$filter with custom function in angular - angularjs

Question

How I can create a $filter with custom function to determining if match ?
This is a json sample with structure:
$scope.routes =[
{
"id": 0,
"name": "Rosa Barnes",
"origin": [
{
"address": [
{
"locality": "Madrid",
"country": "ES"
}
]
}
]
},
{
"id": 1,
"name": "Wright Montoya",
"origin": [
{
"address": [
{
"locality": "London",
"country": "UK"
}
]
}
]
},
{
"id": 2,
"name": "Pearson Johns",
"origin": [
{
"address": [
{
"locality": "London",
"country": "UK"
}
]
}
]
}
];
I want a $filter that passing one country, match in origin.address.country , is possible?
I proved this code but not works:
$scope.routesToShow = $filter('filter')($scope.routes, {origin.address.country: "UK"});
Here there are a demo:
http://jsfiddle.net/86U29/43/
Thanks

Answer 1

What you need is a custom filter. It also looks like you might need to tweak your data structure too.
Take a look at this:
app.filter('CustomFilter', function () {
function parseString(propertyString) {
return propertyString.split(".");
}
function getValue(element, propertyArray) {
var value = element;
propertyArray.forEach(function (property) {
value = value[property];
});
return value;
}
return function (input, propertyString, target) {
var properties = parseString(propertyString);
return input.filter(function (item) {
return getValue(item, properties) == target;
});
}
});
You could use this filter, like this:
$scope.routesToShow = $filter('CustomFilter')($scope.routes, 'origin.address.country', 'UK');
Here is an updated jsfiddle (notice the updated data structure): http://jsfiddle.net/moderndegree/86U29/44/
You can find more information on creating custom filters here:
https://docs.angularjs.org/guide/filter#creating-custom-filters

Answer 2

Simply.
var filter = function(obj){
return obj.origin[0].address[0].country === 'UK';
}
$scope.routesToShow = $filter('filter')($scope.routes, filter);