How to query a database using parameters from a JSON into a PHP?

Question

I have been trying to use JSON in my iPhone app to retrieve information from my SQL database with the use of PHP. I can successfully send a JSON and return it to the iPhone app. My problem occurs when I try to take the parameters in from the JSON array that was sent. It fails and I am not sure what is wrong with my statements. Any suggestions would be great.

function connect($config) {
   try {
     $conn = new \PDO('mysql:host=localhost;dbname=authorization',
                    $config['username'],
                    $config['password']);

     $conn->setAttribute(\PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

     return $conn;              

  }catch(Exception $e){
    return false;
  }
}

function query($conn, $body) {

if (isset($body['authorization'])) {

        // Put parameters into local variables
        $authorization_code = $body['authorize'];
        $device_id = $body['device'];
    }
$stmt = $conn->prepare("SELECT uses_remaining FROM phone_authorization WHERE authorization_code = $authorization_code");
$stmt->execute();
$results = $stmt->fetchAll();
return $results ? $results : false;
}

if ($_SERVER['HTTP_METHOD'] === 'postValues'){ 

    $body;

    if($_POST == null){

        $handle  = fopen('php://input', 'r');
        $rawData = fgets($handle);
        $body = json_decode($rawData);
    }
    else{
        $body == $_POST;
    }

    $conn = connect($config);

    $row = query($conn, $body);

    if ($row) {
        echo json_encode($row);
    }
    else
    {
        $data['value1'] = $row;
        $data['value2'] = "test";
        echo json_encode($data);
    }

}
else {

    $data['error'] = 'The Service you asked for was not recognized';
    echo json_encode($data);
}

Thanks

Osama Jetawe · Accepted Answer · 2013-09-25 19:23:57Z

0

If your $body is json variable you have to decode it to php Use json_decode

function query($conn, $body) {

if (isset($body['authorization'])) {

        // Put parameters into local variables

        $authorization_code =json_decode($body['authorize'], true);
        $device_id = json_decode($body['device'], true);
    }
$stmt = $conn->prepare("SELECT uses_remaining FROM phone_authorization WHERE authorization_code = $authorization_code");
$stmt->execute();
$results = $stmt->fetchAll();
return $results ? $results : false;
}

if ($_SERVER['HTTP_METHOD'] === 'postValues'){ 

    $body;

    if($_POST == null){

        $handle  = fopen('php://input', 'r');
        $rawData = fgets($handle);
        $body = json_decode($rawData);
    }
    else{
        $body == $_POST;
    }

    $conn = connect($config);

    $row = query($conn, $body);

    if ($row) {
        echo json_encode($row);
    }
    else
    {
        $data['value1'] = $row;
        $data['value2'] = "test";
        echo json_encode($data);
    }

}
else {

    $data['error'] = 'The Service you asked for was not recognized';
    echo json_encode($data);
}

answered Sep 25, 2013 at 19:23

Osama Jetawe

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5 Comments

Osama Jetawe Over a year ago

tell me if this help you or not

jmr1706 Over a year ago

So this didn't help any. I did test how my if($_POST == null) was working and it does appear the $_POST is null. So now why wouldn't I be able to access the contents of $body? I thought after decoding I could just $content = $body['authorize']. I assign the JSON in my app with a key value pair like so: [dictionary setValue:code forKey:@"authorize"];

jmr1706 Over a year ago

Ok this is what I was doing wrong it appears my notation is wrong. Even when you decode a json it is still in object notation. So if I do this $body->{'authorization'} it will work. Answer from stackoverflow.com/questions/9192602/php-access-json-decode-data

Osama Jetawe Over a year ago

use $json = file_get_contents('php://input'); $data = json_decode($json, true); to get the data from request

jmr1706 Over a year ago

If you are saying to replace what is inside of if($_POST == null){ with that it doesn't seem to do anything(except reduce a line of code). Unless I have misunderstood.

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