Remove elements from an array of items

Question

Remove elements from an array of items

Object {Results:Array[3]}
Results:Array[3]
[0-2]
    0:Object
           id=1     
           name: "Rick"
           upper:"0.67"
    1:Object
           id=2     
           name:'david'
           upper:"0.46"
    2:Object
           id=3     
           name:'ashley'
           upper:null

In the code snippet above, there is an array of objects displayed along with a variable called delete_id.

delete_id = 1,2

The delete_id variable holds values that correspond to the ids of objects that need to be deleted from the array. The expected output after deletion is shown below:

Object {Results:Array[1]}
Results:Array[3]
[0]
    0:Object
           id=3     
           name:'ashley'
           upper:null

I am seeking assistance on how to achieve this functionality. I have attempted using the following function, but it only deletes the first id in the delete_id variable. For example, if delete_id = 2,3 then only 2 will be deleted. However, I want both 2 and 3 to be removed.

function removeID(delete_id) {
    tabledata = tabledata.filter(function (obj) {
        return delete_id.indexOf(obj.id);
    });

javascript arrays loops object underscore.js

Answer 1

Answer №1

To convert the string delete_id into an array of numeric IDs, you can utilize the functions .split() and .map(). Afterwards, employ .filter() to perform the necessary cleanup.

var users = [
  {
    id: 1,
    name: "Alice",
    score: "0.75"
  }, {
    id: 2,
    name: "Bob",
    score: "0.60"
  }, {
    id: 3,
    name: "Charlie",
    score: null
  }
];

var delete_id = "1,2";

var exclude = delete_id.split(',').map(Number);

users = users.filter(function(user) {
  return exclude.indexOf(user.id) == -1;
});

console.log(users);

Answer 2

To convert the string delete_id into an array of numeric IDs, you can utilize the functions .split() and .map(). Afterwards, employ .filter() to perform the necessary cleanup.

var users = [
  {
    id: 1,
    name: "Alice",
    score: "0.75"
  }, {
    id: 2,
    name: "Bob",
    score: "0.60"
  }, {
    id: 3,
    name: "Charlie",
    score: null
  }
];

var delete_id = "1,2";

var exclude = delete_id.split(',').map(Number);

users = users.filter(function(user) {
  return exclude.indexOf(user.id) == -1;
});

console.log(users);

Answer 3

Answer №2

function searchObject(arr, val, key) {
            var result = arr.filter(function (object) {
                if (object[key] === val || object[key] == val)
                    return object;
            });
            return result;
        };

To begin with, locate the specific object in the array(tabledata) where value=1(delete_id), and the key in JSON is 'id'

    var foundObject = searchObject(tabledata, delete_id, 'id');

Retrieve the index of that object in the array

    var indexOfObject = tabledata.indexOf(foundObject[0]);

Remove the object from the array based on its index

    tabledata.splice(indexOfObject, 1);

Answer 4

function searchObject(arr, val, key) {
            var result = arr.filter(function (object) {
                if (object[key] === val || object[key] == val)
                    return object;
            });
            return result;
        };

To begin with, locate the specific object in the array(tabledata) where value=1(delete_id), and the key in JSON is 'id'

    var foundObject = searchObject(tabledata, delete_id, 'id');

Retrieve the index of that object in the array

    var indexOfObject = tabledata.indexOf(foundObject[0]);

Remove the object from the array based on its index

    tabledata.splice(indexOfObject, 1);

Answer 5

Answer №3

To successfully execute the code, it is necessary to modify the removeID function so that it checks if the index is equal to -1.

function removeID(delete_id) {
  tabledata = tabledata.filter(function(obj) {
    return delete_id.indexOf(obj.id) === -1;  //ensure index is compared with -1
  });
}
var tabledata = [{
  id: 1,
  name: "Rick",
  upper: "0.67"
}, {
  id: 2,
  name: 'david',
  upper: "0.46"
}, {
  id: 3,
  name: 'ashley',
  upper: null
}];

var ids = [1,2];  //defined as an array for clarity
removeID(ids);
console.log(tabledata);

Answer 6

To successfully execute the code, it is necessary to modify the removeID function so that it checks if the index is equal to -1.

function removeID(delete_id) {
  tabledata = tabledata.filter(function(obj) {
    return delete_id.indexOf(obj.id) === -1;  //ensure index is compared with -1
  });
}
var tabledata = [{
  id: 1,
  name: "Rick",
  upper: "0.67"
}, {
  id: 2,
  name: 'david',
  upper: "0.46"
}, {
  id: 3,
  name: 'ashley',
  upper: null
}];

var ids = [1,2];  //defined as an array for clarity
removeID(ids);
console.log(tabledata);

Answer 7

Answer №4

Assuming delete_id is an array of integers, you can use the following code to filter out ids that are present in delete_id.

This code snippet will create a new array called filtered which will only contain objects from tabledata where the id is not found in delete_id. The original tabledata will remain unaffected.

var filtered = tabledata.filter(function(obj) {
   return !(obj.id in delete_id)
})

Answer 8

Assuming delete_id is an array of integers, you can use the following code to filter out ids that are present in delete_id.

This code snippet will create a new array called filtered which will only contain objects from tabledata where the id is not found in delete_id. The original tabledata will remain unaffected.

var filtered = tabledata.filter(function(obj) {
   return !(obj.id in delete_id)
})

Remove elements from an array of items

Answer №1

Answer №2

Answer №3

Answer №4

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