Python Twitter 工具 - 处理 twitter 404 和 401 API 错误答案

【问题标题】：Python Twitter Tools - handling twitter 404 and 401 API errorsPython Twitter 工具 - 处理 twitter 404 和 401 API 错误
【发布时间】：2015-10-25 08:46:12
【问题描述】：

我在查询 Twitter API 时遇到 404 和 401 错误，我的拉朋友的脚本由于未处理异常而中断。已研究并尝试添加以下 try/except 规定，但未记录异常，并且 for 循环停止。

first = 0
last = 15
while last < len(list)+16: #while last group of 15 items is lower than number of items in list#
    for item in list[first:last]: #parses twitter IDs in the list by groups of 15#
        try:
            results = twitter.friends.ids(skip_status="true",include_user_entities="false",count ="5000",user_id=item) #API query#
            results = str(results)
            text_file = open("output.txt", "a") #creates empty or opens current txt output / change path to desired output#
            text_file.write(str(item) + "," + results + "\n") #adds twitter ID, resulting friends list, and a line skip to the txt output#
            text_file.close()
            break
        except ValueError:
            text_file = open("output.txt", "a") #creates empty or opens current txt output / change path to desired output#
            text_file.write(str(item) + "," + "ERROR" + "\n") #adds twitter ID, resulting friends list, and a line skip to the txt output#
            text_file.close()
    print "Succesfully processed users " + str(list[first:last]) #returns recently processed group of 15 users#
    first = first + 15 #updates list navigation to move on to next group of 15#
    last = last + 15
    time.sleep(1000) #suspends activities for 1000 seconds to respect rate limit#

还研究过基于 http 响应标头构建 if 语句，但我不明白如何插入 Python Twitter 工具“response.headers.get('h')”来做到这一点。处理和记录这些异常并让脚本继续提取数据的最佳方法是什么？

【问题讨论】：

你不想break你在做什么！
谢谢，已经删除了中断和“ValueError”，现在它可以正常工作，直到出现错误。然后该错误被正确记录，但 for 循环停止，而不是继续下一个.. 我正在阅读异常.. 我应该添加一个 else 语句来继续循环吗？
尝试在异常处理程序的最后一行之后添加continue。
完美运行，谢谢！

标签： python api twitter error-handling exception-handling

【解决方案1】：

删除try: 块末尾的break 并在except: 块的末尾添加一个continue。即：

first = 0
last = 15
while last < len(list)+16: #while last group of 15 items is lower than number of items in list#
    for item in list[first:last]: #parses twitter IDs in the list by groups of 15#
        try:
            results = twitter.friends.ids(skip_status="true",include_user_entities="false",count ="5000",user_id=item) #API query#
            results = str(results)
            text_file = open("output.txt", "a") #creates empty or opens current txt output / change path to desired output#
            text_file.write(str(item) + "," + results + "\n") #adds twitter ID, resulting friends list, and a line skip to the txt output#
            text_file.close()
            # break is not needed here!
        except ValueError:
            text_file = open("output.txt", "a") #creates empty or opens current txt output / change path to desired output#
            text_file.write(str(item) + "," + "ERROR" + "\n") #adds twitter ID, resulting friends list, and a line skip to the txt output#
            text_file.close()
            continue # You have dealt with the exception so don't stop
    print "Succesfully processed users " + str(list[first:last]) #returns recently processed group of 15 users#
    first = first + 15 #updates list navigation to move on to next group of 15#
    last = last + 15
    time.sleep(1000) #suspends activities for 1000 seconds to respect rate limit#

从 cmets 转移到保存。

【讨论】：