CODEWITHSUNDEEP
phparraysreturn-type
keenProgrammer
keenProgrammer

Reputation: 438

Returning type array not displaying contents in PHP

Background:

Creating a website that displays currency rates from various countries around the world.

I have a function that retrieves the rate and the title of the currency from an XML document:

function get_rate(SimpleXMLElement $xml, $x) {

    $currency['rate'] = $xml->channel->item[$x]->description;

    preg_match('/([0-9]+\.[0-9]+)/', $currency['rate'], $matches);
    $rate = $matches[0];

    $title['rate'] = $xml->channel->item[$x]->title;
    $title = explode('/', $title['rate']);
    $title = $title[0];

    return array(
            $currency[0],
            $title[1]
        );

//    echo $rate . ' ' . $title . '<br />';
}

If I remove the return array an uncomment the echo statement below the function works as expected, but I'm trying to improve this method by using a return type. This method returns both the rate and its title so presume returning an array is the best solution.

I use the following function to echo out the results from functions like the one listed above.

function get_currency_rate($feed) {

    $xml = new SimpleXmlElement(file_get_contents($feed));

    $rate = get_date($xml);
    echo $rate['date'] . '<br />';

    $vars = parse_url($feed, PHP_URL_QUERY);
    parse_str($vars);
    switch ($x) {
        case 15:
            get_rate($xml, 15); //EUR 15
            echo get_rate($xml, $x);  
            get_rate($xml, 56); //USD 56
            echo get_rate($xml, $x); 
            break;
        case 16:
            get_rate($xml, 16); //GBP 16
            get_rate($xml, 15); //EUR 15
            break;
        case 56: default :
            get_rate($xml, 15); // EUR 15
            get_rate($xml, 56); // USD 56
            break;
    }
}

I'm working on case 15 of the switch statement at the moment. The 2 echo calls in case 15 display Array Array, which clearly suggests that I'm not accessing the array variable correctly.

I've tried replacing case 15 with the following which I thought would be more syntactically correct to no avial. This also displays Array Array like the above.

case 15:
        get_rate($xml, 15); //EUR 15
        echo get_rate($xml, $currency[0]); 
        get_rate($xml, 56); //USD 56
        echo get_rate($xml, $title[1]); 
        break;

Is the problem because I am declaring different variable names in the get_rate function?

Upvotes: 2

Views: 265

Answers (2)

Niet the Dark Absol
Niet the Dark Absol

Reputation: 324620

Use var_dump(get_rate($xml,$x)); and see if that helps you find out what variables you have.

Upvotes: 0

AlishahNovin
AlishahNovin

Reputation: 1932

When you echo a non-string, non-numeric type, you will only print out the object type. This is standard behavior across all languages.

It's essentially asking to create a string out of an object, and for more complex types, it's hard to figure out how you turn them into a string, so it just outputs the type.

To get the behavior you want, you'll instead want to echo out the strings in the array, and so reference them directly, like so: 

$rate = get_rate($xml, $x);
echo $rate[0] . ' '. $rate[1];

Also, as a side-fact, to see everything that's inside an object, you can also use:

 var_dump(  get_rate($xml, $x));

This won't be formatted in the prettiest way, but it will allow you to "look into" your object.

Upvotes: 1

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