How can I remove an object from an array?
I wish to remove the object that includes name Kristian from someArray. For example:
someArray = [{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"}];
I want to achieve:
someArray = [{name:"John", lines:"1,19,26,96"}];
You can use several methods to remove item(s) from an Array:
//1 someArray.shift(); // first element removed //2 someArray = someArray.slice(1); // first element removed //3 someArray.splice(0, 1); // first element removed //4 someArray.pop(); // last element removed //5 someArray = someArray.slice(0, a.length - 1); // last element removed //6 someArray.length = someArray.length - 1; // last element removed
If you want to remove element at position x, use:
someArray.splice(x, 1);
Or
someArray = someArray.slice(0, x).concat(someArray.slice(-x));
Reply to the comment of @chill182: you can remove one or more elements from an array using Array.filter, or Array.splice combined with Array.findIndex (see MDN), e.g.
// non destructive filter > noJohn = John removed, but someArray will not change
let someArray = getArray();
let noJohn = someArray.filter( el => el.name !== "John" );
log("non destructive filter > noJohn = ", format(noJohn));
log(`**someArray.length ${someArray.length}`);
// destructive filter/reassign John removed > someArray2 =
let someArray2 = getArray();
someArray2 = someArray2.filter( el => el.name !== "John" );
log("", "destructive filter/reassign John removed > someArray2 =",
format(someArray2));
log(`**someArray2.length ${someArray2.length}`);
// destructive splice /w findIndex Brian remains > someArray3 =
let someArray3 = getArray();
someArray3.splice(someArray3.findIndex(v => v.name === "Kristian"), 1);
someArray3.splice(someArray3.findIndex(v => v.name === "John"), 1);
log("", "destructive splice /w findIndex Brian remains > someArray3 =",
format(someArray3));
log(`**someArray3.length ${someArray3.length}`);
// Note: if you're not sure about the contents of your array,
// you should check the results of findIndex first
let someArray4 = getArray();
const indx = someArray4.findIndex(v => v.name === "Michael");
someArray4.splice(indx, indx >= 0 ? 1 : 0);
log("", "check findIndex result first > someArray4 (nothing is removed) > ",
format(someArray4));
log(`**someArray4.length (should still be 3) ${someArray4.length}`);
function format(obj) {
return JSON.stringify(obj, null, " ");
}
function log(...txt) {
document.querySelector("pre").textContent += `${txt.join("\n")}\n`
}
function getArray() {
return [ {name: "Kristian", lines: "2,5,10"},
{name: "John", lines: "1,19,26,96"},
{name: "Brian", lines: "3,9,62,36"} ];
}
<pre> **Results** </pre>
The clean solution would be to use Array.filter:
var filtered = someArray.filter(function(el) { return el.Name != "Kristian"; });
The problem with this is that it does not work on IE < 9. However, you can include code from a Javascript library (e.g. underscore.js) that implements this for any browser.
I recommend using lodash.js or sugar.js for common tasks like this:
// lodash.js
someArray = _.reject(someArray, function(el) { return el.Name === "Kristian"; });
// sugar.js
someArray.remove(function(el) { return el.Name === "Kristian"; });
in most projects, having a set of helper methods that is provided by libraries like these is quite useful.
How about this?
$.each(someArray, function(i){
if(someArray[i].name === 'Kristian') {
someArray.splice(i,1);
return false;
}
});
Your “array” as shown is invalid JavaScript syntax. Curly brackets {} are for objects with property name/value pairs, but square brackets [] are for arrays – like so:
someArray = [{name:"Kristian", lines:"2,5,10"}, {name:"John", lines:"1,19,26,96"}];
In that case, you can use the .splice() method to remove an item. To remove the first item (index 0), say:
someArray.splice(0,1);
// someArray = [{name:"John", lines:"1,19,26,96"}];
If you don’t know the index but want to search through the array to find the item with name “Kristian” to remove you could to this:
for (var i =0; i < someArray.length; i++)
if (someArray[i].name === "Kristian") {
someArray.splice(i,1);
break;
}
EDIT: I just noticed your question is tagged with “jQuery”, so you could try the $.grep() method:
someArray = $.grep(someArray,
function(o,i) { return o.name === "Kristian"; },
true);
ES2015
let someArray = [
{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"},
{name:"Kristian", lines:"2,58,160"},
{name:"Felix", lines:"1,19,26,96"}
];
someArray = someArray.filter(person => person.name != 'John');
It will remove John!
