Javascript：如何在使用 window.location.href = url 导航到的页面上捕获错误
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Javascript: How to catch error on page navigated to using window.location.href = url
提问by MegaMatt
I am using a REST service to generate a CSV file that I want to prompt the user to download. An example of the service is below:
我正在使用 REST 服务生成一个 CSV 文件，我想提示用户下载该文件。该服务的示例如下：
https://localhost:8444/websvc/exportCSV?viewId=93282392
https://localhost:8444/websvc/exportCSV?viewId=93282392
To prompt the user to download the file, I use this code:
为了提示用户下载文件，我使用以下代码：
window.location.href = exportUrl, where exportUrlwould be a URL like the one above.
window.location.href = exportUrl,exportUrl上面的 URL在哪里。
This works great if there are no errors on the server when executing the service. The file download prompt appears, the page doesn't refresh, and all is well.
如果在执行服务时服务器上没有错误，这很有效。出现文件下载提示，页面不刷新，一切正常。
However, if there isan error, I'm getting a nasty HTTP Status 500 page, which is no good for user experience. What I'd like to do is catch any error on the resulting page, and throw up a more friendly error without leaving the current page. I tried:
然而，如果是一个错误，我越来越讨厌的HTTP状态500页，这对于用户体验不好。我想做的是捕获结果页面上的任何错误，并在不离开当前页面的情况下抛出一个更友好的错误。我试过：
try {
window.location.href = exportUrl;
}
catch (e) {
alert(e);
}
But that doesn't seem to change the behavior at all. Does anyone have any ideas on how to handle this?
但这似乎根本没有改变行为。有没有人对如何处理这个有任何想法？
Thanks very much.
非常感谢。
采纳答案by daniel0mullins
Catching an error like that will only catch a JavaScript error. That's not what you're experiencing here. Your server is returning a status code of 500. You need to make sure that everything is good BEFORE you send your users there.
捕获这样的错误只会捕获 JavaScript 错误。这不是你在这里经历的。您的服务器正在返回状态代码500。在将用户派往那里之前，您需要确保一切正常。
To do that you could effectively 'ping' the URL with Ajax to ensure that it won't return a 500 error.
为此，您可以使用 Ajax 有效地“ping”该 URL，以确保它不会返回 500 错误。
Something like the following:
类似于以下内容：
var xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
if(xhr.readyState == 4 && xhr.status == 200) {
window.location.href = exportUrl;
}
}
xhr.open('head',exportUrl);
xhr.send(null);
This would do a HEAD request to the URL to ensure that there are no nasty server errors waiting.
这将对 URL 执行 HEAD 请求，以确保没有等待的令人讨厌的服务器错误。
Of course, if in the process of actually GENERATING the CSV your server throws an error - it would still return a 500.
当然，如果在实际生成 CSV 的过程中，您的服务器抛出错误 - 它仍然会返回 500。
A more robust way would be to get the data via Ajax, build a data URL via base64encodeand then set window.location.hrefto that data URL.
更可靠的方法是通过 Ajax 获取数据，通过构建数据 URL base64encode，然后设置window.location.href为该数据 URL。
By doing that, you could also ensure that the Ajax didn't return a 500 and you got the data you were expecting in the response.
通过这样做，您还可以确保 Ajax 没有返回 500 并且您在响应中获得了您期望的数据。
Hope this helps!
希望这可以帮助！
