Question

我尝试使用实现为字典的数据库来搜索用户输入的值，如果找到该值，则会显示该值，否则会显示错误消息。＆＃39; UUC＆＃39;或者＆＃39; uum＆＃39;等等是指大学校园的缩写。

用户应该能够输入上述值之一，并且应该显示任何匹配项。但是，这在我的代码中不起作用，我无法找到问题。

def all_from(datab,university):
l = len(datab)
k = 1
i = 0
while k <= 1:
    s = datab[k]
    the_university = s[2]
    if university == the_university:
        i = i + 1
        print datab[k]
    k = k + 1
if i == 0:
    print 'Nobody goes to University there!'

datab = dict()
datab[1] = ['walker','matthew','uuc',1]
datab[2] = ['stewart','rory','uum',2]
datab[3] = ['toner','kevin','qub',4]
datab[4] = ['hughes','johnny','uuj',1]
datab[5] = ['douglas','kenny','uuc', 3]
datab[6] = ['hooks', 'peter','qub',1]
datab[7] = ['campbell','vicky','uuj',2]
datab[8] = ['crooks','laura','uum',4]
datab[9] = ['irwin','emma','uuc',3]
datab[10] = ['patterson','steve','uuc',1]

university = (raw_input('Enter the University here: '))
all_from(datab,university)

Answer 1

您可以简化public class Player : INotifyPropertyChanged { protected void OnPropertyChanged([CallerMemberName] string caller = "") { PropertyChanged?.Invoke(this, new PropertyChangedEventArgs(caller)); } public event PropertyChangedEventHandler PropertyChanged; public Player( int high) { HighScore = high; } private int _highScore; public int HighScore { get { return _highScore; } set { _highScore = value; OnPropertyChanged(); } } }功能：

   public class AllPlayers : INotifyPropertyChanged
{
    public ObservableCollection<Player> Players { get; set; }

    public AllPlayers()
    {
        Players = new ObservableCollection<Player>();
    }

    protected void OnPropertyChanged([CallerMemberName] string caller = "")
    {
        PropertyChanged?.Invoke(this, new PropertyChangedEventArgs(caller));
    }

    public event PropertyChangedEventHandler PropertyChanged;

    public void ChangeScore(int playerIndex, int highScore)
    {
        Players[playerIndex].HighScore = highScore;
    }

    private void UpdateBiggestScore()
    {
        BiggestScore = (from player in Players select player.HighScore).Max();
    }

    private int _biggestScore;
    public int BiggestScore
    {
        get { return _biggestScore; }
        set
        {
            _biggestScore = value;

            OnPropertyChanged();
        }
    }
}

http://www.codeskulptor.org/#user42_wjXMwt73xI_2.py

Answer 2

我的贡献。我觉得将学生安排为列表中的项目，将不同的字段作为字典键是一种更好的数据结构。

def find_all(datab,uni):
    for student in datab:
        if student['uni'] == uni:
            print(student)

datab = []
datab.append( {'lastname':'walker', 'firstname':'matthew','uni':'uuc','year':1})
datab.append( {'lastname':'stewart','firstname':'rory','uni':'uum','year':2})
datab.append( {'lastname':'toner','firstname':'kevin','uni':'qub','year':4})
datab.append( {'lastname':'hughes','firstname':'johnny','uni':'uuj','year':1})

uni = input('Enter the uni here: ')
find_all(datab,uni)

如果您想搜索所有密钥，那么您可以更改 if student['uni'] == uni行到

for key, value in student.items():
    if value == uni:

另一个版本：这将允许您输入一个非常简单的查询。

def find_all(datab,query):
    query = query.split('OR')
    for search_terms in query:
        for student in datab:
            for key, value in student.items():
                if value == search_terms.strip():
                    print(student)



datab = list()
datab.append( {'lastname':'walker', 'firstname':'matthew','uni':'uuc','year':'1'})
datab.append( {'lastname':'stewart','firstname':'rory','uni':'uum','year':'2'})
datab.append( {'lastname':'toner','firstname':'kevin','uni':'qub','year':'4'})
datab.append( {'lastname':'hughes','firstname':'johnny','uni':'uuj','year':'1'})

query = input('Enter your query here: ')
find_all(datab,query)

这是我的代码运行两个简单的查询

Enter the uni here: uum OR uuc
{'firstname': 'rory', 'lastname': 'stewart', 'uni': 'uum', 'year': 2}
{'firstname': 'matthew', 'lastname': 'walker', 'uni': 'uuc', 'year': 1}

Enter your query here: 1 OR UUM
{'firstname': 'matthew', 'lastname': 'walker', 'uni': 'uuc', 'year': '1'}
{'firstname': 'johnny', 'lastname': 'hughes', 'uni': 'uuj', 'year': '1'}

您只需在项目之间添加单词OR，它将搜索词典中包含任何单词的任何项目。注：我不得不将uni年改成字符串以使其正常工作。

正如我所说。使用像SQLite这样的适当数据库会更好。

python字典 - 如何搜索和显示答案

2 个答案: