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我正在尝试使用一组来阻止用户在以下代码中被重新打印。我设法让 python 接受他的代码而不会产生任何错误，但是如果我让代码在 10 秒的循环中运行，它会继续打印应该已经登录的用户。这是我第一次尝试使用集合，而且我是 python 的完全新手（到目前为止，它是根据我看到的示例构建的，并对它们进行逆向工程。）

下面是我正在使用的代码示例

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)

任何评论都将不胜感激，这对您经验丰富的编码人员来说似乎是常识，但我真的只是通过“Hello world”

亲切的问候 AEA

4

1 回答 1

2

这个：

userset.update(userid_delimiter_results)

应该是这样的：

userset.add(userid_delimiter_results)

为了证明这一点，请尝试userset在每次通话后打印内容。

于 2013-06-09T03:41:36.600 回答