Question

为什么菜单无法启动（由Roman Susi解决）
为什么菜单无法按预期工作（错误如下）
如何解决foo.add代码中的错误？

Traceback (most recent call last): File "C:\Users\User\Desktop\phonedatabase.py", line 81, in <module> openphonedb() File "C:\Users\User\Desktop\phonedatabase.py", line 23, in openphonedb for entry in foo.add(name, number, showtype): TypeError: 'NoneType' object is not iterable

在输入“类型”

import shelve
import string

UNKNOWN = 0
HOME = 1
WORK = 2
FAX = 3
CELL = 4

def openphonedb():
    foo = phonedb()
    print "What would you like to do?"
    print "Add = 1, Lookup = 2, Exit = 3"
    entry = int(raw_input('>> '))
    if entry == 1 :
            namelookup = raw_input('Please enter a name: ')
            for entry in foo.lookup(namelookup):
                    print '%-40s %s (%s)' % (entry.name, entry.number, entry.showtype() )
    elif entry == 2:
            name = raw_input('Name: ')
            number = raw_input('Number: ')
            showtype = input('Type (UNKNOWN, HOME, WORK, FAX, CELL): \n>> ')
            for entry in foo.add(name, number, showtype):           
                    print '%-40s %s (%s)'% (entry.name, entry.number, entry.showtype() )
    elif entry == 3:
            print "Close Successful"
            quit
    else:
            print "Invalid."
            openphonedb()

class phoneentry:
    def __init__(self, name = 'Unknown', number = 'Unknown',
        type = UNKNOWN):
        self.name = name
        self.number = number
        self.type = type

#  create string representation
    def __repr__(self):
        return('%s:%d' % ( self.name, self.type ))

#  fuzzy compare or two items
    def __cmp__(self, that):
        this = string.lower(str(self))
        that = string.lower(that)

        if string.find(this, that) >= 0:
            return(0)
        return(cmp(this, that))

    def showtype(self):
        if self.type == UNKNOWN: return('Unknown')
        if self.type == HOME: return('Home')
        if self.type == WORK: return('Work')
        if self.type == FAX: return('Fax')
        if self.type == CELL: return('Cellular')

class phonedb:
    def __init__(self, dbname = 'phonedata'):
        self.dbname = dbname;
        self.shelve = shelve.open(self.dbname);

    def __del__(self):
        self.shelve.close()
        self.shelve = None

    def add(self, name, number, type = HOME):
        e = phoneentry(name, number, type)
        self.shelve[str(e)] = e

    def lookup(self, string):
        list = []
        for key in self.shelve.keys():
            e = self.shelve[key]
        if cmp(e, string) == 0:
            list.append(e)

        return(list)
#edit
if __name__ == '__main__':
    openphonedb()

Answer 1

也许，你只是忘了打电话给openphonedb（）？添加到最后：

if __name__ == '__main__':
    openphonedb()

此外，“戒烟”在那里做什么？

Answer 2

代码有其他问题，但关于问题的第一部分，将条目作为字符串进行比较（如果用户键入“a”，则不会出现错误）？

def openphonedb():
    foo = phonedb()
    print "What would you like to do?"
    print "Add = 1, Lookup = 2, Exit = 3"
    while True:  # note this
        entry = raw_input('>> ')  # removed int()
        if entry == '1' :
            namelookup = raw_input('Please enter a name: ')
            for entry in foo.lookup(namelookup):
                print '%-40s %s (%s)' % (entry.name, entry.number, entry.showtype() )
        elif entry == '2':
            name = raw_input('Name: ')
            number = raw_input('Number: ')
            showtype = input('Type (UNKNOWN, HOME, WORK, FAX, CELL): \n>> ')
            for entry in foo.add(name, number, showtype):           
                print '%-40s %s (%s)'% (entry.name, entry.number, entry.showtype() )
        elif entry == '3':
            print "Close Successful"
            exit()  # note this also
        else:
            print "Invalid."

如何将用户输入到python数据库？

2 个答案: