Question

尝试在设定的时间值后显示居中的弹出窗口，如下所示：

<script type="text/javascript">
    <!--
    function popup(url) {
    var width = 300;
    var height = 200;
    var left = (screen.width - width) / 2;
    var top = (screen.height - height) / 2;
    var params = 'width=' + width + ', height=' + height;
    params += ', top=' + top + ', left=' + left;
    params += ', directories=no';
    params += ', location=no';
    params += ', menubar=no';
    params += ', resizable=no';
    params += ', scrollbars=no';
    params += ', status=no';
    params += ', toolbar=no';
    newwin = window.open(url, 'windowname5', params);
    if (window.focus) {
        newwin.focus()
    }
    setTimeout(popup('`test/login.html'), 5000);
    return false;
}

//-->
//]]>
</script>

但弹出窗口从不显示。如果我这样写：

    <script type="text/javascript">
        <!--
        function popup(url) {
            var width = 300;
            var height = 200;
            var left = (screen.width - width) / 2;
            var top = (screen.height - height) / 2;
            var params = 'width=' + width + ', height=' + height;
            params += ', top=' + top + ', left=' + left;
            params += ', directories=no';
            params += ', location=no';
            params += ', menubar=no';
            params += ', resizable=no';
            params += ', scrollbars=no';
            params += ', status=no';
            params += ', toolbar=no';
            newwin = window.open(url, 'windowname5', params);
            if (window.focus) {
                newwin.focus()
            }

            return false;

        }
        setTimeout(popup('`test/login.html'), 5000);
        //-->
        //]]>
    </script>

弹出窗口立即显示，但随后我收到无效参数错误。建议？

Answer 1

setTimout将函数作为其第一个参数。 popup(...)不返回函数。试试这个：

setTimeout(function() {popup('`test/login.html');}, 5000)

Answer 2

在第一个示例中，您将setTimeout放在popup()函数的正文中。在调用函数之前，该函数的主体不会运行。因此，您在第二次尝试将setTimeout置于popup()身体之外时是正确的。

另一个问题是您需要将a function or code传递给setTimeout()：

var timeoutID = window.setTimeout(func, delay, [param1, param2, ...]);
var timeoutID = window.setTimeout(code, delay);

像这样：

setTimeout(function() {
    popup('`test/login.html'); // provided that tick-mark is right...
}, 5000);

虽然现有方法会立即执行popup()，但不会为timout函数提供任何功能：

setTimeout(popup('`test/login.html'), 5000);

使用setTimeout延迟居中弹出Javascript

2 个答案: