在python中读取外部sql脚本

时间:2013-10-20 01:21:27

标签: python sql

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三个表'Zookeeper','Handles','Animal'。

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

--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。现在我需要在Python中执行它们。我已经给出并完成了在文件中读取的代码。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。我相信循环这是我应该放置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()

4 个答案:

答案 0 :(得分:72)

您的代码已经包含了一种从指定的sql文件执行所有语句的漂亮方法

# 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

将此包装在一个函数中,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')

你说你感到困惑

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

在Python中,您可以使用称为字符串格式的东西将字符串添加到字符串中。

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%("您要用&#34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,

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

表示是否将%s替换为表变量的值。

(创建于)

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


# for loop example

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

如果您有任何其他问题,请戳我。

答案 1 :(得分:2)

将外部脚本读入python中的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()

答案 2 :(得分:0)

首先确保表存在,如果不存在,创建一个表,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)

答案 3 :(得分:-3)

根据我的说法,这不可能

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
    

    然后通过转换为数据框使用.sql文件