我有这个对象列表 -
var o_list = [{id:1,name:jo},{id:2,name:mo}];
我想提取这样的东西 -
var list = [1,2]
是否已经构建了任何内容或者我们如何实现它。
答案 0 :(得分:2)
你可以映射它
var o_list = [{id:1,name:"jo"},{id:2,name:"mo"}];
var list = o_list.map(x => x.id);
document.body.innerHTML = '<pre>' + JSON.stringify(list, 0, 4) + '</pre>';
答案 1 :(得分:2)
您可以使用过滤器执行此操作 https://docs.angularjs.org/api/ng/filter/filter
var myApp = angular.module('myApp', []);
myApp.controller("myCtrl", ['$scope', '$filter',
function($scope, $filter) {
$scope.values = [{
id: 1,
name: 'asdas'
}, {
id: 2,
name: 'blabla'
}];
// this is how you use filters in your script
console.log($filter('extract')($scope.values));
}
]);
myApp.filter('extract', function() {
return function(input) {
return input.map(x => x.id);
}
});
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script>
<div ng-app="myApp">
<div ng-controller="myCtrl">
<ol ng-repeat="id in values | extract">
<li ng-bind="id"></li>
</ol>
</div>
</div>
答案 2 :(得分:0)
您只能使用javscript forEach
数组方法来获得所需的结果
var o_list = [{id:1,name:'jo'},{id:2,name:'mo'}];
var list =[];
o_list.forEach(function(item){
list.push(item.id)
})
console.log(list)
答案 3 :(得分:0)
from nltk import wordnet as wn
def split_word(self, word):
result = list()
while(len(word) > 2):
i = 1
found = True
while(found):
i = i + 1
synsets = wn.synsets(word[:i])
for s in synsets:
if edit_distance(s.name().split('.')[0], word[:i]) == 0:
found = False
break;
result.append(word[:i])
word = word[i:]
print(result)
答案 4 :(得分:0)
你可以使用这个简单的for循环
var o_list = [{id:1,name:'jo'},{id:2,name:'mo'}];
var s=[];
for (var i=0;i< o_list.length;i++){
s.push(o_list[i].id);
}
答案 5 :(得分:0)
替代你可以使用underscore.js每种方法来获得结果
var o_list = [{id:1,name:'jo'},{id:2,name:'mo'}];
var list =[];
_.each(o_list, function(d){
list.push(d.id)
});