javascript - 比较数组并删除不匹配的值

Question

$scope.locations = [
    { name : "One"},
    { name : "Two"},
    { name : "Three"},
    { name : "India"},
    { name : "Japan"},
    { name : "China"}
];

$scope.tempLocations = [
    { name : "One"},
    { name : "Two"},
    { name : "global"},
];

我有两个数组。如果 location 不包含 tempLocations 中的某些名称，我想将它们从 tempLocation 中删除。在这种情况下，我想删除位置全局

我尝试了以下方法，但不起作用。

for(var i=0;i<$scope.tempLocations.length;i++){
    var index = $scope.tempLocations.indexOf($scope.locations[i]);
    if(index == -1){
        console.log($scope.tempLocations[i]);
        $scope.tempLocations.splice(i,1);
    }
}

Answer 1

我猜你正在找这个

$scope = {}

$scope.locations = [
    { name : "One"},
    { name : "Two"},
    { name : "Three"},
    { name : "India"},
    { name : "Japan"},
    { name : "China"}
];

$scope.tempLocations = [
    { name : "One"},
    { name : "Two"},
    { name : "global"},
];
$scope.tempLocations = $scope.tempLocations.filter(function(x) {
     return $scope.locations.some(function(y) {
          return x.name == y.name
     })
})

document.getElementById("output").innerHTML = JSON.stringify($scope.tempLocations, 0,' ');
console.log($scope);

<pre id="output"></pre>

如果您有许多(100+)位置，请考虑首先将它们转换为“关联数组”，例如:

validLocations = { "One": 1, "Two": 1 ... etc