumber];

The quotes in "$filenumber" turn your key into a string, when the keys to your array are all numbers. You are trying to access $bookLocations["1"] when in fact you want to access $bookLocations[1] -- that is to say, 1 is not the same as "1". Therefore, like others have said, you need to get rid of the quotation marks around the key (and check your variable scope too).

function findfile($filenumber)
{
  global $bookLocations;
  echo $bookLocations[$filenumber];
}

Good-style developers usually avoid global variables. Instead, pass the array to the function as the parameter:

function findfile($files, $filenum)
{
  echo $files[$filenum];
}

$bookLocations is out of scope for your function. If you echo $filenumber you will see that it's in scope because you passed it in by value. However, there is no reference to $bookoLocations.

You should pass in $bookLocations

declaration: function findfile($filenumber, $bookLocations){ call: findfile(1, $bookLocations);

You could also to declare $bookLocations as global, but globals should be avoided if possible.