Using Array.map() on array of objects

Question

I am trying to use Array.map to slice the description property of each object within an array.

docs = docs.map(function(currentValue, index, array){
            currentValue['description'] = currentValue.description.slice(0,10);
            return array;
        });

When I console.log(docs) it appears as though it has worked. However, I can no longer access the properties for some reason.

console.log(docs[0].description); //undefined

It appears that I have turned my array of objects into an array of strings that appear to be objects. Any thoughts?

Answer 1

The callback in .map shouldn't return array -- it should return the new value you want that particular item in the array to have.

docs = docs.map(function(item){
  item.description = item.description.slice(0, 10);
  return item;
});

If all you're doing is transforming each item in the array, it would be more performant to use .forEach instead. .map creates a whole new array , whereas .forEach simply loops through the existing array. It also takes less code.

docs.forEach(function(item){
  item.description = item.description.slice(0, 10);
});

Answer 2

It is because you return array in map instead of currentValue . It should be

docs = docs.map(function(currentValue, index, array){
        currentValue['description'] = currentValue.description.slice(0,10);
        return currentValue;
    });

Answer 3

In that case, what you need to use is forEach() not map() as you are not doing any transformation of the items in the array

docs.forEach(function(currentValue, index, array){
    currentValue['description'] = currentValue.description.slice(0,10);
});

.map() is used to transform each item in an array and return a new object, since in your case you are just changing a property of each item, there is no need to use it.

Answer 4

docs = docs.map(function(currentValue, index, array){
        docs[index]['description'] = currentValue.description.slice(0,10);
        return array;
    });

I think it should be like this.