Getting value of first index within jQuery array using .each()

Question

I have a JSON array returned via ajax that looks like:

"stuff": [["2","66%"], ["3","42%"],...

Problem I want to match the zeroth index of each element in this array to a variable outside of the loop and if it matches, I want to return the percentage next to it.

I don't know the syntax in jQuery for this. Please have a look at my code below and see if that's correct or not:

var percentage = 0;
var stuffarr = jsonobj['stuff'];
var stuffID = jsonobj['stuff_id']
if (!stuffID || 0 === stuffID.length){
   $("#stuff-element").html("--");
}
else {
   var percentage = $.each(stuffarr, function (index, value) {
   if(value[0] == stuffID)
      return value[1]   
   });
}

What is var stuffID = jsonobj['stuff_id']. Is this another array? — Amit Kumar Singh
– Amit Kumar Singh, Commented Aug 31, 2017 at 9:56

Rory McCrossan · Accepted Answer · 2017-08-31 10:08:56Z

2

Firstly, a bit of terminology. The data structure you have is an Object which holds several properties. It has nothing to do with JSON after it has been deserialised.

With regard to your issue, there's no jQuery required as you can use find() to find the item in the array by the stuffID variable's value. Try this:

var obj = {
  "stuff": [
    ["2", "66%"],
    ["3", "42%"]
  ],
  "stuff_id": "3"
}

var percentage = 0;
var stuffArr = obj['stuff'];
var stuffId = obj['stuff_id']

if (!stuffId || 0 === stuffId.length) {
  $("#stuff-element").html("--");
} else {
  percentage = stuffArr.find(function(el) {
    return el[0] == stuffId;
  })[1];
}
  
console.log(percentage);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

edited Aug 31, 2017 at 10:08

answered Aug 31, 2017 at 9:57

Rory McCrossan

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4 Comments

LiverpoolOwen Over a year ago

You could use the find method as you only want to return one

Rory McCrossan Over a year ago

@LiverpoolOwen you're right - thanks. I'd misread the question and though the OP would possibly be searching for multiple.

omrakhur Over a year ago

Also, my bad but I think I shouldn't have re-initialised the percentage variable inside the else statement

Rory McCrossan Over a year ago

@omrakhur yep, you should only define var percentage at the outer scope. I missed that too. Having a slow morning :)

LiverpoolOwen · Accepted Answer · 2017-08-31 10:04:22Z

1

try this

var percentage = stuffarr.find(function (value) {
if(value[0] == stuffID)
   return value[1];  
})[1];

edited Aug 31, 2017 at 10:04

answered Aug 31, 2017 at 9:58

LiverpoolOwen

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2 Comments

omrakhur Over a year ago

Owen, your answer works fine as well, however Rory added a bit of an explanation, so I have accepted that as the correct one. Thank you.

LiverpoolOwen Over a year ago

That is fair enough

Snufkin · Accepted Answer · 2017-08-31 10:33:04Z

The return statement you used returns the result of the function(..) inside the .each(..) function and is not the return value of the .each(..) function. the valid return value is boolean which 'tells' the .each(..) function whether it should continue or not. instead use the following syntax:

var ret = '';
$(arr).each(function () {
    var curr = $(this);
    //console.log(curr); 
    if(curr[0] == stuffID){
        //console.log('>>found: ' + curr[1]);
        ret = curr[1];
        //if found -> break loop
        return false; 
    }
});

comment: you should consider instead of using an inner array data structure use an object, this is a more 'correct' data structure:

// "stuff": [{"id":"2","percent":"66%"}, {"id":"3","percent":"42%"},...
var ret = '';
$(arr).each(function () {
    var curr = this;
    //console.log(curr.id); 
    if(curr.id == stuffID){
        //console.log('>>found: ' + curr.percent);
        ret = curr.percent;
        //if found -> break loop
        return false; 
    }
});

@LiverpoolOwen approach is clean and nice, and if you want to use it combining with the object approach do this:

arr.find(function (value) {
if(value.id == stuffID)
   return value;  
}).percent;

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