Querying a json object with jquery variables

Question

I have a json object I am querying in jquery and having trouble using the values stored in jquery variables. Example:

$tops= json.tops
$color = "blue";

If I console.log($tops.blue); I get the expected result. But, if I console.log($tops.$color); I get 'undefined'.

Can anyone explain what I am doing wrong/what the difference is?

Was json ever parsed? What gets returned when you type in typeof json? — Dom
– Dom, Commented Sep 28, 2015 at 17:29
Why would you expect $tops.blue to be the same as $tops.$color? You haven't assigned $tops.$color. — Jonathan M
– Jonathan M, Commented Sep 28, 2015 at 17:30
You probably mean $color = 'blue' (with quotes), and $tops[$color]. — John Bupit
– John Bupit, Commented Sep 28, 2015 at 17:31
@JohnBupit the code does have quotes just left them out here by mistake, sorry! — Tatiana Frank
– Tatiana Frank, Commented Sep 28, 2015 at 17:33

Community · Accepted Answer · 2017-05-23 12:22:27Z

1

From your console.log example, my thought is that what you are looking for is $tops[$color]. $tops.$color would look for a property of '$color'.

From your code sample, I think you would need $color = 'blue' (blue being a string), as currently it would be the value of a variable named blue.

Also, perhaps this would be helpful: JavaScript property access: dot notation vs. brackets?

edited May 23, 2017 at 12:22

CommunityBot

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answered Sep 28, 2015 at 17:29

ryachza

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6 Comments

Tatiana Frank Over a year ago

I edited my question to include "", I left those out by mistake but they are in my code. I wouldnt use brackets because i"m not accessing an array

ryachza Over a year ago

@TatianaFrank If you look at the link I included, you will see one of the uses is 'selection of properties using variables'. The array access notation treats the object as a dictionary, with property names being the keys. Essentially obj.x == obj['x'].

Tatiana Frank Over a year ago

ahh that makes sense. Thank you!

Tatiana Frank Over a year ago

I was just thrown off by the fact that it works for $tops= json.tops but not the rest... Why is that by the way?

ryachza Over a year ago

@TatianaFrank I'm not sure what you mean by "works" with regard to $tops = json.tops? I presume there exists a property named tops on the json object? The . notation will treat whatever is on the right side of it as an identifier, and look for a property with that name. So .blue will look for property blue, .tops will look for property tops, and .$color will look for property $color. At no point do the variables in scope matter for that expansion.

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