Question

我有这样的代码：

var IFS =
    document.isFullScreen || 
    document.webkitIsFullScreen || 
    document.mozIsFullScreen || 
    document.msIsFullScreen;

确切的问题是||网关将false排序为undefined，因为isFullScreen是布尔值。

我该如何解决这个问题？

Answer 1

如果每个元素都未定义，那么IFS将是未定义的。由于undefined是一个假值，您仍然可以将条件语句作为：

var IFS =
    document.isFullScreen || 
    document.webkitIsFullScreen || 
    document.mozIsFullScreen || 
    document.msIsFullScreen;

if(!IFS){
    console.log('not full screen');
}

如果您仍希望保留false值以防其他变量为undefined，那么您可以选择以下内容：

var IFS =
    document.isFullScreen || 
    document.webkitIsFullScreen || 
    document.mozIsFullScreen || 
    document.msIsFullScreen || 
    false;

Answer 2

undefined是 falsey 值，这意味着||运算符将转换为与false相同的值。

如果您只是想找到未定义的第一个值，您可以尝试这样的事情：

var IFS =
    (typeof document.isFullScreen != "undefined") ? document.isFullScreen :
    (typeof document.webkitIsFullScreen != "undefined") ? document.webkitIsFullScreen :
    (typeof document.mozIsFullScreen != "undefined") ? document.mozIsFullScreen :
    (typeof document.msIsFullScreen != "undefined") ? document.msIsFullScreen : 
    false;

或Niet suggests：

var IFS =
    ('isFullScreen' in document) ? document.isFullScreen :
    ('webkitIsFullScreen' in document) ? document.webkitIsFullScreen :
    ('mozIsFullScreen' in document) ? document.mozIsFullScreen :
    ('msIsFullScreen' in document) ? document.msIsFullScreen : 
    false;

或者您可以使用数组，如下所示：

var arr = ['isFullScreen', 'webkitIsFullScreen', 'mozIsFullScreen', 'msIsFullScreen'];
var IFS = false;
for (var i = 0; i < arr.length; i++) {
    if (arr[i] in document) {
        IFS = document[arr[i]];
        break;
    }
}

或者像这样

var arr = ['isFullScreen', 'webkitIsFullScreen', 'mozIsFullScreen', 'msIsFullScreen'];
var key = arr.filter(function(e) { return e in document; })[0];
var IFS = !!document[key];

或者gate（||）将false排序为undefined

2 个答案: