根据对象的属性删除数组的对象

Question

I have an Array like this: var obj = [{x:4, y:5}, {x:6, y:2}, ...] and I'm trying to delete one of the inside objects (properties) based on the x.

this is How I'm trying to do this:

 obj.forEach(function (child){
    if(child.x === 4){
      obj.destroy(child)
    }
 });

But it's not working and i get

obj.destroy is not a funtion

I also tried obj.splice(child) but it just mess up the array. so what am doing wrong here? Also is there a better way to do this by not having to loop through all of Array property every time?

Answer 1

You can just use filter on the array: eg

 let arrayToFilter = [ {x:4, y:5}, {x:6, y:2}]; const valueToFilter = 4; var filteredArray = arrayToFilter .filter((o) => { return ox !== valueToFilter; }); console.log(filteredArray); 

Answer 2

forEach() works on array.

If obj is an array, you can simply use filter() to remove the unwanted object from the array:

 var obj = [{x:4, y:5}, {x:6, y:2}] obj = obj.filter(c => cx !== 4) console.log(obj); 

Answer 3

You perhaps, have an array as obj because the one you posted in the question is simply invalid syntax.

Moreover, you can use Array#findIndex to get the index of the matching element first, and then splice that index from the array.

 var obj = [{x:4, y:5}, {x:6, y:2}]; var index = obj.findIndex(item => item.x === 4); obj.splice(index, 1); console.log(obj); 

Answer 4

i'm assuming your trying to filter out objects in an array which have an x that matches a given value. If thats the case, you should probably use the filter method.

So assuming thats what you mean you could do the following

obj = obj.filter(function (child){
if(child.x !== 4){
  return obj
}
});
// shorter
obj = obj.filter( child => child.x !== 4 );

In this case, only the objects which do not have the value of 4 will be available to you in the obj variable. And all other objects (assuming there are no other references in your code) will be garbage collected.