Given the array, how do I filter out the array matching layer.id?
nodes = [
{id: '1', layer: {id: 'a'}},
{id: '4', layer: {id: 'b'}},
{id: '2'},
{id: '3'},
]
argument = 'a'
// OUTPUT
[
{id: '4', layer: {id: 'b'}},
{id: '2'},
{id: '3'},
]
This statement somehow clear out 2 and 3 elements from the array because I'm checking the node.layer field.
nodes.filter(node => node.layer && node.layer.id !== id) // TO-DO: How do I flip..?
.map(node => node.id);
Invert the test for node.layer and use || instead of &&.
const nodes = [
{id: '1', layer: {id: 'a'}},
{id: '4', layer: {id: 'b'}},
{id: '2'},
{id: '3'},
];
argument = 'a';
const result = nodes.filter(node => !node.layer || node.layer.id != argument).map(({id}) => id);
console.log(result);
You've already accurately described why your first approach doesn't work. To overcome the problem, step back and realise that you need to keep those elements which either have no layer or have a layer with the same id from the one specified.
This leads to:
1 Given the array, how do I filter out the array matching layer.id?
nodes = [
{id: '1', layer: {id: 'a'}},
{id: '4', layer: {id: 'b'}},
{id: '2'},
{id: '3'},
]
argument = 'a'
console.log(nodes.filter(node => !node.layer || node.layer.id !== argument))(I've left out your use of .map because then the output doesn't match what you showed as desired - but you can put that in too if you prefer.)
I think that you need to add one more condition to the .filter method, like this:
var output = nodes.filter(n => {
if ('layer' in n) {
return n.layer.id !== argument;
}
return n;
});
The return n; will return the 2 and 3 elements in case layer prop doesn't exist.
