Question

我必须在phonegap应用程序中显示动态弹出窗口。我使用Jquery mobile和javascript使弹出窗口动态化，但弹出窗口在整个应用程序中都有白色背景。请帮我以正确的形式得到它。 Html代码：

<div id="popupMsgPage" data-role="popup" data-close-btn="right" data-theme="c" data-overlay-theme="c">
        <div data-role="header" data-position="fixed" >
            <h1 id="popup-head">Delete Page?</h1>
        </div>
        <div role="main" class="ui-content">
            <h3 class="ui-title" id="popup-title">Are you sure you want to delete this page?</h3>
            <p id="popup-content">This action cannot be undone.</p>
            <a href="#" data-rel="back" data-rel-count="1" data-role="button" >Continue</a>
        </div>

Jquery代码：

 function show(head, title, msg) {
        debugger;
        head = head ? head : "";
        title = title ? title : "";
        msg = msg ? msg : "";
        jPages["popupMsg"].find("#popup-head").html(head);
        jPages["popupMsg"].find("#popup-title").html(title);
        jPages["popupMsg"].find("#popup-content").html(msg);
        $.mobile.changePage("#popupMsgPage", {
            role: "dialog"

        });

Answer 1

我认为你正在混合弹出窗口小部件和对话框页面。如果您不想更改页面，请使用如下弹出窗口小部件：

function show(head, title, msg) {
    head = head ? head : "";
    title = title ? title : "";
    msg = msg ? msg : "";
    $("#popupMsgPage").find("#popup-head").html(head);
    $("#popupMsgPage").find("#popup-title").html(title);
    $("#popupMsgPage").find("#popup-content").html(msg);

    $("#popupMsgPage").popup("open");
}

正在使用 DEMO

Jquery移动对话框，无需更改页面

1 个答案: