knockout.js - knockout 未在单选按钮上设置初始真值

Question

名称+复选框的单向绑定(bind)工作正常，但它最初不适用于单选按钮employeeTypeA，尽管它在viewmodel中的值为true，html显示单选按钮未设置，为什么会这样？

   <script type="text/javascript">

        $(function()
        {
            var PersonViewModel = function()
            {
                this.firstName = ko.observable('Lisa');
                this.lastName = ko.observable('T');
                this.isCustomer = ko.observable(true);
                this.employeeTypeA = ko.observable(true);
                this.employeeTypeB = ko.observable(false);
            };

            var personViewModel = new PersonViewModel();
            ko.applyBindings(personViewModel, $('data').get(0));
        });
    </script>

    <div id="data">
        <span data-bind="text: firstName"></span>
        <span data-bind="text: lastName"></span>
        <input type="checkbox" data-bind="checked: isCustomer" title="Is a customer" />
        <input name="x" type="radio" data-bind="checked: employeeTypeA" title="Employee type A" />
        <input name="x" type="radio" data-bind="checked: employeeTypeB" title="Employee type B" />
    </div>

Answer 1

checked从文档中，单选按钮的绑定(bind)工作方式不同:

For radio buttons, KO will set the element to be checked if and only if the parameter value equals the radio button node’s value attribute.

所以你需要改变你的 PersonViewModel像这样:

var PersonViewModel = function()
{
    this.firstName = ko.observable('Lisa');
    this.lastName = ko.observable('T');
    this.isCustomer = ko.observable(true);
    this.employeeType = ko.observable('TypeB');                
};

还有你的单选按钮:

<input name="x" type="radio" data-bind="checked: employeeType" 
       value="TypeA" title="Employee type A" />
<input name="x" type="radio" data-bind="checked: employeeType" 
       value="TypeB" title="Employee type B" />

演示 JSFiddle.

如果您想保留 employeeTypeA和 employeeTypeB您可以引入一个返回类型的计算属性:

this.employeeTypeA = ko.observable(false);
this.employeeTypeB = ko.observable(true);
this.employeeType = ko.computed(function()
{
     if (this.employeeTypeA())
        return 'TypeA';
     if (this.employeeTypeB())
        return 'TypeB';
},this);  

同样在这种情况下，您需要添加 value单选按钮上的属性。

演示 JSFiddle.