今天来一个新技能！！！Python生成某个数据库所有表的models.py文件。

好久不见啊！来个硬核，两步解决以上问题

1.pip install sqlacodegen

2.sqlacodegen --outfile models.py "mysql+pymysql://admin:password@127.0.0.1:3306/zby"

猪：如果没有pymysql,可能还需要:pip install pymysql

(venv) E:\zby >sqlacodegen
You must supply a url

usage: sqlacodegen [-h] [--version] [--schema SCHEMA] [--tables TABLES]
                   [--noviews] [--noindexes] [--noconstraints] [--nojoined]
                   [--noinflect] [--noclasses] [--nocomments]
                   [--outfile OUTFILE]
                   [url]

Generates SQLAlchemy model code from an existing database.

positional arguments:
  url                SQLAlchemy url to the database

optional arguments:
  -h, --help         show this help message and exit
  --version          print the version number and exit
  --schema SCHEMA    load tables from an alternate schema
  --tables TABLES    tables to process (comma-separated, default: all)
  --noviews          ignore views
  --noindexes        ignore indexes
  --noconstraints    ignore constraints
  --nojoined         don't autodetect joined table inheritance
  --noinflect        don't try to convert tables names to singular form
  --noclasses        don't generate classes, only tables
  --nocomments       don't render column comments
  --outfile OUTFILE  file to write output to (default: stdout)

使用方法也很简单，帮助文档很详细！！！

偷偷扔一个：生成Python依赖包文件 requirements.txt

(venv) E:\zby >pip freeze > requirements.txt

猪：以上操作都是在Python开发工具【PyCharm】里面【Terminal】