Unable to print an array in PHP

Question

I am having a bit of trouble with this PHP code. I am not sure why it is printing Array 8 times. From what I see, it should be printing the contents of the array

The goal is to declare an array, using a global, define the array, assign it to another array and display the entire array.

$car_array = array( );
$array = array( );

create_array_cars ();
displayProduct ($array);    

function  create_array_cars (  )  {
    global $car_array;
    $car_array = array( );
    $car_array[] = "ID: 12345" ;
    $car_array[] = "ID: 45678" ;
    $car_array[] = "ID: 67890" ;
    $car_array[] = "ID: 89123" ;
    return $car_array;
    }

function displayProduct ($array) {
    global $array;
    for ($i=0;$i<4;$i++) {
        print "$array<br>"; }
    }

$array = create_array_cars();
print (displayProduct($array));

aseferov · Accepted Answer · 2014-03-11 19:56:37Z

1

function displayProduct ($array) {
foreach ($array as $val) {
    print $val."<br>"; }
}

edited Mar 11, 2014 at 19:56

answered Mar 11, 2014 at 19:27

aseferov

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2 Comments

Amal Over a year ago

@user3367565: Using globals is not a very good coding practice. See my answer for a bit more detailed explanation.

nl-x Over a year ago

what is that global for !? Remove it!

TimWolla · Accepted Answer · 2014-03-11 19:24:43Z

1

When PHP converts an array to a String it simply uses Array. You want to use print_r, implode or something similar.

answered Mar 11, 2014 at 19:24

TimWolla

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Comments

Amal · Accepted Answer · 2014-03-11 19:44:01Z

1

Store the return value of the function in a variable:

$array = create_array_cars();
displayProduct($array);

create_array_cars() returns an array — you need to store it in a variable for later use. displayProduct(); takes an array as its argument and displays the contents inside it. Besides, it doesn't return a value — so we don't need to store the return value.

Stop using globals when you don't need it:

Globals are generally considered to be a bad coding style. You really don't want it here. Your function doesn't even need to deal with external data. You're only creating an array, inside the function, and returning it. So your code can just be:

function create_array_cars() {

    // Create the array
    $car_array = array(
        "ID: 12345",
        "ID: 45678",
        "ID: 67890",
        "ID: 89123"
    );

    // Return it
    return $car_array;
}

Print the array values instead of trying to display an entire array:

Printing any array always prints Array. You need to pick individual values in the array — since you're using a for loop which operates on the basis of numeric indices, you can do that with $array[$i].

Also, I've changed $i < 4 to $i < count($array). This makes your code more dynamic. It'll work even if there are more than 4 elements in the array (not in this case though, as you're manually returning a pre-defined array in the other function).

The modified function will look like:

function displayProduct($array) {
    for ($i = 0; $i < count($array); $i++) {
        print $array[$i] . "<br />";
    }
}

edited Mar 11, 2014 at 19:44

answered Mar 11, 2014 at 19:26

Amal

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2 Comments

Coert Metz Over a year ago

Not that the argument $array for displayProduct is not needed if you use 'global $array'.

user3367565 Over a year ago

I also put the breaks on using globals. The assignment was asking to use globals though, that is why there was globals all over the place :P

Marco De Vecchi · Accepted Answer · 2014-03-11 19:46:50Z

-2

$car_array = array( );
$array = array( );

create_array_cars ();
displayProduct ($array);    

function  create_array_cars ()  {

    $car_array = array( );
    $car_array[] = "ID: 12345" ;
    $car_array[] = "ID: 45678" ;
    $car_array[] = "ID: 67890" ;
    $car_array[] = "ID: 89123" ;
    return $car_array;
    }

function displayProduct ($array) {
    //global $array;

        print_r( $array);
        echo "<bR>"; 

    }

$array = create_array_cars();
displayProduct($array);

not use global. and print "$array
" is not correct... remove this " ". or use print_r($VAR)

answered Mar 11, 2014 at 19:46

Marco De Vecchi

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