在表名上出现 Spring 语法错误答案

【问题标题】：Getting Spring syntax error on table name在表名上出现 Spring 语法错误
【发布时间】：2020-09-21 23:23:53
【问题描述】：

我正在尝试建立一个数据库（目前有 2 个实体）。

第一个包包含第一个：

// Recipe.java

import java.util.List;

import javax.persistence.CascadeType;
import javax.persistence.ElementCollection;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.JoinTable;
import javax.persistence.ManyToMany;
import javax.persistence.OneToMany;

import uk.ac.ic.doc.group33.nutrioapi.user.User;

@Entity
public class Recipe {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private int id;
    private String name;
    private int estimatedMins;
    private String photo;
    @ElementCollection
    private List<String> instructions;
    @OneToMany(mappedBy = "recipe", cascade = CascadeType.ALL, orphanRemoval = true)
    private List<Ingredient> ingredients;

    @ManyToMany(cascade = CascadeType.ALL)
    @JoinTable(name = "user",
    joinColumns = @JoinColumn(name = "user_id", referencedColumnName = "id"),
    inverseJoinColumns = @JoinColumn(name = "recipe_id", referencedColumnName = "id"))
    private List<User> users;

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getEstimatedMins() {
        return estimatedMins;
    }

    public void setEstimatedMins(int estimatedMins) {
        this.estimatedMins = estimatedMins;
    }

    public String getPhoto() {
        return photo;
    }

    public void setPhoto(String photo) {
        this.photo = photo;
    }

    public List<String> getInstructions() {
        return instructions;
    }

    public void setInstructions(List<String> instructions) {
        this.instructions = instructions;
    }

    public List<Ingredient> getIngredients() {
        return ingredients;
    }

    public void setIngredients(List<Ingredient> ingredients) {
        this.ingredients = ingredients;
    }
}

// RecipeController.java

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.PostMapping;
import org.springframework.web.bind.annotation.RequestBody;
import org.springframework.web.bind.annotation.RestController;

@RestController
public class RecipeController {

    @Autowired
    private RecipeRepository recipeRepository;

    @GetMapping("/recipes")
    public Iterable<Recipe> getAllRecipes() {
        return recipeRepository.findAll();
    }

    @PostMapping("/recipes/add")
    public String addRecipe(@RequestBody Recipe recipe) {
        recipeRepository.save(recipe);
        return "success";
    }

}

// RecipeRepository.java

import org.springframework.data.repository.CrudRepository;

public interface RecipeRepository extends CrudRepository<Recipe, Integer> {
}

该包中还有一些辅助类（Ingredient 和 Unit，都工作正常）。可以毫无问题地创建此表。

另一个包：

// User.java

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.ManyToMany;

import uk.ac.ic.doc.group33.nutrioapi.recipe.Recipe;

@Entity
public class User {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private int id;
    private String email;

    @ManyToMany(mappedBy = "users")
    private List<Recipe> recipes;

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }
}

// UserController.java

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.PostMapping;
import org.springframework.web.bind.annotation.RequestBody;
import org.springframework.web.bind.annotation.RestController;

@RestController
public class UserController {

    @Autowired
    private UserRepository userRepository;

    @GetMapping("/users")
    public Iterable<User> getAllUsers() {
        return userRepository.findAll();
    }

    @PostMapping("/users/add")
    public String addRecipe(@RequestBody User user) {
        userRepository.save(user);
        return "success";
    }

}

// UserRepository.java

import org.springframework.data.repository.CrudRepository;

public interface UserRepository extends CrudRepository<User, Integer> {
}

在这些上面的包中，我有主要的 API 类：

// NutrioApiApplication.java

import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;

// Main application class

@SpringBootApplication
public class NutrioApiApplication {

    public static void main(String[] args) {
        SpringApplication.run(NutrioApiApplication.class, args);
    }

