如何更新数组中对象的所有值？

Question

How do I change all salary values and increase them by a percentage, when each salary is a property of an object, with each object being stored in an array? Eg increasing by 10 percent and I have to round the result up to the nearest integer:

raiseSalary([
    { name: "Ali", salary: 3000 },
    { name: "Rob", salary: 2000 },
    { name: "Adam", salary: 4500 },
], 10)

The above call should return:

[
   { name: 'Ali', salary: 3300 },
   { name: 'Rob', salary: 2200 }, 
   { name: 'Adam', salary: 4950 }
]

This is the code that I have written:

function raiseSalary(arr, raise) {
    if (arr.length === 0) {
        return [{}];
    } else {
        const raiseArray = arr.map((salaryObj) => {
            return (salaryObj.salaryObj / 100) * 10;
        });
    }
}

Answer 1

The map method should do what you want. Note the destructuring of salaryObj to avoid manipulating the original object.

const percent = 10;
const multiplier = 1 + (percent / 100);
const salaries = [
  { name: "Ali", salary: 3000 },
  { name: "Rob", salary: 2000 },
  { name: "Adam", salary: 4500 },
];
const newSalaries = salaries.map((salaryObj) => {
  return {
    ...salaryObj,
    salary: salaryObj.salary * multiplier,
  };
});

console.log(newSalaries);

Answer 2

Presented below is one possible way to achieve the desired objective.

Code Snippet

 const addRaise = (arr, pcnt) => ( arr.map( ({ salary, ...rest }) => ({ ...rest, salary: ( +salary * (+pcnt + 100) / 100 ) }) ) ); /* // method to raise salary by "pcnt" percent const addRaise = (arr, pcnt) => ( // iterate over array "arr" arr.map( // destructure to access "salary" ({ salary, ...rest }) => ({ ...rest, // all other props retained as-is // salary updated to be 10% more than current salary: ( +salary * (+pcnt + 100) / 100 ) }) ) ); */ const dataArr = [{ name: "Ali", salary: 3000 }, { name: "Rob", salary: 2000 }, { name: "Adam", salary: 4500 }, ]; console.log( 'adding raise of 10%...\n', 'new array:\n', addRaise(dataArr, 10) );

Explanation

Comments added to the snippet above.

Answer 3

You're almost there. Just a few points

• Just as you return something when the array has zero elements, you have to return something when it has elements; in this case the modified array
• Retain the objects by using their keys and providing the data, otherwise, you end up with an array of just numbers.
• Since your aim is to raise salary by raise then instead of multiplying by the raise , in this case 10 , multiply by 100+raise
• You do not have to hard-code raise since you're passing it to the function; it allows you to pass 5 , or 15 or some other raise value without changing your function.

 const input = [{ name: "Ali", salary: 3000 }, { name: "Rob", salary: 2000 }, { name: "Adam", salary: 4500 }], increment = 10; function raiseSalary(arr, raise) { if (arr.length === 0) { return []; } else { return arr.map((salaryObj) => { return ({name:salaryObj.name, salary:(salaryObj.salary / 100) * (100+raise)}); }); } } console.log( raiseSalary(input, increment) );

Alternatively ...

Just return the modified array without testing for elements, and you can also use object destructuring.

 const input = [{ name: "Ali", salary: 3000 }, { name: "Rob", salary: 2000 }, { name: "Adam", salary: 4500 }], increment = 10; function raiseSalary(arr, raise) { return arr.map(({name,salary}) => { return ({name, salary: salary/100*(100+raise)}); }); } console.log( raiseSalary(input, increment) ); console.log( raiseSalary([], increment) );

Answer 4

Here's a simple approach with map and Object spreading:

 const updateSalaries = (people, pct) => people .map (({salary, ...rest}) => ({ ...rest, salary: Math .round (salary * (100 + pct) / 100) })) const people = [{name: "Ali", salary: 3000}, {name: "Rob", salary: 2000}, {name: "Adam", salary: 4500}] console .log (updateSalaries (people, 10))

 .as-console-wrapper {max-height: 100% !important; top: 0}

For each element, we destructure the input to extract the salary from the rest of the object, and then create a new result containing the rest, but with a salary calculated from the original salary and the percentage adjustment.

Answer 5

let userDetail=[
        { name: "Ali", salary: 3000 },
        { name: "Rob", salary: 2000 },
        { name: "Adam", salary: 4500 },
    ];
function getPercentage(num, percentage){
   return num * (percentage / 100);
}
userDetail=userDetail.map(x=> {
    return{
        ...x,
        salary:x.salary+getPercentage(x.salary,10)
    }
        })
console.log(userDetail)