Json Parsing error using JSON.parse()

Question

I have developed an api which will post some data in json format to be used in an android app. However I am getting json parsing error. I am new to this whole json thing so unable to understand what the error means.

This is my json encoded output that the php backend generates

   {
    "data": [
        {
            "id": "2",
            "name": "Rice",
            "price": "120",
            "description": "Plain Rice",
            "image": "6990_abstract-photo-2.jpg",
            "time": "12 mins",
            "catagory": "Lunch",
            "subcat": ""
        }
    ]
}{
    "data": [
        {
            "id": "4",
            "name": "Dal",
            "price": "5",
            "description": "dadadad",
            "image": "",
            "time": "20 mins",
            "catagory": "Dinner",
            "subcat": ""
        }
    ]
}{
    "data": [
        "catagory": "Soup"
    ]
}

This is the error the online json parser gives

SyntaxError: JSON.parse: unexpected non-whitespace character after JSON data at line 2 column 1 of the JSON data

What is actually wrong here? Could you please provide me with the correct json output for the following data?

There are 3 json objects in the examnple above, if you want to parse all of this you will need to wrap them in [] — Andy Gee
– Andy Gee, Commented Oct 20, 2014 at 6:53
Your json code is invalid, because those are 3 different json objects. Can you show us your php code to generate this json result? — Jerodev
– Jerodev, Commented Oct 20, 2014 at 6:53
I have added the php code please help me do the needed modification — Rick Roy
– Rick Roy, Commented Oct 20, 2014 at 6:54

Andy Gee · Accepted Answer · 2014-10-20 06:57:09Z

2

This should clear it up

    $main = array();
    while($row = $result->fetch(PDO::FETCH_ASSOC)){
        $cat    = $row['category'];
        $query1 = "SELECT * FROM item WHERE catagory='$cat'";       //Prepare login query
        $value  = $DBH->query($query1);
        if($row1 = $value->fetch(PDO::FETCH_OBJ))
        {
            $main[] = array('data'=>array($row1));
        }
        else
        {
            $main[] = array('data'=>array('catagory'=>$row['category']));
        }
    }
    echo json_encode($main);

answered Oct 20, 2014 at 6:57

Andy Gee

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Comments

Kevin · Accepted Answer · 2014-10-20 07:02:08Z

1

You shouldn't create your json string by hand. Create your array structure, then finally invoke json_encode() at the end.

$data = array();

try
{
    $query = "SELECT category FROM category"; // select category FROM category? what?
    $result= $DBH->query($query);
    while($row = $result->fetch(PDO::FETCH_ASSOC)){
        $cat    = $row['category'];
        $query1 = "SELECT * FROM item WHERE catagory='$cat'";
        $value  = $DBH->query($query1);
        if($value->rowCount() > 0) {
            $data[] = array('data' => $value->fetch(PDO::FETCH_ASSOC));
        }
        else {
            $sub = array('category' => $row['category']);
            $data[] = array('data' => $sub);
        }

    }

    $result->closeCursor();
    $DBH = null;

    echo json_encode($data); // encode at the end
}
catch(PDOException $e)
{
    print $e->getMessage ();
    die();
}

answered Oct 20, 2014 at 7:02

Kevin

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2 Comments

Rick Roy Over a year ago

Both the table name and field name is category :D I was working late at night my sleep deprived brain just couldn't come up with something not silly :D

Kevin Over a year ago

@RickRoy well you better get some rest then, code another time, a well rested brain is good

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