初学者：如何在打开其他弹出窗口之前关闭弹出窗口 [英] beginner: how to close a pop up window before opening an other pop up
问题描述
你好，
我在页面上有几个链接。当一个人点击一个链接弹出一个窗口，然后打开一个图像。
当人们点击另一个链接我想要已经可用的弹出窗口
关闭之前
另一个弹出窗口打开。
i只是不能正确，我总是会收到一条错误消息：
行：23
字符：20
错误：对象不支持此属性或方法
带错误的行是：
vensters [x] .close（）;
这里是来源：
======================== source ====================
< html>
< head>
< script style =" text / javascript">
//''teller = 1''：没有弹出窗口打开;
//''teller = 0''：弹出窗口打开;
var beller;
teller = 0;
//带弹出窗口的数组;
var vensters = new Array（3 ）;
vensters [0] =" win1& ;;
vensters [1] =" win2";
vensters [2] =" win3";
函数openWindowPlaatje（num）{
var x;
//如果弹出窗口已经打开，请在打开新弹出窗口之前将其关闭
up;
if（teller == 1）{
for（x = 0; x< vensters.length; x ++）{
if（vensters [x]！= num）{
vensters [x] .close（）; //此行给出错误信息;
}
}
}
//打开选中的弹出窗口;
if（num ==" win1"）{
win1 = window.open（" plaatje1.html"，""， ''top = 250，left = 250，
width = 238，height = 255''）;
teller = 1;
}
if（num ==" win2"）{
win2 = window.open（" plaatje2.html"，" "，" top = 250，left = 250，width = 50，height = 50"）;
teller = 1;
}
if（num ==" win3"）{
win3 = window.open（" plaatje3.html"，""，" top = 250，左= 250，宽= 150，高= 150'）
;
出纳员= 1;
}
}
< / script>
< / head>
< body>
< a href =" javascript ：openWindowPlaatje（''win1''）" > plaatje 1< / a>< br>
< a href =" javascript ：openWindowPlaatje（''win2''）" > plaatje 2< / a>< br>
< a href =" javascript ：openWindowPlaatje（''win3''）" > plaatje 3< / a>< br>
< / body>
< / html>
================================================== ===
hello,
i have a few links on a page. when one clicks a link an pop up window with
an image opens.
when people click an other link i want the already availible pop up window
to close before
the other pop up window opens.
i just can''t get it right, i always get an error message:
line: 23
char: 20
error: object doesn''t support this property or method
the line with the error is:
vensters[x].close();
here is the source:
======================== source ====================
<html>
<head>
<script style="text/javascript">
// ''teller=1'': no pop up window open;
// ''teller=0'': pop up window open;
var teller;
teller = 0;
// array with pop up windows;
var vensters = new Array(3);
vensters[0] = "win1";
vensters[1] = "win2";
vensters[2] = "win3";
function openWindowPlaatje(num) {
var x;
// if an pop up is already open, close it before opening a new pop
up;
if(teller == 1){
for(x = 0; x < vensters.length; x++){
if(vensters[x] != num ){
vensters[x].close(); // this line gives error message;
}
}
}
// open the selected pop up;
if(num=="win1"){
win1=window.open("plaatje1.html","",''top=250,left= 250,
width=238,height=255'');
teller = 1;
}
if(num=="win2"){
win2=window.open("plaatje2.html","","top=250,left= 250,width=50,height=50");
teller = 1;
}
if(num=="win3"){
win3=window.open("plaatje3.html","","top=250,left= 250,width=150,height=150")
;
teller = 1;
}
}
</script>
</head>
<body>
<a href="javascript: openWindowPlaatje(''win1'')" >plaatje 1</a><br>
<a href="javascript: openWindowPlaatje(''win2'')" >plaatje 2</a><br>
<a href="javascript: openWindowPlaatje(''win3'')" >plaatje 3</a><br>
</body>
</html>
================================================== ===
推荐答案
Madame Blablavatsky写道：
Madame Blablavatsky wrote:
你好，
我有几个页面上的链接。当一个人点击链接弹出窗口时会打开一个图像。
当人们点击其他链接时，我希望已经可用的弹出窗口
关闭之前
另一个弹出窗口窗口打开。
hello,
i have a few links on a page. when one clicks a link an pop up window with
an image opens.
when people click an other link i want the already availible pop up window
to close before
the other pop up window opens.
