Question 258845 in Javascript Question posted in
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Filter array objects javascript

I want to filter an array of objects based on a specific condition. Specifically, I want to exclude the object with an ID equal to a given value and all objects that appear before the first object that doesn’t satisfy this condition. How can I achieve this using a single loop?

Here’s an example array:

const arr = [{ id: 10 }, { id: 20 }, { id: 30 }, { id: 40 }, { id: 50 }, { id: 60 }];

Let’s say I have a target ID:

const myID = 25;

I want to filter the array to get the following result:

const result = [{ id: 20 }, { id: 30 }, { id: 40 }, { id: 50 }, { id: 60 }];
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3

Answers


  1. 0

    You can achieve this using a combination of filter and find array methods. Here’s how you can do it:

    const arr = [{id: 10}, {id: 20}, {id: 30}, {id: 40}, {id: 50}, {id: 60}];
const myID = 25;

const result = arr.filter(item => item.id > myID || item.id === arr.find(i => i.id !== myID).id);

console.log(result);

    This will give you the desired result:

    [
  { id: 20 },
  { id: 30 },
  { id: 40 },
  { id: 50 },
  { id: 60 }
]
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  2. 0 
    const arr = [{id: 10}, {id: 20}, {id: 30}, {id: 40}, {id: 50}, {id: 60}]

const myID = 25;

// const result = [{id: 20}, {id: 30}, {id: 40}, {id: 50}, {id: 60}]

let result = [];
let lessOne ={id : Number.MIN_VALUE} ;

for(let obj of arr) {
  if(obj.id > myID) {
    result.push(obj)
  } else {
    if(obj.id > lessOne.id) {
      lessOne = obj;
    }
  }
}

result.unshift(lessOne);


console.log(result);
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  3. 0 
    const arr = [{ id: 10 }, { id: 20 }, { id: 30 }, { id: 40 }, { id: 50 }, { id: 60 }];
const myID = 25;

const index = arr.findIndex(obj => obj.id !== myID);
const result = index !== -1 ? arr.slice(index) : [];

console.log(result);
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