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做web 推送怎么隐藏这个ajax请求

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Jun 06, 2016 pm 08:40 PM

phpweb

使用的是comet的技术,一个“长连接” ajax请求60秒请求就会超时重新发起一个ajax
这个ajax请求能在console中隐藏吗？
一般的网站这个请求好像都看不到,是怎么做到的？

回复内容：

https://github.com/walkor/web-msg-sender
web推送可以看下这个，非ajax，console中看不到

如果是xhr请求，console开启log的情况下，无论如何都不可能不显示出来

不过用jsonp的方式实现就看不到了，因为它不是一个xhr请求，而是动态加载一个script标签

注意不要直接用jQuery的jsonp实现，因为你的接口本来是ajax方式实现的，那么说明肯定是同源的请求，如果直接改成jQuery的jsonp，实质上还是通过xhr发出的，jQuery对同源jsonp做了特殊处理，并不是按照动态添加script标签的方式进行的

做web 推送怎么隐藏这个ajax请求

websocket. 第一次看得见，以后都看不见。

短链接肯定都可以看到，可以试试websocket IE10一下不支持

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