PHP - including attribute in shortcode

Question

This may seem like a Wordpress problem but I think it's PHP specific rather than Wordpress. My problem is in the // Code block I think. Here is my code:

function display_app_rating( $atts ) {

// Attributes
extract( shortcode_atts(
    array(
        'app' => '',
    ), $atts )
);

// Code
//return '"http://itunes.apple.com/lookup?id='.($app).'"';
$json_url = '"http://itunes.apple.com/lookup?id='.($app).'"';
$json = file_get_contents($json_url);
$result = json_decode($json, TRUE);
foreach ($result['results'] as $key => $value) {
    return '<p class="appstore rating">Average rating '.$value['averageUserRating'].' out of 5 from '.$value['userRatingCount']. ' users.</p><p class="appstore price">Current Price '.$value['currency'].$value['price'].'</p>';
}
}
add_shortcode( 'apprating', 'display_app_rating' );

If I hardcode an app ID as the $app variable it works fine and if I comment my code out and uncomment the return... line it returns the correct URL. My question is how can I get the $app variable appended to the URL and working via my shortcode which is

[apprating app="439438619"]

Any help appreciated.

$json_url = "itunes.apple.com/lookup?id=$app" should work. You have single quotes around the string which doesn't inline the variable. I have no experience with wordpress, can you verify that $app is set correctly from the extract? — tbddeveloper
– tbddeveloper, Commented Sep 16, 2014 at 20:32
Hi Hammerstein, it doesn't work unfortunately having tried it. When I simply return that line as it is displayed it inlines the URL correctly with the variable inside the quotes as part of the URL. — Paul Brown
– Paul Brown, Commented Sep 16, 2014 at 20:51

adam_bear · Accepted Answer · 2014-09-16 20:26:59Z

3

I don't think you want to return after each iteration- push the results into a variable and just return once.

$result = json_decode($json, TRUE);
$block = '';
foreach ($result['results'] as $key => $value) {
    $block .='<p class="appstore rating">Average rating '.$value['averageUserRating'].' out of 5 from '.$value['userRatingCount']. ' users.</p><p class="appstore price">Current Price '.$value['currency'].$value['price'].'</p>';
}

return $block;

answered Sep 16, 2014 at 20:26

adam_bear

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4 Comments

Paul Brown Over a year ago

Thank you Adam, I may be misunderstanding but I appreciate your input and no doubt your code will improve mine in that block. The problem I have though is getting my $app variable as defined by the shortcode into the URL which happens before this?

adam_bear Over a year ago

It sounds like you should split this function up... try separating this into get_app_rating(), set_app_url(), and display_app_rating() or similar.

Paul Brown Over a year ago

I think between you all you've got this working for me thank you very much. I added your block return idea but didn't need to create individual functions.

adam_bear Over a year ago

Right on =) The reason I suggest individual functions is it makes testing easier- something to consider moving forward.

Community · Accepted Answer · 2017-05-23 12:25:37Z

1

Using extract() is not recommended. To check for query values in the URL, use $_GET.

The following first checks if the URL contains ?app_id=something, if so the result will output it. If the URL doesn't contain it, the shortcode attribute will be used [apprating app="something"]. Otherwise, it prints an error message:

add_shortcode( 'apprating', 'shortcode_so_25877611' );

function shortcode_so_25877611( $atts, $content )
{
    $args = shortcode_atts( 
        array(
            'app'   => '',
        ), 
        $atts
    );

    $result = 'No app id.';
    $app = (int) $args['app'];

    if( !empty( $_GET['app_id'] ) )
        $result = (int) $_GET['app_id'];
    elseif( $app )
        $result = $app;

    $json_url = "http://itunes.apple.com/lookup?id=$result";

    return $return;
}

The test shortcode on a post was as follows, and then appending the query var to the URL:

With id: [apprating app="1234"]

Without: [apprating]

_{PS: as noted by Adam, your return must be outside the foreach loop.}

edited May 23, 2017 at 12:25

CommunityBot

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answered Sep 16, 2014 at 20:37

brasofilo

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3 Comments

Paul Brown Over a year ago

Thank you very much for this. Can you assist further if possible. Where does the rest of my code fit with this? e.g. the $json_url = '"http://itunes.apple.com/lookup?id='.($app).'"'; $json = file_get_contents($json_url); etc I can see the returning of the $app variable but it doesn't appear attached to the URL?

brasofilo Over a year ago

I've added an example. You were using the quotes wrong, learn the difference between single and double quotes in PHP, it's important.

Paul Brown Over a year ago

Thank you, I have a working version incorporating your comments. I did think I was using quotes correctly but I wasn't because I was approaching it incorrectly.

Paul Brown · Accepted Answer · 2014-09-19 15:25:00Z

0

This is the code I finished up using. It's a mixture of the various responses. Seems to work fine. Guidance appreciated.

add_shortcode( 'myrating', 'display_my_app_rating_89' );
function display_my_app_rating_89( $atts, $content )
{
$args = shortcode_atts( 
    array(
        'app'   => '',
    ), 
    $atts
);

$result = 'No app id.';
$app = (int) $args['app'];

if( !empty( $_GET['app_id'] ) )
    $result = (int) $_GET['app_id'];
elseif( $app )
    $result = $app;

$json_url = "http://itunes.apple.com/lookup?id=$result";
//normal PHP version commented out and wordpress version activated - appears to be Wordpress best practice.
//$json = file_get_contents($json_url);
//wordpress version
$json = wp_remote_retrieve_body (wp_remote_get($json_url));
$result = json_decode($json, true);
$block = '';
foreach ($result['results'] as $key => $value) {
$block .='<span> ... whatever I wanted to return from JSON URL plus my HTML here ... </span>';
}
return $block;
}

answered Sep 19, 2014 at 15:25

Paul Brown

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3 Comments

Paul Brown Over a year ago

I have a supplementary question. One of the fields returned by the JSON URL contains \n for line breaks. This function strips them out and returns a simple block of text. I would like to retain the line breaks - so I guess convert \n to <br> tags. Is there a simple edit to achieve this?

Paul Brown Over a year ago

I have tried these methods - stackoverflow.com/questions/5946114/replace-r-n-with-br but with no luck inserting after the line $json = wp_remote_retrieve_body (wp_remote_get($json_url));

Paul Brown Over a year ago

Got it - all the sample code was using double quotes, I needed to switch to single for whatever reason $json = str_replace(array('\r\n','\r','\n'),'<br/>', $json);

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