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I have this javascript object:

{"Vanilla": 36, "Chocolate": 50, "Stracciatella": 24, "Coffee": 18}

to which I would like to "add" this array:

["Vanilla", "Amarena", "Chocolate", "Pistachio"]

The result I want is this:

{
    "Vanilla": 37,
    "Chocolate": 51,
    "Stracciatella": 24,
    "Coffee": 18,
    "Amarena": 1,
    "Pistachio": 1
}

This is what I have tried:

var me = ["Vanilla", "Amarena", "Chocolate", "Pistachio"]
var all = {
    "Vanilla": 36,
    "Chocolate": 50,
    "Stracciatella": 24,
    "Coffee": 18
}
var keys = Object.keys(all);
for (var i = 0; i < me.length; i++) {
    if (me[i] in keys) {
        console.log("keys:" + keys + " - " + me[i] + " in keys");
    } else {
        console.log("keys:" + keys + " - " + me[i] + " not in keys");

    };
}

console.log(keys);

I have this javascript object:

{"Vanilla": 36, "Chocolate": 50, "Stracciatella": 24, "Coffee": 18}

to which I would like to "add" this array:

["Vanilla", "Amarena", "Chocolate", "Pistachio"]

The result I want is this:

{
    "Vanilla": 37,
    "Chocolate": 51,
    "Stracciatella": 24,
    "Coffee": 18,
    "Amarena": 1,
    "Pistachio": 1
}

This is what I have tried:

var me = ["Vanilla", "Amarena", "Chocolate", "Pistachio"]
var all = {
    "Vanilla": 36,
    "Chocolate": 50,
    "Stracciatella": 24,
    "Coffee": 18
}
var keys = Object.keys(all);
for (var i = 0; i < me.length; i++) {
    if (me[i] in keys) {
        console.log("keys:" + keys + " - " + me[i] + " in keys");
    } else {
        console.log("keys:" + keys + " - " + me[i] + " not in keys");

    };
}

console.log(keys);

• javascript
• jquery

Share Improve this question Follow edited Nov 15, 2015 at 12:32 Andreas Louv 47.1k1313 gold badges107107 silver badges126126 bronze badges asked Nov 15, 2015 at 11:55 Ken AraKen Ara 4311 silver badge77 bronze badges 1

• 1 Wele to Stack Overflow! Please take the tour, have a look around, and read through the help center, in particular How do I ask a good question? Any code related to your question must be in your question, not just linked. Links rot, making the question and its answers useless to people in the future, and people shouldn't have to follow some random link to help you. If the question doesn't make sense and can't be answered without the link, it's not appropriate for this site. Instead, put the minimum plete example in the question. – T.J. Crowder Commented Nov 15, 2015 at 12:02

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4 Answers 4

Sorted by: Reset to default 4

This works

the first part of the code in your fiddle

var me = ["Vanilla", "Amarena", "Chocolate", "Pistachio"]
var all = {
    "Vanilla": 36,
    "Chocolate": 50,
    "Stracciatella": 24,
    "Coffee": 18
}

Now the answer:

me.forEach(function(key) { // forEach iterates through an array 
    all[key] = (all[key] || 0) + 1; // create new item if it doesn't exist
});

The forEach() method executes a provided function once per array element.

For a more detailed description of .forEach see MDN documentation where the above description came from

as this question was tagged jQuery, lets add some jQuery code, just for laughs

$.each(me, function(unusedIndex, key) {
    all[key] = (all[key] || 0) + 1;exist
});

jQuery.each() A generic iterator function, which can be used to seamlessly iterate over both objects and arrays. Arrays and array-like objects with a length property (such as a function's arguments object) are iterated by numeric index, from 0 to length-1. Other objects are iterated via their named properties. - source http://api.jquery./jquery.each/

Here's me playing with myself - http://jsfiddle/7ephp3f4/1/

The following code allows you to increment the values within your object stock based on the keys in your array stockUpdate:

var stockUpdate = ["Vanilla", "Amarena", "Chocolate", "Pistachio"];
var stock = {
    "Vanilla": 36,
    "Chocolate": 50,
    "Stracciatella": 24,
    "Coffee": 18
};

for (var i=0; i < stockUpdate.length; i++) {
    var stockType = stockUpdate[i];
    if (stock[stockType] === undefined) {
        stock[stockType] = 0;
    }
    stock[stockType]++;
}

you are almost there,

var me = ["Vanilla", "Amarena", "Chocolate", "Pistachio"]
var all = {
    "Vanilla": 36,
    "Chocolate": 50,
    "Stracciatella": 24,
    "Coffee": 18
}
var keys = Object.keys(all);
for (var i = 0; i < me.length; i++) {
    if (keys.indexOf(me[i]) >= 0) {
        console.log("keys:" + keys + " - " + me[i] + " in keys \n");
        all[me[i]]++;
    } else {
        console.log("keys:" + keys + " - " + me[i] + " not in keys \n");
        all[me[i]] = 1;
    };
}

console.log(all);

you can also use reduce to "merge" the newItems into items

var items = {"Vanilla": 36, "Chocolate": 50, "Stracciatella": 24, "Coffee": 18}
var newItems = ["Vanilla", "Amarena", "Chocolate", "Pistachio"];

var result = newItems.reduce(function(pre, curr) {
        // use double == to check when the item is either undefined or null
        pre[curr] = (pre[curr] == null? 0 : pre[curr]) + 1;
        return pre;
// pass items as the initial data to reduce
}, items);

console.log(result);

The reduce() method applies a function against an accumulator and each value of the array (from left-to-right) to reduce it to a single value.

in this case, I reduce the newItems list, from left to right, calcuate the count in the items object, and return the calculated items.

It is a good moment to use http://underscorejs:

var me = ["Vanilla", "Amarena", "Chocolate", "Pistachio"];

var all = {
  "Vanilla": 36,
  "Chocolate": 50,
  "Stracciatella": 24,
  "Coffee": 18
};

_.each(me, function(name) {
  if (_.has(all, name)) {
    all[name]++;
  } else {
    all[name] = 1;
  }
});

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