Question

我的控制器中有这样的代码：

var forms = $('form'); // my selector is a bit more complex.
var form = forms[0];

这给了我表格。从这个我怎么得到FormController，以便我可以查看$ pristine，$ invalid等属性？

var formController = // how to get this from form?
var ispristine = formController.$pristine;

我实际上有多个子表单，我必须得到它们的原始状态。

由于

Answer 1

我认为你可以这样做来获取表单的“FormController”属性：

    // Get hold of the form in your DOM
    var forms = $('form');
    var form = forms[0];

    // Get hold of the form's scope object, which is injected with an object containing these properties.
    // Note that 'formName is the name of the form element in html. (<form name='formName'>....</form>
    var formScope = $(form).scope().formName;
    var ispristine = formScope.$pristine;

请注意，仅当表单元素具有名称属性时才有效！

获取表单的FormController

1 个答案: