首页 > 解决方案 > SQLITE3 FROM 表选择列 WHERE 布尔语句

问题描述

我尝试了 3 种不同的 sqlite3 语句变体来选择数据:

    cursor.execute('SELECT * FROM users WHERE username = ?', (username,))
    cursor.execute('''SELECT * FROM users WHERE username = ?;''', (username,))
    cursor.execute('SELECT * FROM users WHERE username = "monkey1" ')

这些陈述的参考文献来自1 2。然而,他们都没有工作。我怀疑我正在做一些非常愚蠢的事情,但似乎无法弄清楚这一点。

我希望能够打印出用户名“monkey”的数据。感谢任何帮助指出我的愚蠢错误。

import sqlite3
import datetime


def get_user(connection, rows='all', username=None ):
    """Function to obtain data."""
    #create cursor object from sqlite connection object
    cursor = connection.cursor() 
    if rows == 'all':
        print("\nrows == 'all'")
        cursor.execute("SELECT * FROM users")
        data = cursor.fetchall()
        for row in data:
            print(row)

    if rows == 'one':
        print("\nrows == 'one'")
        cursor.execute('SELECT * FROM users WHERE username = ?', (username,))
        #cursor.execute('''SELECT * FROM users WHERE username = ?;''', (username,))
        #cursor.execute('SELECT * FROM users WHERE username = "monkey1" ')
        data = cursor.fetchone()
        print('data = ',data)

    cursor.close()
    return data



def main():
    database = ":memory:"

    table = """ CREATE TABLE IF NOT EXISTS users (
                    created_on  TEXT    NOT NULL UNIQUE,
                    username    TEXT    NOT NULL UNIQUE,
                    email       TEXT    NOT NULL UNIQUE
                 ); """

    created_on = datetime.datetime.now()
    username   = 'monkey'
    email      = 'monkey@gmail'

    created_on1 = datetime.datetime.now()
    username1   = 'monkey1'
    email1      = 'monkey1@gmail'

    # create a database connection & cursor
    conn = sqlite3.connect(database)
    cursor = conn.cursor() 

    # Insert data
    if conn is not None:
        # create user table
        cursor.execute(table)
        cursor.execute('INSERT INTO users VALUES(?,?,?)',(
            created_on, email, username))
        cursor.execute('INSERT INTO users VALUES(?,?,?)',(
            created_on1, email1, username1))
        conn.commit()
        cursor.close()
    else:
        print("Error! cannot create the database connection.")

    # Select data
    alldata = get_user(conn, rows='all')
    userdata = get_user(conn, rows='one', username=username )
    print('\nalldata = ', alldata)
    print('\nuserdata = ', userdata)
    conn.close()


main()

标签: pythonsqlite

解决方案


您的表定义具有按顺序排列的字段,created_on, username, email但您将数据插入为created_on, email, username. 因此第一行的用户名是'monkey@gmail'

避免这种错误的一个好方法是在 INSERT 语句中指定列,而不是依赖于正确获取原始表定义的顺序:

INSERT INTO users (created_on, email, username) VALUES (?,?,?)

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