使用geolocation API的navigator.geolocation.getCurrentPosition()如何处理否定响应?
When using geolocation API's navigator.geolocation.getCurrentPosition() how to deal with a negative response?
表示出现错误时调用第二个回调函数.但是，当用户通过取消请求来选择不透露他的位置时，该函数永远不会被触发.
It says that the second callback function is called when there is an error. However when user chooses not to reveal his location by cancelling the request that function is never fired.
似乎 getCurrentPosition() 无限期地等待答案.(至少在 Firefox 4 中)
It seems that getCurrentPosition() waits for an answer indefinitely. (at least in Firefox 4)
我如何知道用户何时按下取消(或否等)
How can I know when user presses cancel (or no etc.)
有什么想法吗?
见下方编辑
你是对的，当用户拒绝位置请求时，错误处理程序应该触发.传递给错误处理程序的错误对象应包含错误代码和消息，让您知道用户拒绝了请求.但是，从位置请求对话框中选择选项 Not Now 时，我在 FF4 中没有看到这一点.
See edit below
You are correct, the error handler should fire when a user denies the location request. The error object passed into the error handler should contain an error code and message letting you know the user denied the request. However, I'm not seeing this in FF4 when selecting the option Not Now from the location request dialogue.
在 Chrome 中，API/回调完全按预期工作，但在 Chrome 中没有第三个选项.
In Chrome, the API/callbacks work exactly as expected, but in Chrome there is no 3rd option.
编辑
啊，好吧，我在 FF4 中的这种行为中发现了一个小怪癖.在普通模式下(非隐私浏览)，用户将看到 3 个选项:
EDIT
Ahhh okay I found a little quirk in the behavior of this in FF4. In normal mode (not private browsing), the user will be presented 3 options:
Never share 会正确触发错误处理程序，但 Not Now 不会.
这是什么意思以及如何处理?
Never share triggers the error handler correctly, but Not Now does not.
What does this mean and how to handle it?
好吧，看起来如果用户点击Not Now，您将不会得到响应.因此，我将设置一个超时来检查将由其中一个处理程序设置的标志.如果未设置此标志(意味着处理程序未在分配的时间内触发)，您可以执行以下两种操作之一:
Well, it looks like if the user hits Not Now, you aren't going to get a response. Therefore, I would set a timeout which checks a flag that would be set by one of the handlers. If this flag is not set (meaning the handlers didn't fire in the allotted time), you can do one of two things:
选项 2 的可用性可能很差(而且很烦人)，因此最好假设他们暂时拒绝并在下次访问该网站时再次询问(礼貌！).
Option 2 is probably bad usability (and annoying), so it is probably best to assume they denied temporarily and ask them again (politely!) the next time they visit the site.
我创建了一个 JsFiddle 来玩这个 API:
I created a JsFiddle to play around with this API:
http://jsfiddle.net/7yYpn/11/
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