Question

我想在点击按钮时打开只有一个窗口。即使我多次单击该按钮，我只想要一个窗口（刷新的窗口）。

这是我的代码：

CODE

<!DOCTYPE html>
<html>
<body>

<p>Click the button to open a new browser window.</p>

<button onclick="myFunction()">Try it</button>

<script>
function myFunction() {
    const width = 800;
    const height = 500;
    const left = window.screen.width/2 - width/2;
    const top = window.screen.height/2 - height/2;
    const windowFeatures = `width=${width},height=${height},status,resizable,left=${left},top=${top},screenX=${left},screenY=${top}`;

    window.open("https://www.w3schools.com","",windowFeatures);
}
</script>

</body>
</html>

请使用仅限Javascript

帮助我更新上述问题的代码

Answer 1

将新窗口对象存储在变量中，并在单击按钮时检查该值。如果定义刚刚重新加载该窗口，则打开新窗口。刷新新窗口仅在域名相同时才起作用，否则由于跨源框架而抛出错误DOMException。

＆＃13;

<!DOCTYPE html>
<html>

<body>

    <p>Click the button to open a new browser window.</p>

    <button onclick="myFunction()">Try it</button>

    <script>
        let newWindow;
        function myFunction() {
            if (newWindow) {
                newWindow.location.reload(true);
            } else {
                const width = 800;
                const height = 500;
                const left = window.screen.width / 2 - width / 2;
                const top = window.screen.height / 2 - height / 2;
                const windowFeatures = `width=${width},height=${height},status,resizable,left=${left},top=${top},screenX=${left},screenY=${top}`;

                newWindow = window.open("https://www.w3schools.com", "", windowFeatures);
            }
        }
    </script>

</body>

</html>

＆＃13;

如何只打开一个弹出窗口并将其限制为仅使用JavaScript（没有jQuery）？

1 个答案: