navigator.onLine 检查网络状态的方法返回true or false
if(navigator.onLine){ console.log("链接到网络") return true }else{ console.log("当前网络无链接") return false }
posted on 2018-12-07 09:36 niuben 阅读(250) 评论(0) 编辑 收藏 举报
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