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这可能是一个非常简单的调试问题（我没有单独编码）我有一个循环代码来解析一个抓取的 xml，这个解析发生在一个 5 分钟的循环中，但不会从一个循环返回到下一个循环的重复项，因为将用户 ID 存储在集合中的结果，如果用户 ID 已存在于用户集中，则脚本跳到 xml 的下一行。我想将此脚本的结果输出为 RSS，并且我有一种潜在的方法可以这样做，但是我首先需要将数据存储为某种变量。

我试图这样做，但是每次我这样做时，我似乎都会遇到存储在集合中的最后一个用户 ID 的问题。我没有向您提供我损坏的代码，而是附上了一个工作代码的示例，其中不包括我将结果打印定义为变量的散列尝试。

import mechanize
import urllib
import json
import re
import random
import datetime
from sched import scheduler
from time import time, sleep

######Code to loop the script and set up scheduling time

s = scheduler(time, sleep)
random.seed()

def run_periodically(start, end, interval, func):
    event_time = start
    while event_time < end:
        s.enterabs(event_time, 0, func, ())
        event_time += interval + random.randrange(-5, 45)
    s.run()

###### Code to get the data required from the URL desired
def getData():  
    post_url = "URL OF INTEREST"
    browser = mechanize.Browser()
    browser.set_handle_robots(False)
    browser.addheaders = [('User-agent', 'Firefox')]

######These are the parameters you've got from checking with the aforementioned tools
    parameters = {'page' : '1',
              'rp' : '250',
              'sortname' : 'roi',
              'sortorder' : 'desc'
             }
#####Encode the parameters
    data = urllib.urlencode(parameters)
    trans_array = browser.open(post_url,data).read().decode('UTF-8')

    xmlload1 = json.loads(trans_array)
    pattern1 = re.compile('>&nbsp;&nbsp;(.*)<')
    pattern2 = re.compile('/control/profile/view/(.*)\' title=')
    pattern3 = re.compile('<span style=\'font-size:12px;\'>(.*)<\/span>')

##### Making the code identify each row, removing the need to numerically quantify the     number of rows in the xmlfile,
##### thus making number of rows dynamic (change as the list grows, required for looping function to work un interupted)

    for row in xmlload1['rows']:
        cell = row["cell"]

##### defining the Keys (key is the area from which data is pulled in the XML) for use in the pattern finding/regex

        user_delimiter = cell['username']
        selection_delimiter = cell['race_horse']

        if strikeratecalc2 < 12 : continue;

##### REMAINDER OF THE REGEX DELMITATIONS

        username_delimiter_results = re.findall(pattern1, user_delimiter)[0]
        userid_delimiter_results = (re.findall(pattern2, user_delimiter)[0])
        user_selection = re.findall(pattern3, selection_delimiter)[0]

##### Code to stop duplicate posts of each user throughout the day

    userset = set ([])
    if userid_delimiter_results in userset: continue;

##### Printing the results of the code at hand

        print "user id = ",userid_delimiter_results
        print "username = ",username_delimiter_results
        print "user selection = ",user_selection
        print ""

##### Code to stop duplicate posts of each user throughout the day  part 2 (udating set to add users already printed to the ignore list)

    userset.update(userid_delimiter_results)

    getData()

    run_periodically(time()+5, time()+1000000, 300, getData)

我在尝试生成变量时遇到的问题（我试图将其生成为数组）是代码以某种方式丢失了最后一个 userset.update(userid_delimiter_results) 这导致每次运行时都会重复提要中的最后一个条目代码，因为根据“用户集”，有问题的用户 ID 没有被记录。任何使我能够将此代码的结果作为变量输出的简单方法将不胜感激。亲切的问候 AEA

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1 回答 1

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我通过将打印部分变为;

    arrayna = [arrayna1, arrayna2, arrayna3, arrayna4]

    arraym1 = "user id = ",userid_delimiter_results

然后为了克服每次运行循环 arrayna 的面貌

my_array = [] # Create an empty list

print(my_array)

所以你的代码可能看起来像：

这有效:)

于 2013-06-13T22:22:14.450 回答