pass php array to javascript array

Question

I'm looking for the best way to pass php array to javascript.

I'm making a RPG, and when they login, I'd like to return their saved data from database and store in an array on javascript side:

To get data, I do:

$.getJSON("php/CRUD.php", {"_functionToRun" : ""},
    function (returned_data) {
        game.data.dataArray.push(returned_data.split(" "));
        log("1: " + game.data.dataArray); //output: `1: 120,mymap2` 
        log("2: " + game.data.dataArray[0]); //output: `2: 120,mymap2`
    }
);

PHP:

    $qry = 
        'SELECT * 
        FROM userstats
        WHERE id_user_fk ="' . $_SESSION['userid'] . '"
        LIMIT 1';

    $result = $mysqli->query($qry) or die(mysqli_error($mysqli));

    while ($row = $result->fetch_assoc()) {
        $message =  $row['experience'] . $row['levelname'];
    }   

 echo json_encode($message);

1) Is this the best way to get a set of values into a js array?

2) Why can't I access a certain data element of game.data.dataArray using game.data.dataArray[0].

These give the same output of 1: 120,mymap2

        log("1: " + game.data.dataArray);
        log("2: " + game.data.dataArray[0]);

Should returned_data.split(" ") split the returned string into two array elements?

EDIT: Okay I've done echo json_encode($message); and it returns with quotes, but still returns same results for game.data.dataArray and game.data.dataArray[0]

1: "120,mymap2" 
2: "120,mymap2"

I've also changed the function to $.getJSON

Changed again to

    $qry = 
        'SELECT * 
        FROM userstats
        WHERE id_user_fk ="' . $_SESSION['userid'] . '"';

    $result = $mysqli->query($qry) or die(mysqli_error($mysqli));

    while ($row = $result->fetch_assoc()) {
        $array[] =  $row['experience'] . $row['levelname'];
        die(json_encode($array[]));
    }

and JS is:

$.getJSON("php/CRUD.php", {"_functionToRun" : ""},
    function (returned_data) {
        game.data.dataArray.push(returned_data.split(" "));
        log("1: " + game.data.dataArray);
        log("2: " + game.data.dataArray[0]);
    }
);

Outputs:

Uncaught melonJS: level <br />
<font size='1'><table class='xdebug-error' dir='ltr' border='1' cellspacing='0' cellpadding='1'>
<tr><th align='left' bgcolor='#f57900' colspan="5"><span style='background-color: #cc0000; color: #fce94f; font-size: x-...<omitted>...nd

The php you posted does not output anything, what exactly are you echoing? — jeroen
– jeroen, Commented Apr 26, 2014 at 1:54

jeroen · Accepted Answer · 2014-04-26 02:10:06Z

2

If you use $.getJSON and you provide valid json, you don't need to parse anything, so you can change your php to:

if ($row = $result->fetch_assoc()) {
    echo json_encode($row);
    exit;    // You're limiting the number of rows to 1, so there is no need to continue
}

And the javascript:

$.getJSON("php/CRUD.php", {"_functionToRun" : ""},
    function (returned_data) {
        // you can push `returned_data` to an array / add it to an object,
        // but then you need to keep track of its index
        log("1: " + returned_data.experience); //output: `1: 120` 
        log("2: " + returned_data.levelname); //output: `2: mymap2`
    }
);

answered Apr 26, 2014 at 2:10

jeroen

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2 Comments

user3871 Over a year ago

Had no idea you can do returned_data.experience, and access the returned data's properties. Is that because you're returning it as JSON, a js readable object?

jeroen Over a year ago

@Growler Even better; even if you use the "normal" $.ajax() method, jQuery will make an intelligent guess and parse your json. Although if you want to do it correctly, you use dataType: 'json' or $.getJSON. You just need to make sure no other output is sent back apart from the json (like warnings, etc.).

Praveen Kumar Purushothaman · Accepted Answer · 2014-04-26 01:56:12Z

2

What you need to do is, change this code:

while ($row = $result->fetch_assoc()) {
    $message =  $row['experience'] . $row['levelname'];
}

To make it as an array:

while ($row = $result->fetch_assoc()) {
    $message[] =  $row['experience'] . $row['levelname'];
}

So, $message now becomes an array. You need to output this to the client. You can do that using:

die(json_encode($message));

The reason is, in your code, it will output only the last one.

answered Apr 26, 2014 at 1:56

Praveen Kumar Purushothaman

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1 Comment

Praveen Kumar Purushothaman Over a year ago

What's the response of the PHP page?

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