json object accessing data

Question

I have returned data from php (using json):

"[4, [{
    "quantity": "1",
    "name": "item 1",
    "price": "2",
    "currency": "GBP",
    "description": "item 1 description"
}, {
    "quantity": "1",
    "name": "item 1",
    "price": "2",
    "currency": "GBP",
    "description": "item 1 description"
}]]"

Then how do I access the data. It used to return use the second part:

    [{
        "quantity": "1",
        "name": "item 1",
        "price": "2",
        "currency": "GBP",
        "description": "item 1 description"
    }, {
        "quantity": "1",
        "name": "item 1",
        "price": "2",
        "currency": "GBP",
        "description": "item 1 description"
    }]

which I used jQuery.parseJson to turn into an object but I can't seem to figure out how to add in the second part the number 4

help would be great

my php code is like this

            $array['items'] = array();      
            $arrayOfMyNumbers = array();

            $total = 0;
            for( $i = 0; $i<2; $i++ ) {
                $foo = new StdClass();
                $foo->quantity = "1";
                $foo->name =  "item 1";
                $foo->price = "2";
                $foo->currency = "GBP";
                $foo->description = "item 1 description";;
                $arrayOfMyNumbers[] = $foo;
                $total += ($foo->price*$foo->quantity);
            }

            $returnArray = array();
            array_push($returnArray,$total,$arrayOfMyNumbers);

so I thought I'd be able to access that from javascript:

            jsonArrayResponse =  (jsonObj);
            if(typeReq =="button"){
            console.log("return is "+(jsonArrayResponse));

but it's how I get the 4 and then the second array part so I can convert it to an object.

Finally solve it!!!!

If you look at the original data you can see leading and trailing quotes. These were causing the json decode to fail. So I put in some regex and it work:

            success: function (jsonObj) {
            console.log(jsonObj);
            jsonArrayResponse =  (jsonObj);
            var someStr = jsonObj.data.replace(/^"(.*)"$/, '$1');
            console.log("some str is "+someStr);

            parsed = JSON.parse(someStr);
            console.log("this should be 4"+parsed[1]);

definitely one of these ones where it's very easy to miss. Thanks to Devlin who helped me track it down.

Matt Spinks · Accepted Answer · 2017-12-20 20:12:45Z

1

If all you need to do is add the 4 (that's in the first part) to the second json object, you will just need to create a new array, add 4, and add the second json object:

var secondJsonObject = [{
        "quantity": "1",
        "name": "item 1",
        "price": "2",
        "currency": "GBP",
        "description": "item 1 description"
    }, {
        "quantity": "1",
        "name": "item 1",
        "price": "2",
        "currency": "GBP",
        "description": "item 1 description"
    }];

var newJsonObject = [4, secondJsonObject];

answered Dec 20, 2017 at 20:12

Matt Spinks

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6 Comments

user1616338 Over a year ago

I added some of the code. I already was trying this as two parts of an array - I think?

Matt Spinks Over a year ago

I guess I'm not really clear on what problem you're having, and what you are trying to do.

user1616338 Over a year ago

I'm creating an object in php (from a databbase) and then passing it back to javascript. Where I have the total amount - the two prices added together =4 and the objects themselves. This is then bundled together and sent to the payment system.

Matt Spinks Over a year ago

What does jsonObj look like?

user1616338 Over a year ago

jsonObj in console is data : "[4,[{"quantity":"1","name":"item 1","price":"2","currency":"GBP","description":"item 1 description"},{"quantity":"1","name":"item 1","price":"2","currency":"GBP","description":"item 1 description"}]]"

|

devlin carnate · Accepted Answer · 2017-12-20 21:03:03Z

0

how do I access the data?

You'll still use JSON.parse().

var json = '[4, [{"quantity": "1","name": "item 1","price": "2","currency": "GBP","description": "item 1 description"}, {"quantity": "1","name": "item 1","price": "2","currency": "GBP","description": "item 1 description"}]]';

var parsed = JSON.parse(json);
console.log(parsed[0]);  //4
console.log(parsed[1][0].quantity);
console.log(parsed[1][1].quantity);

Or, put another way:

for (var i = 0; i < parsed.length; i++) {
  console.log(parsed[i]);  //4
  for (var j = 0; j < parsed[i].length; j++) {
    var obj = parsed[i][j];
    console.log(obj);
    console.log(obj.quantity);
  }
}

Demo: https://jsfiddle.net/zephyr_hex/2na2j317/

Update

OP states there is an error regarding "Unexpected token , in JSON at position 1". This can occur if the received response is not a string. JSON.stringify() will allow you to then JSON.parse().

var json2 = JSON.stringify([4, [{"quantity": "1","name": "item 1","price": "2","currency": "GBP","description": "item 1 description"}, {"quantity": "1","name": "item 1","price": "2","currency": "GBP","description": "item 1 description"}]]);

var parsed2 = JSON.parse(json2);

edited Dec 20, 2017 at 21:03

answered Dec 20, 2017 at 20:23

devlin carnate

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10 Comments

user1616338 Over a year ago

I tired this but I got Unexpected token o in JSON at position 1 which is why I'm here

devlin carnate Over a year ago

@user1616338 : That's because your JSON string has line feeds in it. Put the JSON string on one line. You also need to wrap the string in single quotes, not double quotes. See my demo here: jsfiddle.net/zephyr_hex/2na2j317

user1616338 Over a year ago

I can't see where line feeds would have come from the code generates the string - what I'm showing is the console.log output?

devlin carnate Over a year ago

@user1616338 : I just updated my question showing you the problem.

user1616338 Over a year ago

okay so I solve it - your idea of the validator gave me a clue. The whole return had quotes around it so it wan't being decoded correctly. Once I added a bit of regex to it to remove the leading and trailing quote it worked! Thanks you really helped me

|

Wranorn · Accepted Answer · 2017-12-20 20:33:06Z

0

Rough Pass:

$json_array = [];
$json_array[] = 4;
$json_array[] = $first_array;
$json_array[] = $second_array;
return json_encode($json_array);

Note: There is NO guarantee you'll get the array values/objects in the order you added them.

answered Dec 20, 2017 at 20:33

Wranorn

8089 silver badges19 bronze badges

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