根据官方文档,在flask 0.11.x中出现了cli.command这种装饰器
@app.cli.command('initdb')
def initdb_command():
"""Initializes the database."""
init_db()
print 'Initialized the database.'
此时在命令行中设置了FLASK_APP=flaskr后(flaskr为app的名字),输入设置好的命令就可以直接执行函数,官方示例如下
>flask initdb
Initialized the database.
可是我自己操作的时候就会出现
>flask initdb
Usage: flask [OPTIONS] COMMAND [ARGS]...
Error: No such command "initdb".
而当我使用另一种操作方法就能成功
>python -m flask initdb
Initialized the database.
在官方文档(http://flask.pocoo.org/docs/0.11/cli/)中这样提到过这个使用方法
After installation of Flask you will now find a flask script installed into your virtualenv. If you don’t want to install Flask or you have a special use-case you can also use python -m flask to accomplish exactly the same.
我在Windows、Ubuntu上分别测试过python3.5、python2.7.11,均出现同样情况
现在我想知道为什么flask initdb这种用法不能成功?
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源码如下
flaskr.py:
# all the imports
import os
import sqlite3
from flask import Flask, request, session, g, redirect, url_for, abort, render_template, flash
# create our little application :)
app = Flask(__name__)
app.config.from_object(__name__)
# Load default config and override config from an environment variable
app.config.update(dict(
DATABASE=os.path.join(app.root_path, 'flaskr.db'),
SECRET_KEY='development key',
USERNAME='admin',
PASSWORD='default'
))
app.config.from_envvar('FLASKR_SETTINGS', silent=True)
def connect_db():
"""Connects to the specific database."""
rv = sqlite3.connect(app.config['DATABASE'])
rv.row_factory = sqlite3.Row
return rv
def get_db():
"""Opens a new database connection if there is none yet for the
current application context.
"""
if not hasattr(g, 'sqlite_db'):
g.sqlite_db = connect_db()
return g.sqlite_db
@app.teardown_appcontext
def close_db(error):
"""Closes the database again at the end of the request."""
if hasattr(g, 'sqlite_db'):
g.sqlite_db.close()
def init_db():
db = get_db()
with app.open_resource('schema.sql', mode='r') as f:
db.cursor().executescript(f.read())
db.commit()
@app.cli.command('initdb')
def initdb_command():
"""Initializes the database."""
init_db()
print ('Initialized the database.')
if __name__ == '__main__':
app.run()
1. python xxx.py
2. python -m xxx.py
These are two ways to load py files:
1 is called direct operation. If you use flask xxx, it belongs to this method;
2 is equivalent to import, which is called starting as a module.
Different ways of loading py files will affect the sys.path attribute.
Use python -m xxx. Please note that there will be an extra '' in the system path, and the python interpreter will be able to find your "registered" command in the current path!
Reference: http://www.tuicool.com/articles/jMzqYzF
The problem has been solved because flaskr.py was placed under the flaskr folder in the virtualenv folder.
Move the file to the upper-level folder to solve the problem.