models

1.安装SQLAcodegen pip install sqlacodegen 2、使用sqlacodegen生成案列 sqlacodegen mysql://root:123456@127.0.0.1:3306/test > models.py --tables 指定数据表名称 --outfile 指定输出文件名称 3.如果python3 会报错 No module named 'MySQLdb' 这个时候安装pymysql。 然后在sqlacodegen 的__init__.py文件里加上 import pymysql pymysql.install_as_MySQLdb() 来源： https://www.cnblogs.com/xcsg/p/11918273.html

AutoMapper初始化 在global.axax的Application_Start中使用AutoMapperConfiguration.Configure(); using AutoMapper; using System; using System.Collections.Generic; using System.Linq; using System.Web; namespace BaseAsset.Api.Mappings { public class AutoMapperConfiguration { public static void Configure() { Mapper.Initialize(x => { //DomainToViewModelMappingProfile文件将被实例化并添加到配置中。 x.AddProfile<DomainToViewModelMappingProfile>(); }); } } } using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; using AutoMapper; using BaseAsset.Api.Models.Assets;

django ORM基本配置 django中遵循 Code Frist 的原则，即：根据代码中定义的类来自动生成数据库表 1.修改project数据库配置 （1）settigs.py里面 默认 DATABASES = { 'default': { 'ENGINE': 'django.db.backends.sqlite3', 'NAME': os.path.join(BASE_DIR, 'db.sqlite3'), } } 修改为mysql数据库： DATABASES = { 'default': { 'ENGINE': 'django.db.backends.mysql', 'NAME': 'mysql', #数据库名字 'USER': 'root', #账号 'PASSWORD': '123456', #密码 'HOST': '192.168.43.128', #IP 'PORT': '3306', #端口 } } （2）把模块改成pymysql 修改project目录下的init.py import pymysql pymysql.install_as_MySQLdb() 2.创建数据库表结构文件 对应app目录下的models.py （1）生成一个简单的数据库表： from django.db import models class UseInfo(models.Model

from django.db import models from django.contrib.auth.models import AbstractUser # Create your models here. # 用户表 class UserInfo(AbstractUser): phone = models.BigIntegerField(null=True)因为我们继承了django的user表所以需要在setting.py中配置 AUTH_USER_MODEL = "app01.UserInfo" 来源： https://www.cnblogs.com/hellozizi/p/11885579.html

I'm now in the design phase of application. Mostly doing stuff on the paper to get the insight of the application models, define what I will need and so on. And I found there one problem, which I don't know how to solve, or be more specific, I don't know how to implement it to keep it dry. So here is the deal: I have application where will be three different types of users/accounts: Regular account - Account has basic profile with basic rights in the application, User with regular account can see everyone's public profile, can modify own profile, can have only 1 profile picture, can search,

# 在ORM中设置自关联字段 class Comment(models.Model): user = models.ForeignKey(to='UserInfo') article = models.ForeignKey(to='Article') content = models.CharField(max_length=255) create_time = models.DateField(auto_now_add=True) parent = models.ForeignKey(to='self', null=True) 注意:自关联,关联的是当前表的主键ID字段!!!! 上表中parent对应的是Comment.pk. 切记,若关联错误,则会查询出错. 来源： https://www.cnblogs.com/hellozizi/p/11879115.html

I am new to Laravel 4. I wanted to know if it is possible to generate Models , Views and Controllers from existing database? I Googled and found https://github.com/JeffreyWay/Laravel-4-Generators But it allow to generate migration script, model, views and controllers by providing resource name where as i want to reverse engineering of the same in which by command line i want to create models, views and controllers from the existing database. Rokib php artisan generate:model dbtablename it will create individual model from your existing database. In this case you won't need the generate

I have a Conversation model and a view that displays this model. This model is fetched from the server without any problem (the url property works fine then), and the view is rendered. However, when I attempt to destroy the model in a function of the view, I get the error 'A "url" property or function must be specified', even though when I display said url right before the destroy call, it is exactly the url it should be. Here is the code for the model: MessageManager.models.Conversation = Backbone.Model.extend({ defaults: { uid: '', title: '', messages: [], users: [], dateUpdated: null, isNew