如何使用JS检查某人当前是否在网站上（在线）？

发布于 2025-02-12 19:46:21 字数 631 浏览 1 评论 0原文

我正在建立一个带有聊天等的网站等。问题是我无法正确定义某人当前在网站上，这是因为我无法设置他们的活动。

我写了一个函数，该函数由html文件中的身体标签调用＆lt; body onload =“ changestatus（true）” onunload =“ changestatus（false）（false）”＆gt;，但正如我所说的那样，这似乎不是工作。

当第一个用户的状态设置为在线值时，第二个用户的状态立即更改为离线。当我尝试在网站上的输入中键入某些内容时，它也将状态更改为离线，并不重要。

我目前正在考虑仅使用会议就将此部分写在PHP中，但是也许有人知道如何在JS中执行此操作？

function changeStatus(event) {
  if (event) {
    db.ref("status/" + username + "/" + receiver).set({
      act: "online"
    });
  } else {
    db.ref("status/" + username + "/" + receiver).set({
      act: "offline"
    });
  }
}

原文

I am building a website with chat etc. The problem is I cannot correctly define if someone is currently on the website which cause I cannot set their activity.

I wrote a function which is invoked by the body tag in HTML file <body onload="changeStatus(true)" onunload="changeStatus(false)"> but as I said it seems not to work.

When the status of the first users sets to the online value, the second one immediately changes to offline. When I try to type something in the input on the website it also changes the status to offline, it doesn't matter who is typing.

I am currently thinking of writing this part in PHP just using the session, but maybe someone knows how to do this in JS?

function changeStatus(event) {
  if (event) {
    db.ref("status/" + username + "/" + receiver).set({
      act: "online"
    });
  } else {
    db.ref("status/" + username + "/" + receiver).set({
      act: "offline"
    });
  }
}

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终弃我 2025-02-19 19:46:21

听起来您正在尝试构建一个存在系统，在这种情况下，我建议您在管理在场，如何OnDisconnect working，检测连接状态 href =“ https://firebase.google.com/docs/database/web/web/offline-capabilities#section-sample-sample” rel =“ nofollow noreferrer”>示例存在app app 。

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彡翼 2025-02-19 19:46:21

您是否正在尝试确定页面是否已加载？如果是这样，那么一个简单的jQuery文档加载将执行

https：// 。

  <script src="https://code.jquery.com/jquery-1.9.1.min.js"></script>
    <script>
    $( window ).on( "load", function() {
        console.log( "window loaded" );
    });
//or
$(function() {
    console.log( "ready!" );
});
    </script>

https://api.jquery.com/mouseover/

https://api.jquery.com/click/

我希望这会有所帮助

Are you trying to determine if the page has loaded? If so then a simple jquery document load on will do that

https://learn.jquery.com/using-jquery-core/document-ready/

  <script src="https://code.jquery.com/jquery-1.9.1.min.js"></script>
    <script>
    $( window ).on( "load", function() {
        console.log( "window loaded" );
    });
//or
$(function() {
    console.log( "ready!" );
});
    </script>

if your trying to capture if the user is on the page then mouse over or mouse click might be better

https://api.jquery.com/mouseover/

https://api.jquery.com/click/

I hope this helps

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