Jquery ajax完成回调没有响应201 | ajax完成回调没有响应201

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问题描述

我有一个ajax帖子：

I have an ajax post as such:

$.post("/api/v1/payment_methods/create_credit_card", values)
.done (response) ->
  console.log("GOOD JOB")
.fail (response) ->
  console.log("Adas")

响应是201，但是，完成了似乎没有抓住它，相反它会失败。我认为201会被认为是成功的，并且会被完成捕获。关于它为什么不起作用的任何想法？

The response is a 201, however, done doesn't seem to be capturing it and instead it's going to fail. I thought 201 would of been considered a success and would of been captured by done. Any ideas of why it wouldn't be working?

注意：上面的代码是在coffeescript中，它并没有真正影响问题，但解释了我的语法

Note: The above code is in coffeescript, which doesn't really affect the question but explains my syntax

推荐答案

所以我们弄清楚出了什么问题，JSON.parse抛出了一个语法错误 - 所以发送的值不是有效的JSON格式。海报无法在chrome中看到语法错误，但是firebug显示错误。

So we figured out what was wrong, JSON.parse was throwing a Syntax Error - so the values sent in isnt in a valid JSON format. The poster wasnt able to see the Syntax error in chrome, but firebug showed the error.

这表明只要Javascript抛出异常，响应可能仍然是200， 201,202等 - 但由于语法错误，会触发fail（）函数。

This indicates that whenever Javascript will throw an exception, the response might still be 200, 201, 202 etc. - but because of the syntax error the fail() function would be triggered.

EDIT - 你也应该使用一个不同的响应，许多人使用200 - OK，但Id建议在这种情况下使用202 - ACCEPTED。

EDIT - Also you should probably use a different response, many use 200 - OK, but Id recommend to use 202 - ACCEPTED in this case.

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