【题解】Counting Email in a Database (Using Databases with Python)

吐槽：一开始没仔细看题，然后就憨了= = 以及听课真的很重要。

题目：Counting Organizations

This application will read the mailbox data (mbox.txt) and count the number of email messages per organization (i.e. domain name of the email address) using a database with the following schema to maintain the counts.

CREATE TABLE Counts (org TEXT, count INTEGER)

When you have run the program on mbox.txt upload the resulting database file above for grading.

If you run the program multiple times in testing or with dfferent files, make sure to empty out the data before each run.

You can use this code as a starting point for your application: http://www.py4e.com/code3/emaildb.py.

The data file for this application is the same as in previous assignments: http://www.py4e.com/code3/mbox.txt.

Because the sample code is using an UPDATE statement and committing the results to the database as each record is read in the loop, it might take as long as a few minutes to process all the data. The commit insists on completely writing all the data to disk every time it is called.

The program can be speeded up greatly by moving the commit operation outside of the loop. In any database program, there is a balance between the number of operations you execute between commits and the importance of not losing the results of operations that have not yet been committed.

我的解法：

import sqlite3

conn = sqlite3.connect('emaildb.sqlite')
cur = conn.cursor()

cur.execute('DROP TABLE IF EXISTS Counts')

cur.execute('''
CREATE TABLE Counts (org TEXT, count INTEGER)''')

fname = input('Enter file name: ')
if (len(fname) < 1): fname = 'mbox.txt'
fh = open(fname)
for line in fh:
    if not line.startswith('From: '): continue
    pieces = line.split()
    email = pieces[1].split('@')
    org = email[1]
    cur.execute('SELECT count FROM Counts WHERE org = ? ', (org,))
    row = cur.fetchone()
    if row is None:
        cur.execute('''INSERT INTO Counts (org, count)
                VALUES (?, 1)''', (org,))
    else:
        cur.execute('UPDATE Counts SET count = count + 1 WHERE org = ?',
                    (org,))
    conn.commit()

# https://www.sqlite.org/lang_select.html
sqlstr = 'SELECT org, count FROM Counts ORDER BY count DESC LIMIT 10'

for row in cur.execute(sqlstr):
    print(str(row[0]), row[1])

cur.close()

我想说的：

1，这题让求的是电子邮件的域名，也就是题目中的org，就是邮箱后缀，类似@126.com，@qq.com之类的东西，不是电子邮件名字，别搞错了。

2，注意一下 conn.commit() 就跟生成按钮一样，不写的话不会更改内容。具体其他部分的代码也可用注释一部分去看看效果，如果不懂的话。