2017-05-27 53 views
1

我有一个代码片段角UI格:只显示所选对象

$scope.users = [ 
{ name: "abc", age: 10, location: 'Nagpur' }, 
{ name: "bcd", age: 30, location: 'Chennai' }, 
{ name: "efr", age: 29, location: 'Chennai' }, 
{ name: "abc", age: 25, location: 'Bangalore' }, 
{ name: "abc", age: 27, location: 'Vizag' } 
]; 
$scope.gridOptions.data = $scope.users; 

在电网如何显示仅具有名=“ABC”的用户?

+0

检查下面的答案,标记为接受,如果它帮助你 – Vivz

回答

0

你必须从你的阵列,它的名字过滤掉所有的值是“ABC”

angular.forEach($scope.users,function(val){ 
    if(val.name=='abc') 
    $scope.abcUsers.push(val); 
}); 

$scope.gridOptions.data = $scope.abcUsers; 
+0

尽管此代码可以回答这个问题,提供关于如何解决问题和/或为何解决问题的附加背景可以提高答案的长期价值。 – Badacadabra