小编Tyl*_*rey的帖子

如何修复:使用名称创建 bean 时出错:通过字段表达的不满足的依赖关系

我正在尝试使用 hibernate 设置 spring rest api。尝试使用我设置的 userRespository 时出现此错误

org.springframework.beans.factory.UnsatisfiedDependencyException: Error creating bean with name 'userController': Unsatisfied dependency expressed through field 'userService'; nested exception is org.springframework.beans.factory.UnsatisfiedDependencyException: Error creating bean with name 'userServiceImpl': Unsatisfied dependency expressed through field 'userRepository'; nested exception is org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'userRepository': Invocation of init method failed; nested exception is java.lang.IllegalArgumentException: Not a managed type: class com.potholeapi.models.User
    at org.springframework.beans.factory.annotation.AutowiredAnnotationBeanPostProcessor$AutowiredFieldElement.inject(AutowiredAnnotationBeanPostProcessor.java:596) ~[spring-beans-5.1.8.RELEASE.jar:5.1.8.RELEASE]
    at org.springframework.beans.factory.annotation.InjectionMetadata.inject(InjectionMetadata.java:90) ~[spring-beans-5.1.8.RELEASE.jar:5.1.8.RELEASE]
    at org.springframework.beans.factory.annotation.AutowiredAnnotationBeanPostProcessor.postProcessProperties(AutowiredAnnotationBeanPostProcessor.java:374) ~[spring-beans-5.1.8.RELEASE.jar:5.1.8.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.populateBean(AbstractAutowireCapableBeanFactory.java:1411) ~[spring-beans-5.1.8.RELEASE.jar:5.1.8.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:592) ~[spring-beans-5.1.8.RELEASE.jar:5.1.8.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:515) ~[spring-beans-5.1.8.RELEASE.jar:5.1.8.RELEASE] …
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java spring spring-data spring-data-jpa spring-boot

3
推荐指数
1
解决办法
4万
查看次数

Express 和 Mongoose:无法读取 Postman 中未定义的属性“名称”

我正在尝试根据对localhost:3000/api/student的 POST 请求创建和插入在以下模型中定义的“用户”json 文档。使用邮递员发送 POST 请求时遇到以下错误:

TypeError: Cannot read property 'name' of undefined
    at module.exports.makeStudent (/home/tyler/Dropbox/Projects/Curricula/API/controllers/student.js:17:22)
    at Layer.handle [as handle_request] (/home/tyler/Dropbox/Projects/Curricula/node_modules/express/lib/router/layer.js:95:5)
    at next (/home/tyler/Dropbox/Projects/Curricula/node_modules/express/lib/router/route.js:131:13)
    at Route.dispatch (/home/tyler/Dropbox/Projects/Curricula/node_modules/express/lib/router/route.js:112:3)
    at Layer.handle [as handle_request] (/home/tyler/Dropbox/Projects/Curricula/node_modules/express/lib/router/layer.js:95:5)
    at /home/tyler/Dropbox/Projects/Curricula/node_modules/express/lib/router/index.js:277:22
    at Function.process_params (/home/tyler/Dropbox/Projects/Curricula/node_modules/express/lib/router/index.js:330:12)
    at next (/home/tyler/Dropbox/Projects/Curricula/node_modules/express/lib/router/index.js:271:10)
    at Function.handle (/home/tyler/Dropbox/Projects/Curricula/node_modules/express/lib/router/index.js:176:3)
    at router (/home/tyler/Dropbox/Projects/Curricula/node_modules/express/lib/router/index.js:46:12)
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以下是我认为的相关文件

API/模型/student.js

var mongoose = require('mongoose')
var studentSchema = new mongoose.Schema({
    name: {type: String, required: true},
    password: {type: String, required: true},
    classes: [Number]
    });

mongoose.model('Student', studentSchema);
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API/控制器/student.js

var mongoose = require('mongoose');
var …
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javascript mongoose mongodb node.js express

2
推荐指数
1
解决办法
6218
查看次数

我的表达是非法的,我不知道为什么

当我编译这行时,74出现是一个非法的表达开始,为什么会这样?我完全无法弄清楚这一点,任何帮助都将非常感谢谢谢.

import java.awt.*;
import javax.swing.*;

public class NumericKeypadPanel2 extends JPanel
{
    public NumericKeypadPanel2()
    {
        String num = " ";


        JPanel panel_main = new JPanel();
        panel_main.setLayout(new BorderLayout());

        JPanel keypad = new JPanel();
        keypad.setLayout(new GridLayout (4, 3));
        keypad.setBorder (BorderFactory.createLineBorder (Color.black, 3));

        JButton b1 = new JButton ("1");
        JButton b2 = new JButton ("2");
        JButton b3 = new JButton ("3");
        JButton b4 = new JButton ("4");
        JButton b5 = new JButton ("5");
        JButton b6 = new JButton ("6");
        JButton b7 = new JButton …
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java syntax expression actionlistener

0
推荐指数
1
解决办法
71
查看次数