JSON mapping in Javascript

Question

So I have this JSON

https://bitcoinpayflow.com/orders{"order":{"bitcoin_address":"1NwKSH1DJHhobCeuwxNqdMjK5oVEZBFWbk"}}

No I want to reference the bitcoin_address

So first I strip away the string at the beginning

        var stripped = data.substring(33);
        alert(stripped);
                var btc = stripped.orders.bitcoin_address;
        alert(btc);

I get the first alert, but not the second. Any idea why?

James Allardice · Accepted Answer · 2012-04-26 08:41:29Z

2

Because stripped is still just a string. You need to parse it into an object. You can use the native JSON.parse method to do this:

var stripped = JSON.parse(data.substring(33));

Also, you are referencing the orders property, which doesn't exist. It's order.

Note that JSON.parse is not supported by older browsers. You can use this polyfill to make sure it's always available.

answered Apr 26, 2012 at 8:41

James Allardice

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1 Comment

Joseph Over a year ago

Thanks! I did use 'order' before. This was just hastily rewritten for stackoverflow. I didn't downvote? I'm waiting till the time limit passes so I can tick.

Denis Ermolin · Accepted Answer · 2012-04-26 08:42:16Z

1

The easiest way to decode json - string - use eval

var bitcoins = eval('(' + json_string + ')');

And access bitcoins['order']['bitcoin_address'] But it a bit unsafe. Upper method is more safer.

answered Apr 26, 2012 at 8:42

Denis Ermolin

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6 Comments

James Allardice Over a year ago

No, the easiest way to deal with a JSON string is to use JSON.parse. You should not need to use eval for this.

Kirstein Over a year ago

Using eval is generally a bad idea. Try to avoid that if possible.

Denis Ermolin Over a year ago

Omg i know it but it is works too. If i write this method that it doenst mean that i use it

James Allardice Over a year ago

@DenisErmolin - Then why promote bad practice?

Denis Ermolin Over a year ago

I didnt promote bad practice. I comented that your way is safer to use. He must know all possible methods and bad too.

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Nick · Accepted Answer · 2012-04-26 08:51:22Z

1

I would use JSON.parse as follows.

$.post('php/whatever.php',{data:dS},function(res){
    var o=JSON.parse(res);
    var bitcoins=o.order.bitcoinaddress;
},"text");

answered Apr 26, 2012 at 8:51

Nick

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2 Comments

Denis Ermolin Over a year ago

Why you are using "text" instead of "json" ?

Nick Over a year ago

That's a really good question, Denis. I started using 'text' to avoid the need for 'intelligent guess' and because I guess the model I used initially also used 'text'. I'm not sure what the practical difference would be if I made it 'json'. My initial impression is that it might remove the need for the parse step. Thwt's how the second last example appears on api.jquery.com/jQuery.post. I might need to update a heap of code! It works fine as I posted, though.

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