更正：
你没有验证弹出关闭是否存在
if（teller == 1）{
for（x = 0; x< vensters.length; x ++）{
if（vensters [x] ！= num&& vensters [x]&&！vensters [x] .closed）{
vensters [x] .close（）; //这一行给出了错误信息;
}
}
}
最好的方法是所有弹出窗口有相同的名字（这里：''赢''）
< html>
< head>
< script style =" text / javascript">
var win = false;
function pop（page，attrib）{
//如果popup''win''存在或者没有关闭：关闭它
if（win || !! win.closed）win.close（）;
win = window.open（页面，''''，attrib）;
}
函数openWindowPlaatje（num） {
if（num == 1）
pop（" plaatje1.html"，" top = 250，left = 250，width = 238，he ight = 255）;
if（num == 2）
pop（" plaatje2.html"，" top = 250，left = 250，width = 50， hei ght = 50"）;
if（num == 3）
pop（" plaatje3.html"，" top = 250，left = 250，width = 150，他ight = 150）;
}
< / script>
< / head>
< body>
< a href =" plaatje1.html" target =" myWin"
onclick =" openWindowPlaatje（1）; return false; > plaatje 1< / a>< br>
< a href =" plaatje2.html" target =" myWin"
onclick =" openWindowPlaatje（2）; return false; > plaatje 2< / a>< br>
< a href =" plaatje3.html" target =" myWin"
onclick =" openWindowPlaatje（3）; return false; > plaatje 3< / a>< br>
< / body>
< / html>
如果javascript被禁用，
访问者将在新的正常窗口中获得正确的页面
-
Stephane Moriaux et son [moins] vieux Mac
correction :
you didn''t verify if popup to close was existing
if(teller == 1){
for(x = 0; x < vensters.length; x++){
if(vensters[x] != num && vensters[x] && !vensters[x].closed){
vensters[x].close(); // this line gives error message;
}
}
}
the best way would be all popups have same name (here : ''win'')
<html>
<head>
<script style="text/javascript">
var win=false;
function pop(page,attrib) {
// if popup ''win'' exists or ins''t closed : close it
if(win || !!win.closed) win.close();
win = window.open(page,'''',attrib);
}
function openWindowPlaatje(num) {
if(num==1)
pop("plaatje1.html","top=250,left=250,width=238,he ight=255");
if(num==2)
pop("plaatje2.html","top=250,left=250,width=50,hei ght=50");
if(num==3)
pop("plaatje3.html","top=250,left=250,width=150,he ight=150");
}
</script>
</head>
<body>
<a href="plaatje1.html" target="myWin"
onclick="openWindowPlaatje(1); return false;" >plaatje 1</a><br>
<a href="plaatje2.html" target="myWin"
onclick="openWindowPlaatje(2); return false;" >plaatje 2</a><br>
<a href="plaatje3.html" target="myWin"
onclick="openWindowPlaatje(3); return false;" >plaatje 3</a><br>
</body>
</html>
if javascript is disabled,
visitor will get the right page in a new normal window
--
Stephane Moriaux et son [moins] vieux Mac
感谢您的帮助，它现在正在工作，它应该工作
blablavatksy
thanks for your help, it is now working as it should
blablavatksy
函数pop（page，attrib）{
//如果弹出''win''存在或者ins 关闭它：关闭它
if（win || !! win.closed）win.close（）;
win = window.open（page，''''，attrib）;
}
function pop(page,attrib) {
// if popup ''win'' exists or ins''t closed : close it
if(win || !!win.closed) win.close();
win = window.open(page,'''',attrib);
}
你好Stephane Moriaux，
我发现它是一个非常聪明的解决方案，但我还有一个问题：
双感叹号是什么意思？
if（win || !! win.closed）win.close（）;
起初我认为这是一个错误，但当我omi它给了它一个
错误信息。
我可以知道它找到它的含义。
谢谢
Hello Stephane Moriaux,
I find it a very clever solution but i still have one question:
what is the meaning of the double exclamation mark?
if(win || !!win.closed) win.close();
At first I thought it to be a mistake, but when I omitted it, it gave an
error message.
I can knowwhere find the meaning of it.
thanks
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