Question

我有一个JSON对象，如下所示

[{
"id": 1,
"firstName": "Sam",
"middleName": "poller",
"lastName": "Aniston",
"address": "New York City",
}, {
"id": 2,
"firstName": "Apple",
"middleName": null,
"lastName": "Jolie",
"address": "Beverley Hills",
}, {
"id": 3,
"firstName": "Anna",
"middleName": "mary",
"lastName": "Dobrev",
"address": "London",
}]

我使用select as，

在视图中填充此数据

  <div >
  <select  ng-model="invigilator" ng-options="invigilator.id as  (invigilator.firstName+' '+invigilator.middleName+' '+invigilator.lastName) for invigilator in invigilatorList"  ng- click="getinvigilator(invigilator)" class="form-control"> 
                                <option value="">Select Invigilator</option>

                            </select></div>

但我得到选项，

Sam poller Aniston
Apple null Jolie
Anna mary dobrev

如何从中间名中删除该null并仅显示Apple Jolie。

Answer 1

使用ternary operator：

<select ng-options="… as (invigilator.firstName + ' ' + (invigilator.middleName !== null ? (invigilator.middleName + ' ') : '') + invigilator.lastName) for …"></select>

由于表达式变得非常复杂，您还可以（并且最好）在函数中移动此代码：

<select ng-options="… as getFullName(invigilator) for …"></select>

// In the controller
$scope.getFullName = function (invigilator) {
    if (invigilator.middleName === null) {
        return invigilator.firstName + ' ' + invigilator.lastName;
    }

    return invigilator.firstName + ' ' + invigilator.middleName + ' ' + invigilator.lastName;
};

Answer 2

使用array.join [invigilator.firstName,invigilator.middleName,invigilator.lastName].join(' ')的简单方法。这样，如果其值为null，则无需担心任何值。

示例演示：http://plnkr.co/edit/JTczvhMaNEeujrWnothP?p=preview

  <select ng-model="invigilator" ng-options="invigilator.id as  [invigilator.firstName,invigilator.middleName,invigilator.lastName].join(' ') for invigilator in data">
    <option value="">Select Invigilator</option>
  </select>

从angularjs中删除null选择选项

2 个答案: