Flask中多APP应用以及admin后台系统

Posted 2022-12-14 pythonywy

一.多APP

from werkzeug.wsgi import DispatcherMiddleware
from werkzeug.serving import run_simple
from flask import Flask, current_app
app1 = Flask('app01')
app2 = Flask('app02')

@app1.route('/index')
def index():
    return "app01"

@app2.route('/index2')
def index2():
    return "app2"

# http://localhost:5000/index
# http:/localhost:5000/sec/index2
dm = DispatcherMiddleware(app1, 
    '/sec': app2,
)

if __name__ == "__main__":
    run_simple('localhost', 5000, dm)

二.flask-admin

安装

pip3 install flask_admin

简单使用

from flask import Flask
from flask_admin import Admin

app = Flask(__name__)
#将app注册到adminzhong 
admin = Admin(app)

if __name__=="mian":

    app.run()
#访问
#127.0.0.1:5000/admin端口，会得到一个空白的页面

将表模型注册到admin中

#在将表注册之前应该对app进行配置
SQLALCHEMY_DATABASE_URI = "mysql+pymysql://root:@127.0.0.1:3307/py9api?charset=utf8mb4"
SQLALCHEMY_POOL_SIZE = 5
SQLALCHEMY_POOL_TIMEOUT = 30
SQLALCHEMY_POOL_RECYCLE = -1

#导入models文件的中的表模型
from flask_admin.contrib.sqla import ModelView
from api.models import Stock,Product,Images,Category,Wxuser,Banner

admin.add_view(ModelView(Stock, db.session))
admin.add_view(ModelView(Product, db.session))

admin.add_view(ModelView(Category, db.session))

如果有个字段是图片指端

#配置上传文件的路径
#导入from flask_admin.contrib.fileadmin import FileAdmin
from flask_admin.contrib.fileadmin import FileAdmin,form
file_path = op.join(op.dirname(__file__), 'static')
admin = Admin(app)
admin.add_view(FileAdmin(file_path, '/static/', name='文件'))

#如果有个字段要是上传文件重写该方法的modleView类，假设imgae_url是文件图片的字段
class ImagesView(ModelView):

    form_extra_fields = 
        'image_url': form.ImageUploadField('Image',
                                          base_path=file_path,
                                          relative_path='uploadFile/'
                                          )
    

admin.add_view(ImagesView(Images, db.session))