javascript - 停止在额外点击时重新加载子页面

Question

单击按钮时，将打开一个子窗口。然后，如果再次单击它，它将重新加载该页面。即使重新加载父页面，我也试图避免这种情况发生。只要子页面“测试”打开，我希望点击事件中的条件不对子窗口执行任何操作并发出警告。这可能吗？谢谢

$('#send').click(function(){

   if(*child page already open*){
    alert('already open');
 }else{
        window.open("test.html","Ratting","width=550,height=170,0,status=0,");
      }
});

Answer 1

您需要测试子窗口是否存在 并且仍然打开：

var childWindow;

$('#send').click(function(){
    if (childWindow && !childWindow.closed) {
        alert('already open');
    } else {
        childWindow = window.open("test.html","Ratting","width=550,height=170,0,status=0,");
    }
});

Answer 2

var childPage = null;
$('#send').click(function(){
  if(childPage && !childPage.closed){
    alert('already open');
  } else{
    childPage = window.open("test.html","Ratting","width=550,height=170,0,status=0,");
  }
});