Question

createDataBase1 = sqlite3.connect('PythonDatabase_1.db')
queryCurs1 = createDataBase1.cursor()
def createtable():
    queryCurs1.execute("""CREATE TABLE Scores
    (id INTEGER PRIMARY KEY, Name TEXT, Class TEXT, Score INTERGER)""")

def addPupil1(Name, Class, Score):
    queryCurs1.execute("""INSERT INTO Scores (Name, Class, Score)
    VALUES(?,?,?)""",(Name, Class, Score,))

def main1():
    createtable1()
    addPupil1
    (Name, Class, Score)
    createDataBase1.commit()

    queryCurs1.execute("SELECT * FROM Scores ORDER BY Score DESC")

    title = ["id","Name", "Class", "Score"]
    k = 0
    for i in queryCurs1:
        print "\n"
        for j in i:
            print title[k],
            print j
            if k < 3: k += 1
            else: k = 0

    queryCurs1.close()
    if __name__ == '__main__':main()

这用于打印结果现在它没有什么问题？它用于以列表格式显示结果，前面的标题现在没有错误

Answer 1

我冒昧地改变你的代码。检查它是否适合您。我更喜欢使用PEP 8代码样式。我还将一个通用记录插入数据库（将其更改为您需要的任何内容）。

另一方面，我还更改了表的create语句，因为一旦创建了表（运行命令两次），它就会显示一个错误，因为表已经创建了。

代码如下所示：

import sqlite3

created_db = sqlite3.connect('PythonDatabase_1.db')
cursor = created_db.cursor()

def create_table():
    cursor.execute("""CREATE TABLE IF NOT EXISTS Scores
    (id INTEGER PRIMARY KEY, Name TEXT, Class TEXT, Score INTERGER)""")

def add_pupil(name, klass, score):
    cursor.execute("""INSERT INTO Scores (Name, Class, Score)
    VALUES(?,?,?)""",(name, klass, score,))

def main():
    create_table()

    name = "John Doe"
    klass = "Math"
    score = 100

    add_pupil(name, klass, score)
    created_db.commit()

    cursor.execute("SELECT * FROM Scores ORDER BY Score DESC")

    title = ["id","Name", "Class", "Score"]
    k = 0

    for i in cursor:
        print "\n"
        for j in i:
            print title[k],
            print j
            if k < 3: k += 1
            else: k = 0

    cursor.close()

if __name__ == '__main__':
    main()

希望它有所帮助，

sqlite数据库不打印数据库的内容

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