Question

try:
    r = requests.get(url, params={'s': thing})
except requests.ConnectionError, e:
    print e #should I also sys.exit(1) after this?

这是对的吗？有没有更好的方法来构建它？这会涵盖我的所有基础吗？

Answer 1

查看请求exception docs。简而言之：

如果出现网络问题（例如DNS失败，拒绝连接等），请求将引发 ConnectionError 例外。

如果罕见的无效HTTP响应，请求将引发 HTTPError 例外。

如果请求超时，则会引发 Timeout 异常。

如果请求超出配置的最大重定向数，则会引发 TooManyRedirects 异常。

请求明确提出的所有异常都来自 requests.exceptions.RequestException 。

要回答您的问题，您展示的内容将不涵盖您的所有基础。您只能捕获与连接相关的错误，而不是那些超时的错误。

捕获异常时该怎么做才能完全取决于脚本/程序的设计。退出是否可以接受？你能继续再试一次吗？如果错误是灾难性的并且您无法继续，那么是的，对sys.exit()的调用是有序的。

您可以捕获基类异常，它将处理所有情况：

try:
    r = requests.get(url, params={'s': thing})
except requests.exceptions.RequestException as e:  # This is the correct syntax
    print e
    sys.exit(1)

或者你可以单独捕捉它们并做不同的事情。

try:
    r = requests.get(url, params={'s': thing})
except requests.exceptions.Timeout:
    # Maybe set up for a retry, or continue in a retry loop
except requests.exceptions.TooManyRedirects:
    # Tell the user their URL was bad and try a different one
except requests.exceptions.RequestException as e:
    # catastrophic error. bail.
    print e
    sys.exit(1)

正如Christian所指出的那样：

如果您希望http错误（例如401 Unauthorized）引发异常，您可以致电Response.raise_for_status。如果响应是http错误，那将引发HTTPError。

一个例子：

try:
    r = requests.get('http://www.google.com/nothere')
    r.raise_for_status()
except requests.exceptions.HTTPError as err:
    print err
    sys.exit(1)

将打印：

404 Client Error: Not Found for url: http://www.google.com/nothere

Answer 2

另外一个明确的建议。最好从特定的错误到一般的错误堆栈，以获得所需的错误，因此特定的错误不会被普通错误掩盖。

url='http://www.google.com/blahblah'

try:
    r = requests.get(url,timeout=3)
    r.raise_for_status()
except requests.exceptions.HTTPError as errh:
    print ("Http Error:",errh)
except requests.exceptions.ConnectionError as errc:
    print ("Error Connecting:",errc)
except requests.exceptions.Timeout as errt:
    print ("Timeout Error:",errt)
except requests.exceptions.RequestException as err:
    print ("OOps: Something Else",err)

Http Error: 404 Client Error: Not Found for url: http://www.google.com/blahblah

VS

url='http://www.google.com/blahblah'

try:
    r = requests.get(url,timeout=3)
    r.raise_for_status()
except requests.exceptions.RequestException as err:
    print ("OOps: Something Else",err)
except requests.exceptions.HTTPError as errh:
    print ("Http Error:",errh)
except requests.exceptions.ConnectionError as errc:
    print ("Error Connecting:",errc)
except requests.exceptions.Timeout as errt:
    print ("Timeout Error:",errt)     

OOps: Something Else 404 Client Error: Not Found for url: http://www.google.com/blahblah

Answer 3

异常对象还包含原始响应e.response，如果需要查看服务器响应中的错误正文，则该异常很有用。例如：

try:
    r = requests.post('somerestapi.com/post-here', data={'birthday': '9/9/3999'})
    r.raise_for_status()
except requests.exceptions.HTTPError as e:
    print (e.response.text)

尝试/除了使用Python请求模块的正确方法？

3 个答案: