没有返回任何内容时处理JSON解码错误 [英] Handle JSON Decode Error when nothing returned

问题描述

我正在解析json数据.我的解析没有问题，我正在使用simplejson模块.但是某些api请求返回空值.这是我的示例:

I am parsing json data. I don't have an issue with parsing and I am using simplejson module. But some api requests returns empty value. Here is my example:

{
"all" : {
    "count" : 0,
    "questions" : [     ]
    }
}

这是我解析json对象的代码段:

This is the segment of my code where I parse the json object:

 qByUser = byUsrUrlObj.read()
 qUserData = json.loads(qByUser).decode('utf-8')
 questionSubjs = qUserData["all"]["questions"]

正如我对某些请求所提到的，出现以下错误:

As I mentioned for some requests I get the following error:

Traceback (most recent call last):
  File "YahooQueryData.py", line 164, in <module>
    qUserData = json.loads(qByUser)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/simplejson/__init__.py", line 385, in loads
    return _default_decoder.decode(s)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/simplejson/decoder.py", line 402, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/simplejson/decoder.py", line 420, in raw_decode
    raise JSONDecodeError("No JSON object could be decoded", s, idx)
simplejson.decoder.JSONDecodeError: No JSON object could be decoded: line 1 column 0 (char 0)

处理此错误的最佳方法是什么?

What would be the best way to handle this error?

推荐答案

Python编程中有一个规则，即宽恕比许可容易"(简称EAFP).这意味着您应该捕获异常，而不是检查值的有效性.

There is a rule in Python programming called "it is Easier to Ask for Forgiveness than for Permission" (in short: EAFP). It means that you should catch exceptions instead of checking values for validity.

因此，请尝试以下操作:

Thus, try the following:

try:
    qByUser = byUsrUrlObj.read()
    qUserData = json.loads(qByUser).decode('utf-8')
    questionSubjs = qUserData["all"]["questions"]
except ValueError:  # includes simplejson.decoder.JSONDecodeError
    print 'Decoding JSON has failed'

编辑:由于simplejson.decoder.JSONDecodeError实际上是从ValueError继承的(

EDIT: Since simplejson.decoder.JSONDecodeError actually inherits from ValueError (proof here), I simplified the catch statement by just using ValueError.

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