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我正在尝试打开队列中的所有弹出窗口，以便它们的大小和位置与我的队列数组中有多少项目有关。当弹出窗口启动时，无论是打开 2 个弹出窗口还是 200 个弹出窗口，它们都会完全填满用户屏幕（使用 screen.width 和 screen.height）。我已经计算出了每个弹出窗口的宽度等式，但是无法获得高度或顶部和左侧位置。我相信高度属性与屏幕宽高比（screen.width/screen.height）相关的宽度有关，但我可能错了。

这是代码，感谢您的帮助！

 var queue = [someUrl, someUrl, someUrl, someUrl];

 function popUp(){ //open popups

    var W=screen.width/Math.ceil(Math.sqrt(queue.length));

    for(i=0; i<queue.length; i++) {
        window.open(queue[i],i,'width='+W+',height='???',toolbar=0,
        menubar=0,location=0,status=0,scrollbars=0,resizable=0,left='???'
        ,top='???'');
    }
}

W*i几乎可以工作，left=但是一旦弹出窗口到达屏幕右侧，它就无法重新分配弹出窗口。也许条件语句可以用于if (W*i>screen.width).

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1 回答 1

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looking at the related questions on the side I found the answer (more or less) written in another language. Trick is to first calculate the number of rows and columns your grid needs to have (based on the length of your array), then to do a loop inside a loop, where the first loop iterates your rows and the second your columns.

here's the javascript translation of that other answer

function popUp(){ 

    // calculate number of rows and columns 
    var cols = Math.ceil(Math.sqrt(queue.length));    
    var rows = Math.ceil(queue.length/cols);

    // calculate width and height based on number of cols && rows
    var W = screen.width / cols;
    var H = screen.height / rows;

    var item = 0; //item of the queue array

    for(i=0; i<rows; i++) {
        for(j=0; j<cols; j++) {

            window.open(queue[item],item,'width='+W+',height='+H+',toolbar=0,menubar=0,location=0,status=0,scrollbars=0,resizable=0,left='+W*j+',top='+H*i+'');

            item++; //add 1 to item
            if(item >= queue.length) { break } //break out of loop once item reaches length of queue

        }                                            
    }
}

this loop would always create a perfect grid, which might not always be what you want, if you've got an odd number of items the last row should be missing a few cells, but this loop-ina-loop would create empty pop ups to finish off the grid, this is why we add that last line to break out of the loop once items >= queue.length (i.e. once you've reached the end of your queue)

UPDATE: here's looping in reverse (little easier to read up here than in the comment)

    for(i=rows-1; i>=0; i--) {
        for(j=cols-1; j>=0; j--) {
            //pop up code
        }                                            
    }
于 2012-11-19T03:37:56.980 回答