我有一个函数displaySelectedRole(),我有变量$scope.Role和$scope.rolenames .I需要从$scope.rolename.中删除$scope.role中可用的所有值
$scope.role= ["A","B","C"];
$scope.rolename =["A","B","C","D","E"]我需要拼接这些值并得到$scope.rolename = ["D","E"]
$scope.displaySelectedRole = function(role, index) {
debugger;
$scope.role.splice(RoleNames[index]);
console.log($scope.role);我尝试使用基于索引的拼接,但问题是它在控制台中给出了空数组值。
发布于 2016-10-28 09:29:25
您可以使用差(),它使用一种方法来减除数组:
$scope.role = ["A","B","C"];
$scope.rolename = ["A","B","C","D","E"];
$scope.diff = _.difference($scope.rolename, $scope.role); // ["D","E"]发布于 2016-10-28 09:34:03
您可以使用过滤器
var $scope = {}; // Ignore this line
$scope.role= ["A","B","C"];
$scope.rolename = ["A","B","C","D","E"];
$scope.rolename = $scope.rolename.filter(function(role){
return $scope.role.indexOf(role) === -1;
})
console.log($scope.rolename);
如果您想直接删除它们,可以遍历$scope.role并使用剪接
var $scope = {}; // Ignore this line
$scope.role= ["A","B","C"];
$scope.rolename = ["A","B","C","D","E"];
$scope.role.forEach(function(role){
var index = $scope.rolename.indexOf(role);
if(index !== -1) $scope.rolename.splice(index, 1);
})
console.log($scope.rolename);
注意: Array.filter将返回一个新数组,而array.splice将修改原始数组。
参考文献
https://stackoverflow.com/questions/40302038
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