我可以使用 Javascript 来检测用户是否在另一个打开的选项卡上，而不是我的网站吗？

Question

我想知道用户当前是否在另一个打开的浏览器选项卡上处于活动状态，并且目前没有在我的网站上进行交互。这样我就可以用Javascript把小说HTML写成DOM.

Answer 1

可见度 API - MDN

来自 MDN 的示例

// Set the name of the hidden property and the change event for visibility
var hidden, visibilityChange;
if (typeof document.hidden !== "undefined") { // Opera 12.10 and Firefox 18 and later support
  hidden = "hidden";
  visibilityChange = "visibilitychange";
} else if (typeof document.msHidden !== "undefined") {
  hidden = "msHidden";
  visibilityChange = "msvisibilitychange";
} else if (typeof document.webkitHidden !== "undefined") {
  hidden = "webkitHidden";
  visibilityChange = "webkitvisibilitychange";
}

var videoElement = document.getElementById("videoElement");

// If the page is hidden, pause the video;
// if the page is shown, play the video
function handleVisibilityChange() {
  if (document[hidden]) {
    videoElement.pause();
  } else {
    videoElement.play();
  }
}

// Warn if the browser doesn't support addEventListener or the Page Visibility API
if (typeof document.addEventListener === "undefined" || hidden === undefined) {
  console.log("This demo requires a browser, such as Google Chrome or Firefox, that supports the Page Visibility API.");
} else {
  // Handle page visibility change
  document.addEventListener(visibilityChange, handleVisibilityChange, false);

  // When the video pauses, set the title.
  // This shows the paused
  videoElement.addEventListener("pause", function(){
    document.title = 'Paused';
  }, false);

  // When the video plays, set the title.
  videoElement.addEventListener("play", function(){
    document.title = 'Playing';
  }, false);
}

我可以使用 Javascript 来检测用户是否在另一个打开的选项卡上，而不是我的网站吗？

Can I use Javascript do detect whether a user is on another open tab, rather than my website?

javascript

browser