獲得一個形式

Question

的的FormController我有這樣的代碼在我的控制器：

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

這使我的形式。從這我怎麼弄到FormController，以便我可以檢查像$ pristine，$無效等屬性？

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

我實際上有多個子窗體，我必須得到它們的原始狀態。

感謝

Answer 1

我想，你可以這樣做是爲了獲得一種形式的「的FormController」性質的保持：

// 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; 

請注意，這隻有當表單元素具有名稱屬性！