在python中读取外部sql脚本

Question

我正在学习如何在 python 中执行 SQL（我知道 SQL，而不是 Python）。

我有一个外部 sql 文件。它创建数据并将数据插入到三个表“Zookeeper”、“Handles”、“Animal”中。

然后我有一系列查询要运行表。以下查询位于我在 python 脚本顶部加载的 zookeeper.sql 文件中。前两个的例子是：

--1.1

SELECT ANAME,zookeepid
FROM ANIMAL, HANDLES
WHERE AID=ANIMALID;

--1.2

SELECT ZNAME, SUM(TIMETOFEED)
FROM ZOOKEEPER, ANIMAL, HANDLES
WHERE AID=ANIMALID AND ZOOKEEPID=ZID
GROUP BY zookeeper.zname;

这些都在 SQL 中执行得很好。现在我需要从 Python 中执行它们。我已经获得并完成了读取文件的代码。然后执行循环中的所有查询。

1.1 和 1.2 是我感到困惑的地方。我相信在循环中，这是我应该放入一些东西来运行第一个和第二个查询的行。

result = c.execute("SELECT * FROM %s;" % table);

但是什么？我想我错过了一些非常明显的东西。我认为让我失望的是 % table。在查询 1.1 和 1.2 中，我不是在创建表，而是在查找查询结果。

我的整个python代码如下。

import sqlite3
from sqlite3 import OperationalError

conn = sqlite3.connect('csc455_HW3.db')
c = conn.cursor()

# Open and read the file as a single buffer
fd = open('ZooDatabase.sql', 'r')
sqlFile = fd.read()
fd.close()

# all SQL commands (split on ';')
sqlCommands = sqlFile.split(';')

# Execute every command from the input file
for command in sqlCommands:
    # This will skip and report errors
    # For example, if the tables do not yet exist, this will skip over
    # the DROP TABLE commands
    try:
        c.execute(command)
    except OperationalError, msg:
        print "Command skipped: ", msg


# For each of the 3 tables, query the database and print the contents
for table in ['ZooKeeper', 'Animal', 'Handles']:


    **# Plug in the name of the table into SELECT * query
    result = c.execute("SELECT * FROM %s;" % table);**

    # Get all rows.
    rows = result.fetchall();

    # \n represents an end-of-line
    print "\n--- TABLE ", table, "\n"

    # This will print the name of the columns, padding each name up
    # to 22 characters. Note that comma at the end prevents new lines
    for desc in result.description:
        print desc[0].rjust(22, ' '),

    # End the line with column names
    print ""
    for row in rows:
        for value in row:
            # Print each value, padding it up with ' ' to 22 characters on the right
            print str(value).rjust(22, ' '),
        # End the values from the row
        print ""

c.close()
conn.close()

Answer 1

您的代码已经包含了一种从指定的 sql 文件中执行所有语句的绝妙方法

# Open and read the file as a single buffer
fd = open('ZooDatabase.sql', 'r')
sqlFile = fd.read()
fd.close()

# all SQL commands (split on ';')
sqlCommands = sqlFile.split(';')

# Execute every command from the input file
for command in sqlCommands:
    # This will skip and report errors
    # For example, if the tables do not yet exist, this will skip over
    # the DROP TABLE commands
    try:
        c.execute(command)
    except OperationalError, msg:
        print("Command skipped: ", msg)

将其包装在一个函数中，您可以重复使用它。

def executeScriptsFromFile(filename):
    # Open and read the file as a single buffer
    fd = open(filename, 'r')
    sqlFile = fd.read()
    fd.close()

    # all SQL commands (split on ';')
    sqlCommands = sqlFile.split(';')

    # Execute every command from the input file
    for command in sqlCommands:
        # This will skip and report errors
        # For example, if the tables do not yet exist, this will skip over
        # the DROP TABLE commands
        try:
            c.execute(command)
        except OperationalError, msg:
            print("Command skipped: ", msg)

使用它

executeScriptsFromFile('zookeeper.sql')

你说你被弄糊涂了

result = c.execute("SELECT * FROM %s;" % table);

在 Python 中，您可以使用称为字符串格式化的方法向字符串添加内容。

你有一个带有 %s 的字符串"Some string with %s"，它是其他东西的占位符。要替换占位符，请在字符串后添加 %（“你想用什么替换它”）

例如：

a = "Hi, my name is %s and I have a %s hat" % ("Azeirah", "cool")
print(a)
>>> Hi, my name is Azeirah and I have a Cool hat

有点幼稚的例子，但应该很清楚。

现在，什么

result = c.execute("SELECT * FROM %s;" % table);

意思是，是不是用表变量的值代替了%s。

（创建于）

for table in ['ZooKeeper', 'Animal', 'Handles']:


# for loop example

for fruit in ["apple", "pear", "orange"]:
    print(fruit)
>>> apple
>>> pear
>>> orange

如果您还有其他问题，请戳我。

Answer 2

在python中将外部脚本读入sqlite数据库的一种非常简单的方法是使用executescript()：

import sqlite3

conn = sqlite3.connect('csc455_HW3.db')

with open('ZooDatabase.sql', 'r') as sql_file:
    conn.executescript(sql_file.read())

conn.close()

Answer 3

先确认表存在，如果不存在，创建表然后按照步骤操作。

import sqlite3
from sqlite3 import OperationalError

conn = sqlite3.connect('Client_DB.db')
c = conn.cursor()

def execute_sqlfile(filename):
    
    c.execute("CREATE TABLE clients_parameters (adress text, ie text)")
    #
    fd = open(filename, 'r')
    sqlFile = fd.readlines()
    fd.close()
    lvalues = [tuple(v.split(';')) for v in sqlFile[1:] ]
    try:
        #print(command)
        c.executemany("INSERT INTO clients_parameters VALUES (?, ?)", lvalues)
    except OperationalError as msg:
        print ("Command skipped: ", msg)

execute_sqlfile('clients.sql')

print(c.rowcount)

Answer 4

在我看来，这是不可能的

解决方案：

1. 在mysql服务器上导入.sql文件

2. 之后

   import mysql.connector
   import pandas as pd
   

   然后您通过转换为数据框来使用 .sql 文件