How do you do an eval within the scope of a json?

Question

I have a json in the form of { a:1, b:10, c:43 } for example.

I wish to perform eval( "(a+b-5)*c" ) but applying it to the json, not the place where the json and the formula is.

Attempted this, trying to utilize the scope, but wouldn't find a.

var z = { a:1, b:10, c:43, eval:eval };
console.log( z.eval( "a+b" ) );

I do not understand what you mean... but if you mean what I think you do, there'll be no way but to unpack the JSON, do the array, and re-pack it. — Pekka, Oct 05 '13 at 00:04
This: `{ a:1, b:10, c:43 }` array, object, same difference in this case... it'd look something like `result = (myJSON.a+myJSON.b-5)*myJSON.c)` — Pekka, Oct 05 '13 at 00:09
Well, where does the JSON come from? Can you show some code? — Pekka, Oct 05 '13 at 00:10
In no way encouraging this, but: `var json = { a:1, b:10, c:43 }; with (json) { eval( "(a+b-5)*c" ) }` — go-oleg, Oct 05 '13 at 00:12
This json is the sum of several jsons with all (maybe some missing) properties, on which I must apply a formula as a string and I can guarantee that all the variables in that formula exist in the json — Discipol, Oct 05 '13 at 00:12
@go-oleg I would accept your answer should you post it like that. It worked for my purposes. also I can add several evals within that `with`. — Discipol, Oct 05 '13 at 00:20
@Discipol: Looks like vkurchatkin has a very similar answer. — go-oleg, Oct 05 '13 at 00:54

score 2 · Accepted Answer · answered Oct 05 '13 at 00:22

2

You can use the with keyword, but please don't. Using with and eval is not recommended.

var z = { a:1, b:10, c:43 };
with(z) {
    console.log(eval('a+b'));
}

Here's some more info on MDN.

answered Oct 05 '13 at 00:22

DarthJDG

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then how do I scope inside the json without `with`? – Discipol Oct 05 '13 at 00:39
1

You can use ```with``` if you like, the whole approach as it is, is not recommended. If I would need to do something like this, I'd go back and re-think what I want to achieve and if there's a better way to do it. Although I'm not saying I've never done messy things myself due to lack of time or interest. ;) – DarthJDG Oct 05 '13 at 00:44

score 1 · Answer 2 · answered Oct 05 '13 at 00:11

1

Check this out:

function expression (expr) {
    return new Function('obj', 'with (obj) return ' + expr);
}

console.log(expression('(a+b-5)*c')({ a:1, b:10, c:43 }));

answered Oct 05 '13 at 00:11

vkurchatkin

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score 0 · Answer 3 · edited May 23 '17 at 12:29

0

Dont use eval! USE JSON.parse

JSON.parse

JSON.parse vs. eval()

var obj=JSON.parse(json);

alert((obj.a+obj.b-5)*obj.c)

edited May 23 '17 at 12:29

Community

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answered Oct 05 '13 at 00:13

zloctb

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the question is about parsing and evaluating an expression, not parsing json – vkurchatkin Oct 05 '13 at 00:16

score 0 · Answer 4 · answered Oct 05 '13 at 00:16

I guess you want something like:

var yourJSon = '{ "a":1, "b":10, "c":43 }'; 
//Your original JSon String (attention to the standard double quotes)

yourJSon = JSON.parse(yourJSon); 
//this will render your JSon string into a "real" object

var answer = eval("(yourJSon.a + yourJSon.b - 5) * yourJSon.c"); 
//And now you can do the eval, using the object's variable to define the scope of that object.

How do you do an eval within the scope of a json?

4 Answers4

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