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Angularjs 如何从视图中有条件地为ViewModel特性赋值_Angularjs - Fatal编程技术网

Angularjs 如何从视图中有条件地为ViewModel特性赋值

Angularjs 如何从视图中有条件地为ViewModel特性赋值,angularjs,Angularjs,我有以下HTML,它在满足ng show条件时显示错误消息。但是,无论是否显示错误消息,都会发生ng init赋值 <div ng-show="peopleworksForm.submitted && formData['DisclosureQuestion'+($index+1)+'Answer'] == 'true' && formData['DisclosureQuestion'+($index+1)+'OptionalAnswer'] == unde

我有以下HTML,它在满足
ng show
条件时显示错误消息。但是,无论是否显示错误消息,都会发生
ng init
赋值

<div ng-show="peopleworksForm.submitted && formData['DisclosureQuestion'+($index+1)+'Answer'] == 'true' && formData['DisclosureQuestion'+($index+1)+'OptionalAnswer'] == undefined" 
     ng-init="$parent.answerValFailed = true">
   invalid value
</div>

无效值

我想要的是,只有在满足
ng show
条件后,才能在
ng init
内执行操作。有没有一种简单的方法可以做到这一点?

如果使用ng,则在其中使用ng init:)


无效值

希望有帮助:)

与问题无关,只是一个建议:不要在标记中进行“计算”。创建一个函数,例如,
canShow()
shouldShow()
,并将该逻辑放入控制器中。
    <div class="pw-form-inline-error" ng-if="peopleworksForm.submited && formData['DisclosureQuestion'+($index+1)+'Answer'] == 'true' && formData['DisclosureQuestion'+($index+1)+'OptionalAnswer'] == undefined">
            <span ng-init="$parent.answerValFailed = true">
                            invalid value
              </span>
    </div>