CODEWITHSUNDEEP
php
Kevin Smith
Kevin Smith

Reputation: 731

Json encode parameters issue

I'm receiving a error message explaining the following error.

json_encode() expects at most 2 parameters, 3 given</p>

When I make the call to the json_encode function I have all three of the paramters set with accpted values.

I"m trying to figure out why this is because when I do tests on my code I get all accepted values with this function. Any thoughts? I think it 's something to do with the switch statement but I need further verifcation as well as information on what I'm doing wrong. Can someone enlighten me please?

public function output($message, $title, $status)
{
    switch ($status)
    {
        case 'Error':
            array('status' => 'Error');
            break;
        case 'Notice':
            array('status' => 'Notice');
            break;
        case 'Success':
            array('status' => 'Success');
            break;
    }
    echo json_encode($status, $title, $message);
}

Upvotes: 1

Views: 1452

Answers (6)

David B&#233;langer
David B&#233;langer

Reputation: 7438

Here's what I suggest and would work : 

public function output($Message='', $Title='', $Status=''){
    #   We make sure our status is perfect.
    #   We make sure our status will always be what we want and not something different by mistake.
    #   We default to "Error".
    switch(strtoupper($Status)){
        default:
            $Status = 'error';
        break;

        case 'NOTICE':
            $Status = 'notice';
        break;

        case 'SUCCESS':
            $Status = 'success';
        break;
    }

    #   We output the content as JSON
    header('Content-Type: application/json');
    echo json_encode(array(
        'status'    => $Status,
        'title'     => $Title,
        'message'   => $Message
    ));

    #   Done - 0 mean the page end with no error (PHP error !)
    exit(0);
}

Output:

output('this is my message', 'this is my title', 'error');

{
    "status" : "error",
    "title" : "this is my title",
    "message" : "this is my message"
}

Documentations:

Upvotes: 3

Mahesh
Mahesh

Reputation: 320

 public function output($message, $title, $status)
    {
        switch ($status)
        {
            case 'Error':
                array('status' => 'Error');
                break;
            case 'Notice':
                array('status' => 'Notice');
                break;
            case 'Success':
                array('status' => 'Success');
                break;
        }
        echo json_encode(array('status' => $status, 'title' => $title, 
    'message' =>$message));

    }

For more about json_encode refer this json_encode

Upvotes: 2

web2students.com
web2students.com

Reputation: 307

Read, http://php.net/manual/en/function.json-encode.php http://php.net/manual/en/control-structures.switch.php what are you doing with switch?nothing!! what it means? case 'Error': array('status' => 'Error');

I think you want something like following, 

public function output($message, $title, $status)
{
    switch ($status)
    {
        case 'Error':
            array('status' => 'Error');
            break;
        case 'Notice':
            array('status' => 'Notice');
            break;
        case 'Success':
            $output =  $title . $message;
            echo json_encode($output);
            break;
    }

}

Upvotes: 2

Olivier Lerone
Olivier Lerone

Reputation: 151

I think what you are trying to do is encode an array?

public function output($message, $title, $status)
{
    switch ($status)
    {
        case 'Error':
            array('status' => 'Error');
            break;
        case 'Notice':
            array('status' => 'Notice');
            break;
        case 'Success':
            array('status' => 'Success');
            break;
    }
    echo json_encode(array($status, $title, $message));
}
output('messageval', 'titleval', 'statusval');

that will output JSON like:

["statusval", "titleval", "messageval"]

or there is also this:

public function output($message, $title, $status)
{
    switch ($status)
    {
        case 'Error':
            array('status' => 'Error');
            break;
        case 'Notice':
            array('status' => 'Notice');
            break;
        case 'Success':
            array('status' => 'Success');
            break;
    }
    echo json_encode(array('status'=>$status, 'title'=>$title, 'message'=>$message));
}
output('messageval', 'titleval', 'statusval');

which will output something similar to:

{"message":"messageval", "title":"titleval", "status":"statusval"}

Also, your switch block will do nothing since you are not using the array produced by array().

Upvotes: 3

Quentin
Quentin

Reputation: 943220

You can only encode a single data structure. If you have three bits of data that you want to encode, then you must combine them into a single data structure first. For example:

echo json_encode(Array("status" => $status, "title" => $title, "message" => $message));

Upvotes: 3

hek2mgl
hek2mgl

Reputation: 157957

You might looking for something like this:

echo json_encode(array($status, $title, $message));

or, as others suggested, like this:

json_encode(array("status"=>$status, "title"=>$title, "message"=>$message))

Upvotes: 5

Related Questions