我正在尝试在 python django 中创建我自己的第一个项目。到目前为止我所做的是一个简单的商店。当我点击这个时:
item_details.html
<a href="{% url 'buy_item' pk=item.pk %}" class="submit">Kup teraz</a>
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我将被重定向
urls.py
<a href="{% url 'buy_item' pk=item.pk %}" class="submit">Kup teraz</a>
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对于这种形式:
表格.py
path('buy/<int:pk>', ItemCreate.as_view(), name='buy_item')
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有了这样的看法:
视图.py
class BuyForm(forms.ModelForm):
class Meta:
model = Order
fields = (
'item',
'delivery',
'price',
'buyer',)
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模型.py
from django.db import models
from django.utils import timezone
from django.contrib.auth.models import User
class Category(models.Model):
title = models.CharField(max_length=100)
description = models.TextField()
def __str__(self):
return self.title
class Item(models.Model):
seller = models.ForeignKey(User, on_delete=models.CASCADE)
category = models.ForeignKey(Category, on_delete=models.CASCADE)
title …Run Code Online (Sandbox Code Playgroud) 我正在尝试学习 Java Spring 并且完全像这里一样:https : //spring.io/guides/gs/securing-web/
但是我的 IDE 在导入配置时说“无法解析‘安全’符号”。我正在使用相同版本的 spring 并且具有相同的代码。我不知道出了什么问题。
package com.github.SpringSite.config;
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;
import org.springframework.security.config.annotation.web.configuration.EnableWebSecurity;
import org.springframework.security.config.annotation.web.configuration.WebSecurityConfigurerAdapter;
import org.springframework.security.core.userdetails.User;
import org.springframework.security.core.userdetails.UserDetails;
import org.springframework.security.core.userdetails.UserDetailsService;
import org.springframework.security.provisioning.InMemoryUserDetailsManager;
@Configuration
@EnableWebSecurity
public class WebSecurityConfig extends WebSecurityConfigurerAdapter {
@Override
protected void configure(HttpSecurity http) throws Exception {
http
.authorizeRequests()
.antMatchers("/", "/home").permitAll()
.anyRequest().authenticated()
.and()
.formLogin()
.loginPage("/login")
.permitAll()
.and()
.logout()
.permitAll();
}
@Bean
@Override
public UserDetailsService userDetailsService() {
UserDetails user =
User.withDefaultPasswordEncoder()
.username("user")
.password("password")
.roles("USER")
.build();
return new InMemoryUserDetailsManager(user);
}
}
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我的 pom.xml …