对于对象:

<select ng-model="mySelect" ng-options="key as value for (key, value) in object"></select>

See ngOptions ngOptions(optional) – {comprehension_expression=} – in one of the following forms: For array data sources: label for value in array select as label for value in array label group by group for value in array select as label group by group for value in array track by trackexpr For object data sources: label for (key , value) in object select as label for (key , value) in object label group by group for (key, value) in object select as label group by group for (key, value) in object

对你来说，应该是

array = [{ "value": 1, "text": "1st" }, { "value": 2, "text": "2nd" }];

<select ng-options="obj.value as obj.text for obj in array"></select>

更新

随着AngularJS的更新，现在可以用track by表达式为select element的value属性设置实际值了。

<select ng-options="obj.text for obj in array track by obj.value">
</select>

如何记住这些丑陋的东西

对于那些很难记住这种语法形式的人:我同意这不是最简单或最漂亮的语法。这种语法是Python列表推导式的扩展版本，知道这一点有助于我很容易地记住语法。大概是这样的:

Python代码:

my_list = [x**2 for x in [1, 2, 3, 4, 5]]
> [1, 4, 9, 16, 25]

# Let people to be a list of person instances
my_list2 = [person.name for person in people]
> my_list2 = ['Alice', 'Bob']

这实际上与上面列出的第一个语法相同。但是，在<select>中，我们通常需要区分代码中的实际值和<select>元素中显示的文本(标签)。

比如，我们需要人手。Id，但我们不想把Id显示给用户;我们想要显示它的名字。同样，我们对代码中的person.name不感兴趣。这里有as关键字来标记东西。所以它变成了这样:

person.id as person.name for person in people

或者，代替person。Id我们可能需要person实例/引用本身。见下文:

person as person.name for person in people

对于JavaScript对象，同样的方法也适用。只需要记住，对象中的项是用(key, value)对解构的。

这个问题的正确答案由用户frm提供。Adiputra，因为目前这似乎是显式控制选项元素的value属性的唯一方法。

然而，我只是想强调，“select”在这里不是关键字，而只是表达式的占位符。请参考以下列表，“select”表达式的定义以及ng-options指令中可以使用的其他表达式。

问题中描述的select的用法:

ng-options='select p.text for p  in resultOptions'

本质上是错误的。

根据表达式列表，当在对象数组中给出选项时，似乎可以使用trackexpr来指定值，但它仅用于分组。

来自AngularJS的ng-options文档:

array / object: an expression which evaluates to an array / object to iterate over. value: local variable which will refer to each item in the array or each property value of object during iteration. key: local variable which will refer to a property name in object during iteration. label: The result of this expression will be the label for element. The expression will most likely refer to the value variable (e.g. value.propertyName). select: The result of this expression will be bound to the model of the parent element. If not specified, select expression will default to value. group: The result of this expression will be used to group options using the DOM element. trackexpr: Used when working with an array of objects. The result of this expression will be used to identify the objects in the array. The trackexpr will most likely refer to the value variable (e.g. value.propertyName).

在ng-options中选择一个项目可能有点棘手，这取决于您如何设置数据源。

在与它们斗争了一段时间后，我最终用我使用的最常用的数据源制作了一个样本。你可以在这里找到它:

http://plnkr.co/edit/fGq2PM?p=preview

现在要让ng-options工作，需要考虑以下几点:

Normally you get the options from one source and the selected value from other. For example: states :: data for ng-options user.state :: Option to set as selected Based on 1, the easiest/logical thing to do is to fill the select with one source and then set the selected value trough code. Rarely would it be better to get a mixed dataset. AngularJS allows select controls to hold more than key | label. Many online examples put objects as 'key', and if you need information from the object set it that way, otherwise use the specific property you need as key. (ID, CODE, etc.. As in the plckr sample) The way to set the value of the dropdown/select control depends on #3, If the dropdown key is a single property (like in all examples in the plunkr), you just set it, e.g.: $scope.dropdownmodel = $scope.user.state; If you set the object as key, you need to loop trough the options, even assigning the object will not set the item as selected as they will have different hashkeys, e.g.: for (var i = 0, len = $scope.options.length; i < len; i++) { if ($scope.options[i].id == savedValue) { // Your own property here: console.log('Found target! '); $scope.value = $scope.options[i]; break; } }

您可以在另一个对象$scope.myObject.myProperty中替换相同属性的savedValue。

对我来说，布鲁诺·戈麦斯的答案是最好的答案。

但实际上，您不需要担心设置select options的value属性。AngularJS会处理这个。让我详细解释一下。

请考虑这把小提琴

angular.module('mySettings', []).controller('appSettingsCtrl', function ($scope) {

    $scope.timeFormatTemplates = [{
        label: "Seconds",
        value: 'ss'
    }, {
        label: "Minutes",
        value: 'mm'
    }, {
        label: "Hours",
        value: 'hh'
    }];


    $scope.inactivity_settings = {
        status: false,
        inactive_time: 60 * 5 * 3, // 15 min (default value), that is, 900 seconds
        //time_format: 'ss', // Second (default value)
        time_format: $scope.timeFormatTemplates[0], // Default seconds object
    };

    $scope.activity_settings = {
        status: false,
        active_time: 60 * 5 * 3, // 15 min (default value), that is,  900 seconds
        //time_format: 'ss', // Second (default value)
        time_format: $scope.timeFormatTemplates[0], // Default seconds object
    };

    $scope.changedTimeFormat = function (time_format) {
        'use strict';

        console.log('time changed');
        console.log(time_format);
        var newValue = time_format.value;

        // do your update settings stuffs
    }
});

正如你在fiddle输出中看到的，无论你为选择框选项选择了什么，它都是你的自定义值，或者AngularJS自动生成的0、1、2值，这在你的输出中并不重要，除非你使用jQuery或任何其他库来访问选择组合框选项的值并相应地操作它。

开发人员使用ng-options总是很痛苦的。例如:在select标记中获取空/空的选定值。特别是在ng-options中处理JSON对象时，它变得更加乏味。这里我已经做了一些练习。

目的:通过ng-option迭代JSON对象数组，并设置选中的第一个元素。

数据:

someNames = [{"id":"1", "someName":"xyz"}, {"id":"2", "someName":"abc"}]

在选择标签中，我必须显示xyz和abc，其中xyz必须毫不费力地选择。

HTML:

<pre class="default prettyprint prettyprinted" style=""><code>
    &lt;select class="form-control" name="test" style="width:160px"    ng-options="name.someName for name in someNames" ng-model="testModel.test" ng-selected = "testModel.test = testModel.test || someNames[0]"&gt;
&lt;/select&gt;
</code></pre>

通过上面的代码示例，您可能会摆脱这种夸张。

另一个参考: