Question

我有问题。我分配给两个按钮有两个功能，如果单击则会弹出每个按钮。在任何时候，一次只会弹出一个（或者我想要的）。这是我到目前为止所做的：

var ADD_TESTER_POPUP = null;
var DELETE_TESTER_POPUP = null;

// user clicks on Add Tester
function openAddNewTesterWindow(width, height)
{
    //if no windows are open
    if((ADD_TESTER_POPUP===null || ADD_TESTER_POPUP.closed) && 
       (DELETE_TESTER_POPUP===null || DELETE_TESTER_POPUP.closed)) 
    {
        var left = (screen.width/2)-(width/2);
        var top = (screen.height/3)-(height/3);
        ADD_TESTER_POPUP = window.open("addNewTesterWindow.html", "", "scrollbars=1,
             resizable=1, width="+width+", height="+height+", top="+top+", left="+left);
    }
    else
    {
        //error occurs at this if statement line only
        //if add tester is already open
        if((!ADD_TESTER_POPUP===null || !ADD_TESTER_POPUP.closed) && 
           (DELETE_TESTER_POPUP===null || DELETE_TESTER_POPUP.closed))
        {
            ADD_TESTER_POPUP.focus();
        }
        //if delete tester is open
        else if((ADD_TESTER_POPUP===null || ADD_TESTER_POPUP.closed) && 
                (!DELETE_TESTER_POPUP===null || !DELETE_TESTER_POPUP.closed)) 
        {
            alert("Please close Delete Tester Window before adding tester.");
            DELETE_TESTER_POPUP.focus();
        }
    }
} // openAddnNewTesterWindow(width, height)

// user clicks on Delete Tester
function openDeleteTesterWindow(width, height)
{
    //if no windows are open
    if ((ADD_TESTER_POPUP===null || ADD_TESTER_POPUP.closed) && 
        (DELETE_TESTER_POPUP===null || DELETE_TESTER_POPUP.closed)) 
    {
        var left = (screen.width/2)-(width/2);
        var top = (screen.height/3)-(height/3);
        DELETE_TESTER_POPUP = window.open("deleteTesterWindow.html", "", "scrollbars=1,
            resizable=1, width="+width+", height="+height+", top="+top+", left="+left);
    }
    else
    {
        //if delete tester is already open
        if((ADD_TESTER_POPUP===null || ADD_TESTER_POPUP.closed) && 
           (!DELETE_TESTER_POPUP===null || !DELETE_TESTER_POPUP.closed)) 
        {
            DELETE_TESTER_POPUP.focus();
        }
        //if add tester is already open
        else if((!ADD_TESTER_POPUP===null || !ADD_TESTER_POPUP.closed) && 
                (DELETE_TESTER_POPUP===null || DELETE_TESTER_POPUP.closed)) 
        {
            alert("Please close Add Tester Window before deleting tester.");
            ADD_TESTER_POPUP.focus();
        }
    }
} // openDeleteTesterWindow(width, height)

当“添加新测试器”窗口打开第一个时，如果用户尝试单击“删除测试器”按钮，则会提醒用户。当Delete Tester窗口打开时，如果用户尝试单击Add New Tester按钮，它将提醒用户。

但是，如果我打开Delete Tester窗口第一个（刷新/清理页面），然后单击Add New Tester按钮，则会出现错误，指示Uncaught TypeError: Cannot read property 'closed' of null（这是使用Chrome的调试工具）。我在第一个函数中指出了发生错误的代码。对此有何帮助？

Answer 1

找到答案。这是一个逻辑错误，必须检查是否首先打开弹出窗口。只需在if语句中的else if和else之间进行简单的切换即可解决问题。

var ADD_TESTER_POPUP = null;
var DELETE_TESTER_POPUP = null;

// user clicks on Add Tester
function openAddNewTesterWindow(width, height)
{
    //if no windows are open
    if((ADD_TESTER_POPUP===null || ADD_TESTER_POPUP.closed) && 
       (DELETE_TESTER_POPUP===null || DELETE_TESTER_POPUP.closed)) 
    {
        var left = (screen.width/2)-(width/2);
        var top = (screen.height/3)-(height/3);
        ADD_TESTER_POPUP = window.open("addNewTesterWindow.html", "", "scrollbars=1,
             resizable=1, width="+width+", height="+height+", top="+top+", left="+left);
    }
    else
    {
        //if delete tester is open
        if((ADD_TESTER_POPUP===null || ADD_TESTER_POPUP.closed) && 
                (!DELETE_TESTER_POPUP===null || !DELETE_TESTER_POPUP.closed)) 
        {
            alert("Please close Delete Tester Window before adding tester.");
            DELETE_TESTER_POPUP.focus();
        }
        //if add tester is already open
        else if((!ADD_TESTER_POPUP===null || !ADD_TESTER_POPUP.closed) && 
           (DELETE_TESTER_POPUP===null || DELETE_TESTER_POPUP.closed))
        {
            ADD_TESTER_POPUP.focus();
        }
    }
} // openAddnNewTesterWindow(width, height)

// user clicks on Delete Tester
function openDeleteTesterWindow(width, height)
{
    //if no windows are open
    if ((ADD_TESTER_POPUP===null || ADD_TESTER_POPUP.closed) && 
        (DELETE_TESTER_POPUP===null || DELETE_TESTER_POPUP.closed)) 
    {
        var left = (screen.width/2)-(width/2);
        var top = (screen.height/3)-(height/3);
        DELETE_TESTER_POPUP = window.open("deleteTesterWindow.html", "", "scrollbars=1,
            resizable=1, width="+width+", height="+height+", top="+top+", left="+left);
    }
    else
    {
        //if add tester is already open
        if((!ADD_TESTER_POPUP===null || !ADD_TESTER_POPUP.closed) && 
                (DELETE_TESTER_POPUP===null || DELETE_TESTER_POPUP.closed)) 
        {
            alert("Please close Add Tester Window before deleting tester.");
            ADD_TESTER_POPUP.focus();
        }
        //if delete tester is already open
        else if((ADD_TESTER_POPUP===null || ADD_TESTER_POPUP.closed) && 
           (!DELETE_TESTER_POPUP===null || !DELETE_TESTER_POPUP.closed)) 
        {
            DELETE_TESTER_POPUP.focus();
        }
    }
} // openDeleteTesterWindow(width, height)

弹出窗口的空值

1 个答案: