I need to extract a value from this array and use it inside a variable. but then if I log array I see the user one now is Julian. How do I avoid it?
let array = [{'id':'1','name':'User one','phone': '+51 111 222 333'},'id':'2','name':'User two','phone': '+51 111 222 333'}, 'id':'3','name':'User Three','phone': '+51 111 222 333'}]; I do:
let user = array[0]; user.name = "Julian" You have to copy the object so it has it's own reference in memory so you can use Object.assign to create a new Object
let array = [{'id':'1','name':'User one','phone': '+51 111 222 333'},{'id':'2','name':'User two','phone': '+51 111 222 333'},{ 'id':'3','name':'User Three','phone': '+51 111 222 333'}]; const ob=Object.assign({},array[0]) ob.name="Julian" console.log(array)alternatively you can use the spread operator
let array = [{'id':'1','name':'User one','phone': '+51 111 222 333'},{'id':'2','name':'User two','phone': '+51 111 222 333'},{ 'id':'3','name':'User Three','phone': '+51 111 222 333'}]; const ob={...array[0]} ob.name="Julian" console.log(array) console.log(ob)
78:21 error "ob" is not defined no-undef
You can use the spread operator ... to create a "duplicate" of the object that can be modified without changing the original.
Remember that array contains references to objects. By doing let user = array[0], you are assigning user to the same reference as array[0]. Both point to the same object, and changes done through one can be seen with the other. However, by using ..., you can put all the key-value pairs from the object stored in array[0] into a new object that can be modified independently.
let array = [ {'id':'1', 'name':'User one', 'phone': '+51 111 222 333' }, {'id':'2', 'name':'User two', 'phone': '+51 111 222 333' }, { 'id':'3', 'name':'User Three', 'phone': '+51 111 222 333' } ] let user = {...array[0]} user.name = "Julian" console.log(user) console.log(array[0])
