使用Python分析谷歌浏览器Chrome的网页浏览历史记录

1.代码示例如下：

import os
import sqlite3
import operator
from collections import OrderedDict
import matplotlib.pyplot as plt

def parse(url):
    try:
        parsed_url_components = url.split('//')
        sublevel_split = parsed_url_components[1].split('/', 1)
        domain = sublevel_split[0].replace("www.", "")
        return domain
    except IndexError:
        print "URL format error!"

def analyze(results):

    prompt = raw_input("[.] Type <c> to print or <p> to plot\n[>] ")

    if prompt == "c":
        for site, count in sites_count_sorted.items():
            print site, count
    elif prompt == "p":
        plt.bar(range(len(results)), results.values(), align='edge')
        plt.xticks(rotation=45)
        plt.xticks(range(len(results)), results.keys())
        plt.show()
    else:
        print "[.] Uh?"
        quit()

#path to user's history database (Chrome)
data_path = os.path.expanduser('~')+"\AppData\Local\Google\Chrome\User Data\Default"
files = os.listdir(data_path)

history_db = os.path.join(data_path, 'history')

#querying the db
c = sqlite3.connect(history_db)
cursor = c.cursor()
select_statement = "SELECT urls.url, urls.visit_count FROM urls, visits WHERE urls.id = visits.url;"
cursor.execute(select_statement)

results = cursor.fetchall() #tuple

sites_count = {} #dict makes iterations easier :D

for url, count in results:
    url = parse(url)
    if url in sites_count:
        sites_count[url] += 1
    else:
        sites_count[url] = 1

sites_count_sorted = OrderedDict(sorted(sites_count.items(), key=operator.itemgetter(1), reverse=True))

analyze (sites_count_sorted)

效果展示：

备注：主要用到sqlite和使用第三方python包matplotlib绘图

参考资料来源于：

<<使用Python分析谷歌浏览器Chrome的历史记录>>

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