Move them to an array, sort that array, and then use that array for your purposes. Here’s a solution:
var maxSpeed = {
car: 300,
bike: 60,
motorbike: 200,
airplane: 1000,
helicopter: 400,
rocket: 8 * 60 * 60
};
var sortable = [];
for (var vehicle in maxSpeed) {
sortable.push([vehicle, maxSpeed[vehicle]]);
}
sortable.sort(function(a, b) {
return a[1] – b[1];
});
//[[“bike”, 60], [“motorbike”, 200], [“car”, 300],
//[“helicopter”, 400], [“airplane”, 1000], [“rocket”, 28800]]
Once you have the array, you could rebuild the object from the array in the order you like, thus achieving exactly what you set out to do. That would work in all the browsers I know of, but it would be dependent on an implementation quirk, and could break at any time. You should never make assumptions about the order of elements in a JavaScript object.
var objSorted = {}
sortable.forEach(function(item){
objSorted[item[0]]=item[1]
})
In ES8, you can use Object.entries() to convert the object into an array:
const maxSpeed = {
car: 300,
bike: 60,
motorbike: 200,
airplane: 1000,
helicopter: 400,
rocket: 8 * 60 * 60
};
const sortable = Object.entries(maxSpeed)
.sort(([,a],[,b]) => a-b)
.reduce((r, [k, v]) => ({ …r, [k]: v }), {});
console.log(sortable);
In ES10, you can use Object.fromEntries() to convert array to object. Then the code can be simplified to this:
const maxSpeed = {
car: 300,
bike: 60,
motorbike: 200,
airplane: 1000,
helicopter: 400,
rocket: 8 * 60 * 60
};
const sortable = Object.fromEntries(
Object.entries(maxSpeed).sort(([,a],[,b]) => a-b)
);
console.log(sortable);
We don’t want to duplicate the entire data structure, or use an array where we need an associative array.
Here’s another way to do the same thing as bonna:
var list = {“you”: 100, “me”: 75, “foo”: 116, “bar”: 15};
keysSorted = Object.keys(list).sort(function(a,b){return list[a]-list[b]})
console.log(keysSorted); // bar,me,you,foo