CODEWITHSUNDEEP
javascriptphpjson
Jgunzblazin
Jgunzblazin

Reputation: 135

Accessing PHP JSON response using Javascript

Is this valid approach: I want to keep api key from being accessible via source code so I have been trying to keep it hidden with PHP and use Javascript to display data. (I prefer to use js syntax to display data) I've been able to display data successfully but when I look at the source code I can see the JSON response. Can anyone tell me if this is a valid approach and why not good idea to have json shown in source?

<?php
$apikey = "xxxx";
$data = file_get_contents('http://url?apikey=' . $apikey);
$json = json_decode($data,true);
?>

I then access the response like so:

<script type="text/javascript">

    var data =  <?php echo json_encode($json) ?>;
    $('.in-theaters-soon').append('<p>' + data.movies[0].title + '</p>');

</script>

Upvotes: 1

Views: 60

Answers (3)

Rendition
Rendition

Reputation: 53

You could do something like this if you have the php in a separate file.

Your php file.

<?php
// create a token check to make sure it is being called.
$apikey = "xxxx";
$data = file_get_contents('http://url?apikey=' . $apikey);
echo json_encode($data);
?>

Then query your php file something like this sending a token or something similar.

$.ajax({
    url: url,
    type: 'POST',
    data: {token:token},
    success: function(data){
        var response = $.parseJSON(data);
        for(var x = 0; x < response.length; x++){
            $('.in-theaters-soon').append('<p>' + response[x].title + '</p>');
        }
    },
    cache: false,
    contentType: false,
    processData: false
});

Hope this helps.

Upvotes: 0

koredalin
koredalin

Reputation: 445

Always make some validation of the printed data.

<?php
$apikey = "xxxx";
$data = file_get_contents('http://url?apikey=' . $apikey);
if (is_array($data) && ! empty($data)) {
    /**
     * Do something.
    /**/
}

Upvotes: 1

littleibex
littleibex

Reputation: 1712

You can directly echo the values from PHP since you already have the response in $json. For example:

<div class="in-theaters-soon">
    <p><?php echo $json['movies'][0]['title']; ?></p>
</div>

Upvotes: 3

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