21

我有一个应用程序，在单击链接时会打开一个新窗口。这会生成一个包含 Java 小程序的页面。我遇到的问题是单击相同的链接会重新加载页面，从而重置 Java 应用程序。有什么办法可以捕获这个吗？可以接受的两种解决方案是：

  1. 允许从点击处理程序打开多个窗口
  2. 如果窗口已经打开，则忽略后续请求

为成为 Javascript 新手而道歉 - 这并不是我的主要工作。

附加到处理程序的代码是

function launchApplication(l_url, l_windowName)
{
  var l_width = screen.availWidth;
  var l_height = screen.availHeight;

  var l_params = 'status=1' +
                 ',resizable=1' +
                 ',scrollbars=1' +
                 ',width=' + l_width +
                 ',height=' + l_height +
                 ',left=0' +
                 ',top=0';

  winRef = window.open(l_url, l_windowName, l_params);
  winRef.moveTo(0,0);
  winRef.resizeTo(l_width, l_height);
}

编辑：

感谢您的回复 - 我稍微修改了建议，以便可以通过该功能打开多个 URL。

EDIT2：此代码的另一个版本在Check for a URL open on another window

var g_urlarray = [];

Array.prototype.has = function(value) {
    var i;
    for (var i in this) {
        if (i === value) {
            return true;
        }
    }
    return false;
};


function launchApplication(l_url, l_windowName)
{
  var l_width = screen.availWidth;
  var l_height = screen.availHeight;
  var winRef;

  var l_params = 'status=1' +
                 ',resizable=1' +
                 ',scrollbars=1' +
                 ',width=' + l_width +
                 ',height=' + l_height +
                 ',left=0' +
         ',top=0';
  if (g_urlarray.has(l_url)) {
    winRef = g_urlarray[l_url];
  }
  alert(winRef);
  if (winRef == null || winRef.closed) {
      winRef = window.open(l_url, l_windowName, l_params);
      winRef.moveTo(0,0);
      winRef.resizeTo(l_width, l_height);
      g_urlarray[l_url] = winRef;
  }
}
4

6 回答 6

20

我会这样做 - 基本上将所有引用的打开窗口存储在函数本身上。当函数触发时，检查窗口是否不存在或已关闭 - 如果是这样，启动弹出窗口。否则，请关注该请求的现有弹出窗口。

function launchApplication(l_url, l_windowName)
{
  if ( typeof launchApplication.winRefs == 'undefined' )
  {
    launchApplication.winRefs = {};
  }
  if ( typeof launchApplication.winRefs[l_windowName] == 'undefined' || launchApplication.winRefs[l_windowName].closed )
  {
    var l_width = screen.availWidth;
    var l_height = screen.availHeight;

    var l_params = 'status=1' +
                   ',resizable=1' +
                   ',scrollbars=1' +
                   ',width=' + l_width +
                   ',height=' + l_height +
                   ',left=0' +
                   ',top=0';

    launchApplication.winRefs[l_windowName] = window.open(l_url, l_windowName, l_params);
    launchApplication.winRefs[l_windowName].moveTo(0,0);
    launchApplication.winRefs[l_windowName].resizeTo(l_width, l_height);
  } else {
    launchApplication.winRefs[l_windowName].focus()
  }
}
于 2009-02-09T15:56:47.997 回答
20

您需要执行 2 次测试... 1 检查弹出窗口是否已定义，2 检查是否已关闭。

if(typeof(winRef) == 'undefined' || winRef.closed){
  //create new
  winRef = window.open(....);
} else {
  //it exists, load new content (if necs.)
  winRef.location.href = 'your new url';
  //give it focus (in case it got burried)
  winRef.focus();
}
于 2009-02-09T15:59:10.563 回答
5

您可以在打开新窗口的页面中使用类似的内容：

var newWindow = null;

function launchApplication()
{
  // open the new window only if newWindow is null (not opened yet)
  // or if it was closed
  if ((newWindow == null) || (newWindow.closed))
    newWindow = window.open(...);
}
于 2009-02-09T15:45:33.030 回答
3

工作代码

var newwindow = null;
function popitup(url) {
    if ((newwindow == null) || (newwindow.closed)) {
        newwindow=window.open(url,'Buy','width=950,height=650,scrollbars=yes,resizable=yes');
        newwindow.focus();
    } else {
        newwindow.location.href = url;
        newwindow.focus();    
    }
}
于 2012-05-06T01:21:51.613 回答
2

你可以这样检查：

if(!winref || winref.closed)
{
}
于 2009-02-09T15:46:35.947 回答
0

尝试检查：

if (!winref || winref.closed || !winref.document) { }

于 2011-11-05T02:35:41.670 回答