javascript trying to get 3rd nested array.length and value

Question

I have a generated nested Array which I store some data.

How can i get the 3rd nested array (in this case the array starting with "yellow")

the array looks like this (this is a dynamically generated array):

[
    ["Large", 
       ["yellow", "green", "Blue"],
       ["$55.00", "$55.00", "$55.00"]
     ]
    ["Medium",
       ["yellow", "green", "Blue", "Red"],
       ["$55.00", "$55.00", "$55.00", "$55.00"]
    ]
    ["small",
       ["yellow", "green", "Blue", "Red"],
       ["$55.00", "$55.00", "$55.00", "$55.00"]
    ]
]

I am trying to get to the ["yellow", "green", "Blue"] array's length and loop to get the values

for(i=0; colorNSize.dataArray[0][0][1].length<i; i++){
    alert(colorNSize.dataArray[colorNSize.Sizeindex][0][0][i])// alert's A (which is the char[1] of the second array.
}

alert(colorNSize.dataArray[0][0][1].length) actually alerts the length of "Large" which is "5" is there a limit for nested arrays? Can i still get the real 3rd array nested here?

Chetan S · Accepted Answer · 2010-03-21 18:10:57Z

3

Why does everything have to be an array? This structure makes much more sense to me.

obj = {
    "Large": {
       "yellow": "$55.00",
       "green": "$55.00",
       "Blue": "$55.00"
    },
    "Medium": {
    ...
    ...
]

To get the Large, yellow price, all I have to do is

obj["Large"]["yellow"] //or obj.Large.yellow

answered Mar 21, 2010 at 18:10

Chetan S

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2 Comments

adardesign Over a year ago

Thanks, but I cant call ["Large"] since i don't know the values at run-time

Chetan S Over a year ago

@adardesign: Ok. As a rule of thumb, use arrays when you are representing lists of data you want to process as a batch and objects when you want to lookup fast by a property. You can have arrays nested inside objects or vice versa based on the above rule.

Eli Bendersky · Accepted Answer · 2010-03-21 18:08:13Z

2

Your formatting is probably wrong - does "Large" reside in the same array as the array of colors and then prices? Indentation tells it is, brackets say it's all nested. Have you forgotten a bracket after the array of prices?

Anyway, [0][0][1] already refers to chars of "Large". [0][0] is "Large", [0][1] is the other array.

Consider this simplified example:

>>> arr = [['large', [[1, 2], [3, 4]]]]
[["large", [[1, 2], [3, 4]]]]
>>> arr[0]
["large", [[1, 2], [3, 4]]]
>>> arr[0][0]
"large"
>>> arr[0][1]
[[1, 2], [3, 4]]
>>> arr[0][1][0]
[1, 2]

Can you see what's going on?

edited Mar 21, 2010 at 18:08

answered Mar 21, 2010 at 17:54

Eli Bendersky

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5 Comments

adardesign Over a year ago

Can you please explain? how can i get the 3rd array?

adardesign Over a year ago

in your answer i see the last example.. >>> arr[0][1]... Can you please add a 3rd nest example?

Eli Bendersky Over a year ago

@adardesign: alright, added the 3rd nest. You can have a 4th one too, in my example, just by adding more indexing

adardesign Over a year ago

Correct. My mistake.. May i ask you? is @Chetan correct that objects will make more sense?

Eli Bendersky Over a year ago

@adardesign: it's hard to say without knowing your real application, but the way your data looks he may well be right

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