}

当我添加用户实体时，它可以创建配方表，但不能创建用户表。我在这些方面遇到了很多错误：

GenerationTarget encountered exception accepting command : Error executing DDL "create table user (id int4 not null, email varchar(255), user_id int4 not null, recipe_id int4 not null, primary key (id))" via JDBC Statement

org.hibernate.tool.schema.spi.CommandAcceptanceException: Error executing DDL "create table user (id int4 not null, email varchar(255), user_id int4 not null, recipe_id int4 not null, primary key (id))" via JDBC Statement
    at org.hibernate.tool.schema.internal.exec.GenerationTargetToDatabase.accept(GenerationTargetToDatabase.java:67) ~[hibernate-core-5.4.15.Final.jar!/:5.4.15.Final]
    at org.hibernate.tool.schema.internal.AbstractSchemaMigrator.applySqlString(AbstractSchemaMigrator.java:559) ~[hibernate-core-5.4.15.Final.jar!/:5.4.15.Final]
    at org.hibernate.tool.schema.internal.AbstractSchemaMigrator.applySqlStrings(AbstractSchemaMigrator.java:504) ~[hibernate-core-5.4.15.Final.jar!/:5.4.15.Final]
    at org.hibernate.tool.schema.internal.AbstractSchemaMigrator.createTable(AbstractSchemaMigrator.java:277) ~[hibernate-core-5.4.15.Final.jar!/:5.4.15.Final]
    at org.hibernate.tool.schema.internal.GroupedSchemaMigratorImpl.performTablesMigration(GroupedSchemaMigratorImpl.java:71) ~[hibernate-core-5.4.15.Final.jar!/:5.4.15.Final]
    at org.hibernate.tool.schema.internal.AbstractSchemaMigrator.performMigration(AbstractSchemaMigrator.java:207) ~[hibernate-core-5.4.15.Final.jar!/:5.4.15.Final]
    at org.hibernate.tool.schema.internal.AbstractSchemaMigrator.doMigration(AbstractSchemaMigrator.java:114) ~[hibernate-core-5.4.15.Final.jar!/:5.4.15.Final]
    at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.performDatabaseAction(SchemaManagementToolCoordinator.java:184) ~[hibernate-core-5.4.15.Final.jar!/:5.4.15.Final]
    at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.process(SchemaManagementToolCoordinator.java:73) ~[hibernate-core-5.4.15.Final.jar!/:5.4.15.Final]
    at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:314) ~[hibernate-core-5.4.15.Final.jar!/:5.4.15.Final]
    at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:468) ~[hibernate-core-5.4.15.Final.jar!/:5.4.15.Final]
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:1249) ~[hibernate-core-5.4.15.Final.jar!/:5.4.15.Final]
    at org.springframework.orm.jpa.vendor.SpringHibernateJpaPersistenceProvider.createContainerEntityManagerFactory(SpringHibernateJpaPersistenceProvider.java:58) ~[spring-orm-5.2.6.RELEASE.jar!/:5.2.6.RELEASE]
    at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:365) ~[spring-orm-5.2.6.RELEASE.jar!/:5.2.6.RELEASE]
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.buildNativeEntityManagerFactory(AbstractEntityManagerFactoryBean.java:391) ~[spring-orm-5.2.6.RELEASE.jar!/:5.2.6.RELEASE]
    at java.base/java.util.concurrent.FutureTask.run(FutureTask.java:264) ~[na:na]
    at java.base/java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1128) ~[na:na]
    at java.base/java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:628) ~[na:na]
    at java.base/java.lang.Thread.run(Thread.java:830) ~[na:na]
Caused by: org.postgresql.util.PSQLException: ERROR: syntax error at or near "user"

我相信我遵循了正确链接的 Spring 指南，所以我不确定为什么表名会导致语法错误。

谁能帮忙？

更新：我加了

spring.jpa.hibernate.naming.implicit-strategy=org.hibernate.boot.model.naming.ImplicitNamingStrategyLegacyJpaImpl
spring.jpa.hibernate.naming.physical-strategy=org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl

到 application.properties。

【问题讨论】：

你试过了吗 - stackoverflow.com/questions/40399460/…
@SiriusCurious，是的，我确实尝试了 Table 注释，但它仍然尝试创建一个名为 user 的表！我也试过重命名它，但仍然没有用！

标签： java spring

【解决方案1】：

user 是 PostgreSQL 中的保留字 (https://postgresql.org/docs/8.1/sql-keywords-appendix.html)。这不会发生在您的“集成测试”中，因为它们可能使用内存中的测试数据库。

你能用类似的东西注释你的User 实体吗：

@Entity
@Table(name="my_users_table")
public class User {
// ...
}

【讨论】：

嗨，我刚试过这个，但 PostgreSQL 仍然尝试创建一个名为 user 的表，这很奇怪！

【解决方案2】：

我发现问题出在 Recipe.java 中，在 @JoinTable 中，我仍然将表称为“用户”

【讨论】：