loop through array within object, within another object with jQuery?

Question

I'm trying to loop through an array of values contained within an object which is itself contained in another object.

The object looks like this:

var guitar = {
  high4high5: {
    name: 'high 4th, high 5th',
    tuning: [ [5,4], [-2,3], [2,3], [7,3] ]
  },
  high4low5: {
    name: 'high 4th, low 5th',
    tuning: [ [5,4], [-2,3], [2,3], [7,2] ]
  }
}

I know I can just keep looping with a jQuery each loop like so:

$.each(guitar, function(key, value) {
  console.log('1st loop: ' + key, value);
  $.each(value, function(key, value) {
    console.log('2nd : ' + key, value);
    $.each(value, function(key, value) {
      console.log('3rd : ' + key, value);
    });
  });
});

But obviously this end up looping through everything again and again.

The data I need to get is the 'name' (string) and the 'tuning' (array) of each object.

I assume there's a better way to get what I want than just endless loops!

Probably important to note is that I won't know the name of the object inside the object ('high4high5' etc), but I WILL know that values within this object will always be name: (string) and tuning: (array).

EDIT:

Ok, I figured it out.

$.each(guitar, function(key, value) {
  var tuningName = value.name;
  var tuningArray = value.tuning;
  console.log('name: ' + tuningName);
  $.each(tuningArray, function(key,value) {
    console.log(value);
  });
});

Phew!

If you figured it out, you should post it as an answer rather than editing it into the question. You can earn rep that way, and people are more likely to post other answers if they think that their approach is better. — karim79
– karim79, Commented Aug 29, 2011 at 8:51

Corey Sunwold · Accepted Answer · 2011-08-29 08:45:21Z

7

You really don't need jQuery to do this. You can do it with good old fashioned Javascript:

for(g in guitar) {
    console.log(guitar[g].name);
    console.log(guitar[g].tuning);
}

answered Aug 29, 2011 at 8:45

Corey Sunwold

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Comments

Richard Sweeney · Accepted Answer · 2011-08-30 08:26:13Z

1

I'll say here that I've managed to answer my own question using a jQuery each loop. It's definitely a more verbose way to do it!

$.each(guitar, function(key, value) {
  var tuningName = value.name;
  var tuningArray = value.tuning;
  console.log('name: ' + tuningName);
  $.each(tuningArray, function(key,value) {
    console.log(value);
  });
});

edited Aug 30, 2011 at 8:26

answered Aug 30, 2011 at 7:54

Richard Sweeney

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2 Comments

Dhaivat Pandya Over a year ago

why in the world would you do this instead of normal javascript?

Richard Sweeney Over a year ago

Can't say it's a better answer than Corey's (it's clearly not!), but I'm just learning to walk.

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