小编Kev*_*ann的帖子

我似乎无法弄清楚如何从联系人那里获取用户"我的卡".我正在使用swift开发本机Mac应用程序.

日期选择器似乎只在第一次点击时工作,然后它不会弹出.我猜它与某些交叉变量或函数定义有关,但我无法弄清楚如何修复它.

这是代码:http: //plnkr.co/edit/ridTspMBHE1iphrSobQr？p = preview

HTML

<div ng-controller="ModalDemoCtrl">
    <button class="btn btn-info" ng-click="open_modal()">Edit</button>

    <script type="text/ng-template" id="notificationInput.html">
        <div class="modal-header">
            <h3 class="modal-title">My Modal</h3>
        </div>
        <div class="modal-body">
            <form role="form">
                <div class="form-group">
                    <label for="n_title">Title</label>
                    <input type="text" class="form-control" id="n_title" value="{{selectedNotification.title}}">
                </div>

                <div class="form-group">
                    <label for="n_release">Release</label>
                    <p class="input-group">
                        <input type="text" id="n_release" class="form-control" datepicker-popup="{{format}}" ng-model="dt" is-open="opened" min-date="minDate" max-date="'2015-06-22'" datepicker-options="dateOptions" date-disabled="disabled(date, mode)" ng-required="true" close-text="Close" />
                        <span class="input-group-btn">
                            <button type="button" class="btn btn-default" ng-click="open($event)"><i class="glyphicon glyphicon-calendar"></i></button>
                        </span>
                    </p>
                </div>
            </form>

        </div>
        <div class="modal-footer">
            <button class="btn btn-primary" ng-click="ok()">OK</button>
            <button class="btn …

有没有什么方法可以在静态函数中包含动态变量,因为你不能在动态函数中使用"this".

我想做什么:

public static function convertToDynamicString(pString:String):String
{
    if(pString == "" || pString == null) return "";
    var re:RegExp = /(\{\w+\})/;
    var results:Array = pString.split(re);
    var dynamicString:String = "";
    for each(var pWord:String in results)
    {
        if(pWord.substr(0, 1) == "{") dynamicString += this[pWord.substring(1, (pWord.length - 1))];    
        else dynamicString += pWord;
    }
    return dynamicString;
}

问题:

此["变量名称"]在静态函数中不起作用