You could use array.filter().
e.g.
someArray = [{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"}];
someArray = someArray.filter(function(returnableObjects){
return returnableObjects.name !== 'Kristian';
});
//someArray will now be = [{name:"John", lines:"1,19,26,96"}];
Arrow functions:
someArray = someArray.filter(x => x.name !== 'Kristian')
I have made a dynamic function takes the objects Array, Key and value and returns the same array after removing the desired object:
function removeFunction (myObjects,prop,valu)
{
return myObjects.filter(function (val) {
return val[prop] !== valu;
});
}
Full Example: DEMO
var obj = {
"results": [
{
"id": "460",
"name": "Widget 1",
"loc": "Shed"
}, {
"id": "461",
"name": "Widget 2",
"loc": "Kitchen"
}, {
"id": "462",
"name": "Widget 3",
"loc": "bath"
}
]
};
function removeFunction (myObjects,prop,valu)
{
return myObjects.filter(function (val) {
return val[prop] !== valu;
});
}
console.log(removeFunction(obj.results,"id","460"));
This is a function that works for me:
function removeFromArray(array, value) {
var idx = array.indexOf(value);
if (idx !== -1) {
array.splice(idx, 1);
}
return array;
}
You could also try doing something like this:
var myArray = [{'name': 'test'}, {'name':'test2'}];
var myObject = {'name': 'test'};
myArray.splice(myArray.indexOf(myObject),1);
someArray = jQuery.grep(someArray , function (value) {
return value.name != 'Kristian';
});
Use splice function on arrays. Specify the position of the start element and the length of the subsequence you want to remove.
someArray.splice(pos, 1);
Vote for the UndercoreJS for simple work with arrays.
_.without() function helps to remove an element:
_.without([1, 2, 1, 0, 3, 1, 4], 0, 1);
=> [2, 3, 4]
With ES 6 arrow function
let someArray = [
{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"}
];
let arrayToRemove={name:"Kristian", lines:"2,5,10"};
someArray=someArray.filter((e)=>e.name !=arrayToRemove.name && e.lines!= arrayToRemove.lines)
Simplest solution would be to create a map that stores the indexes for each object by name, like this:
//adding to array
var newPerson = {name:"Kristian", lines:"2,5,10"}
someMap[ newPerson.name ] = someArray.length;
someArray.push( newPerson );
//deleting from the array
var index = someMap[ 'Kristian' ];
someArray.splice( index, 1 );
Although this is probably not that appropriate for this situation I found out the other day that you can also use the delete keyword to remove an item from an array if you don’t need to alter the size of the array e.g.
var myArray = [1,2,3]; delete myArray[1]; console.log(myArray[1]); //undefined console.log(myArray.length); //3 - doesn't actually shrink the array down
If you want to remove all occurrences of a given object (based on some condition) then use the javascript splice method inside a for the loop.
Since removing an object would affect the array length, make sure to decrement the counter one step, so that length check remains intact.
var objArr=[{Name:"Alex", Age:62},
{Name:"Robert", Age:18},
{Name:"Prince", Age:28},
{Name:"Cesar", Age:38},
{Name:"Sam", Age:42},
{Name:"David", Age:52}
];
for(var i = 0;i < objArr.length; i ++)
{
if(objArr[i].Age > 20)
{
objArr.splice(i, 1);
i--; //re-adjust the counter.
}
}
The above code snippet removes all objects with age greater than 20.
This answer
for (var i =0; i < someArray.length; i++)
if (someArray[i].name === "Kristian") {
someArray.splice(i,1);
}
is not working for multiple records fulfilling the condition. If you have two such consecutive records, only the first one is removed, and the other one skipped. You have to use:
for (var i = someArray.length - 1; i>= 0; i--) ...
instead .
