我们如何通过javascript窗口打开一个弹出窗口的中心。打开功能上的屏幕中心变量，以当前选择的屏幕分辨率?
当前回答
由于在多显示器设置中确定当前屏幕中心的复杂性，一个更简单的选择是将弹出窗口置于父窗口的中央。简单地将父窗口作为另一个参数传递:
function popupWindow(url, windowName, win, w, h) {
const y = win.top.outerHeight / 2 + win.top.screenY - ( h / 2);
const x = win.top.outerWidth / 2 + win.top.screenX - ( w / 2);
return win.open(url, windowName, `toolbar=no, location=no, directories=no, status=no, menubar=no, scrollbars=no, resizable=no, copyhistory=no, width=${w}, height=${h}, top=${y}, left=${x}`);
}
实现:
popupWindow('google.com', 'test', window, 200, 100);
其他回答
试着这样做:
function popupwindow(url, title, w, h) {
var left = (screen.width/2)-(w/2);
var top = (screen.height/2)-(h/2);
return window.open(url, title, 'toolbar=no, location=no, directories=no, status=no, menubar=no, scrollbars=no, resizable=no, copyhistory=no, width='+w+', height='+h+', top='+top+', left='+left);
}
由于在多显示器设置中确定当前屏幕中心的复杂性，一个更简单的选择是将弹出窗口置于父窗口的中央。简单地将父窗口作为另一个参数传递:
function popupWindow(url, windowName, win, w, h) {
const y = win.top.outerHeight / 2 + win.top.screenY - ( h / 2);
const x = win.top.outerWidth / 2 + win.top.screenX - ( w / 2);
return win.open(url, windowName, `toolbar=no, location=no, directories=no, status=no, menubar=no, scrollbars=no, resizable=no, copyhistory=no, width=${w}, height=${h}, top=${y}, left=${x}`);
}
实现:
popupWindow('google.com', 'test', window, 200, 100);
下面是上述解决方案的另一个版本……
const openPopupCenter = (url, title, w, h) => {
const getSpecs = (w, h, top, left) => {
return `scrollbars=yes, width=${w}, height=${h}, top=${top}, left=${left}`;
};
const getFirstNumber = (potentialNumbers) => {
for(let i = 0; i < potentialNumbers.length; i++) {
const value = potentialNumbers[i];
if (typeof value === 'number') {
return value;
}
}
};
// Fixes dual-screen position
// Most browsers use window.screenLeft
// Firefox uses screen.left
const dualScreenLeft = getFirstNumber([window.screenLeft, screen.left]);
const dualScreenTop = getFirstNumber([window.screenTop, screen.top]);
const width = getFirstNumber([window.innerWidth, document.documentElement.clientWidth, screen.width]);
const height = getFirstNumber([window.innerHeight, document.documentElement.clientHeight, screen.height]);
const left = ((width / 2) - (w / 2)) + dualScreenLeft;
const top = ((height / 2) - (h / 2)) + dualScreenTop;
const newWindow = window.open(url, title, getSpecs(w, h, top, left));
// Puts focus on the newWindow
if (window.focus) {
newWindow.focus();
}
return newWindow;
}
我的ES6 JavaScript版本。 工作良好的铬和铬双屏幕设置。
function openCenteredWindow({url, width, height}) {
const pos = {
x: (screen.width / 2) - (width / 2),
y: (screen.height/2) - (height / 2)
};
const features = `width=${width} height=${height} left=${pos.x} top=${pos.y}`;
return window.open(url, '_blank', features);
}
例子
openCenteredWindow({
url: 'https://stackoverflow.com/',
width: 500,
height: 600
}).focus();
我的建议是将剩余空间的33%或25%用于顶部位置，而不是像这里发布的其他示例那样使用50%， 主要是因为窗口头文件， 看起来更好，更舒适，
完整的代码:
<script language="javascript" type="text/javascript">
function OpenPopupCenter(pageURL, title, w, h) {
var left = (screen.width - w) / 2;
var top = (screen.height - h) / 4; // for 25% - devide by 4 | for 33% - devide by 3
var targetWin = window.open(pageURL, title, 'toolbar=no, location=no, directories=no, status=no, menubar=no, scrollbars=no, resizable=no, copyhistory=no, width=' + w + ', height=' + h + ', top=' + top + ', left=' + left);
}
</script>
</head>
<body>
<button onclick="OpenPopupCenter('http://www.google.com', 'TEST!?', 800, 600);">click on me</button>
</body>
</html>
看看这句话: Var top =(屏幕。高度- h) / 4;| for 33% -除以3
