在AngularJS指令中转换属性

Question

我正在创建一个选择替换指令,以便根据设计轻松选择样式,而不必总是对一堆标记(即指令为您做到!).

我没有意识到属性不会转移到你放置的位置ng-transclude,只是转到根元素.

我在这里有一个例子:http://plnkr.co/edit/OLLntqMzbGCJS7g7h1j4？p = preview

你可以看到它看起来很棒......但是有一个主要的缺陷.将id和name属性不被转移.其中,你知道,没有name,它没有发布到服务器(这种形式绑定到现有系统,所以AJAXing模型不是一个选项).

例如,这是我开始的:

<select class="my-select irrelevant-class" name="reason" id="reason" data-anything="banana">
    <option value="">Reason for Contact...</option>
    <option>Banana</option>
    <option>Pizza</option>
    <option>The good stuff</option>
    <option>This is an example of a really, really, really, really, really, really long option item</option>
</select>

......这就是我想要的样子:

<div class="faux-select" ng-class="{ placeholder: default == viewVal, focus: obj.focus }">
    <span class="faux-value">{{viewVal}}</span>
    <span class="icon-arrow-down"></span>
    <select ng-model="val" ng-focus="obj.focus = true" ng-blur="obj.focus = false" ng-transclude class="my-select irrelevant-class" name="reason" id="reason" data-anything="banana">
        <option value="">Reason for Contact...</option>
        <option>Banana</option>
        <option>Pizza</option>
        <option>The good stuff</option>
        <option>This is an example of a really, really, really, really, really, really long option item</option>
    </select>
</div>

......但这是实际发生的事情:

<div class="faux-select my-select irrelevant-class" ng-class="{ placeholder: default == viewVal, focus: obj.focus }" name="reason" id="reason" data-anything="banana">
    <span class="faux-value">{{viewVal}}</span>
    <span class="icon-arrow-down"></span>
    <select ng-model="val" ng-focus="obj.focus = true" ng-blur="obj.focus = false" ng-transclude>
        <option value="">Reason for Contact...</option>
        <option>Banana</option>
        <option>Pizza</option>
        <option>The good stuff</option>
        <option>This is an example of a really, really, really, really, really, really long option item</option>
    </select>
</div>

具体来说,问题是nameselect 上没有属性,所以它实际上并没有将数据发送到服务器.

显然,我可以使用预编译阶段来传输name和id属性(这就是我现在正在做的事情),但如果它只是自动传输所有属性以便它们可以添加任何类,任意数据, (ng-)必需,(ng-)禁用属性等等.

我试着transclude: 'element'开始工作,但后来我无法将模板中的其他属性放到它上面.

注意,我在这里看到了帖子:如何将其转换为属性？,但看起来他们只是手动传输数据,我的目标是让它自动传输所有属性.

Answer 1

您可以使用compile函数来访问元素的属性并构建模板.

app.directive('mySelect', [function () {
  return {
    transclude: true,
    scope: true,
    restrict: 'C',
    compile: function (element, attrs) {
        var template = '<div class="faux-select" ng-class="{ placeholder: default == viewVal, focus: obj.focus }">' +
                       '<span class="faux-value">{{viewVal}}</span>' +
                       '<span class="icon-arrow-down entypo-down-open-mini"></span>' +
                       '<select id="' + attrs.id + '" name="' + attrs.name  + '" ng-model="val" ng-focus="obj.focus = true" ng-blur="obj.focus = false" ng-transclude>';
                       '</select>' +
                     '</div>';
        element.replaceWith(template);

        //return the postLink function 
        return function postLink(scope, elem, attrs) {
          var $select = elem.find('select');
          scope.default = scope.viewVal = elem.find('option')[0].innerHTML;

          scope.$watch('val', function(val) {
              if(val === '') scope.viewVal = scope.default;
              else scope.viewVal = val;
          });

          if(!scope.val) scope.val = $select.find('option[selected]').val() || '';
        }
     }
  };
}]);

编译函数返回postLink函数,还有其他方法可以执行此操作,您可以在此处找到更多信息.

这是一个掠夺者