There seems to be an error in your array syntax so assuming you mean an array as opposed to an object, Array.splice is your friend here:
someArray = [{name:"Kristian", lines:"2,5,10"}, {name:"John", lines:"1,19,26,96"}];
someArray.splice(1,1)
Use javascript’s splice() function.
This may help: http://www.w3schools.com/jsref/jsref_splice.asp
You can use map function also.
someArray = [{name:"Kristian", lines:"2,5,10"},{name:"John",lines:"1,19,26,96"}];
newArray=[];
someArray.map(function(obj, index){
if(obj.name !== "Kristian"){
newArray.push(obj);
}
});
someArray = newArray;
console.log(someArray);
You could also use some:
someArray = [{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"}];
someArray.some(item => {
if(item.name === "Kristian") // Case sensitive, will only remove first instance
someArray.splice(someArray.indexOf(item),1)
})
This is what I use.
Array.prototype.delete = function(pos){
this[pos] = undefined;
var len = this.length - 1;
for(var a = pos;a < this.length - 1;a++){
this[a] = this[a+1];
}
this.pop();
}
Then it is as simple as saying
var myArray = [1,2,3,4,5,6,7,8,9]; myArray.delete(3);
Replace any number in place of three. After the expected output should be:
console.log(myArray); //Expected output 1,2,3,5,6,7,8,9
I guess the answers are very branched and knotted.
You can use the following path to remove an array object that matches the object given in the modern JavaScript jargon.
coordinates = [
{ lat: 36.779098444109145, lng: 34.57202827508546 },
{ lat: 36.778754712956506, lng: 34.56898128564454 },
{ lat: 36.777414146732426, lng: 34.57179224069215 }
];
coordinate = { lat: 36.779098444109145, lng: 34.57202827508546 };
removeCoordinate(coordinate: Coordinate): Coordinate {
const found = this.coordinates.find((coordinate) => coordinate == coordinate);
if (found) {
this.coordinates.splice(found, 1);
}
return coordinate;
}
splice(i, 1) where i is the incremental index of the array will remove the object. But remember splice will also reset the array length so watch out for ‘undefined’. Using your example, if you remove ‘Kristian’, then in the next execution within the loop, i will be 2 but someArray will be a length of 1, therefore if you try to remove “John” you will get an “undefined” error. One solution to this albeit not elegant is to have separate counter to keep track of index of the element to be removed.
Returns only objects from the array whose property name is not “Kristian”
var noKristianArray = $.grep(someArray, function (el) { return el.name!= "Kristian"; });
var someArray = [
{name:"Kristian", lines:"2,5,10"},
{name:"John", lines:"1,19,26,96"},
{name:"Kristian", lines:"2,58,160"},
{name:"Felix", lines:"1,19,26,96"}
];
var noKristianArray = $.grep(someArray, function (el) { return el.name!= "Kristian"; });
console.log(noKristianArray);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
Here is an example with map and splice
const arrayObject = [
{ name: "name1", value: "value1" },
{ name: "name2", value: "value2" },
{ name: "name3", value: "value3" },
];
let index = arrayObject.map((item) => item.name).indexOf("name1");
if (index > -1) {
arrayObject.splice(index, 1);
console.log("Result", arrayObject);
}
This Concepts using Kendo Grid
var grid = $("#addNewAllergies").data("kendoGrid");
var selectedItem = SelectedCheckBoxList;
for (var i = 0; i < selectedItem.length; i++) {
if(selectedItem[i].boolKendoValue==true)
{
selectedItem.length= 0;
}
}
If you wanna access and remove object of an array simply you can try something like this.
// inside some function
let someArray = [ {"ColumnName" : "a", "PropertySerno" : 100005,"UpdateType" : 1},
{"ColumnName" : "b", "PropertySerno" : 100202,"UpdateType" : 1,
"ShowRemoveButton" : true} ];
for (let item of someArray) {
delete item.ShowRemoveButton;
}
console.log(item.outputMappingData.Data);
//output will be like that = [ {"ColumnName" : "a", "PropertySerno" : 100005,"UpdateType" : 1},
// {"ColumnName" : "b", "PropertySerno" : 100202,"UpdateType" : 1 }];
