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Java Spring Boot+Hibernate+Postgres-不创建表_Java_Hibernate_Postgresql_Jpa_Spring Boot - Fatal编程技术网

Java Spring Boot+Hibernate+Postgres-不创建表

Java Spring Boot+Hibernate+Postgres-不创建表,java,hibernate,postgresql,jpa,spring-boot,Java,Hibernate,Postgresql,Jpa,Spring Boot,我正在尝试基于实体生成模式表。应用程序正确启动,生成SQL,但没有结果-没有创建任何表。发生了什么?我在纯SpringMVC+HibernateJPA中使用了相同的设置,没有SpringBoot,一切都正常工作 以下是我的pom.xml: <?xml version="1.0" encoding="UTF-8"?> <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/20

我正在尝试基于实体生成模式表。应用程序正确启动,生成SQL,但没有结果-没有创建任何表。发生了什么?我在纯SpringMVC+HibernateJPA中使用了相同的设置,没有SpringBoot,一切都正常工作

以下是我的pom.xml:

<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
    <modelVersion>4.0.0</modelVersion>

    <groupId>com.agileplayers</groupId>
    <artifactId>applicationname</artifactId>
    <version>0.0.1-SNAPSHOT</version>
    <packaging>jar</packaging>

    <name>ApplicationName</name>
    <description>Application Name</description>

    <parent>
        <groupId>org.springframework.boot</groupId>
        <artifactId>spring-boot-starter-parent</artifactId>
        <version>1.3.5.RELEASE</version>
        <relativePath/> <!-- lookup parent from repository -->
    </parent>

    <properties>
        <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
        <java.version>1.8</java.version>
    </properties>

    <dependencies>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-actuator</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-data-jpa</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-security</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-web</artifactId>
        </dependency>

        <dependency>
            <groupId>org.postgresql</groupId>
            <artifactId>postgresql</artifactId>
            <!--<scope>runtime</scope>-->
            <version>9.4-1201-jdbc41</version>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-test</artifactId>
            <scope>test</scope>
        </dependency>
    </dependencies>

    <build>
        <plugins>
            <plugin>
                <groupId>org.springframework.boot</groupId>
                <artifactId>spring-boot-maven-plugin</artifactId>
            </plugin>
        </plugins>
    </build>
</project>
BaseEntity.java:

package com.agileplayers.applicationname.core.domain;

import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.MappedSuperclass;
import java.util.Date;

@MappedSuperclass
public class BaseEntity {
    @Id
    @GeneratedValue
    private int id;
    private Date createdOn;
    public String description;

    public BaseEntity() {
    }

    public int getId() {
        return id;
    }

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

    public Date getCreatedOn() {
        return createdOn;
    }

    public void setCreatedOn(Date createdOn) {
        this.createdOn = createdOn;
    }

    public String getDescription() {
        return description;
    }

    public void setDescription(String description) {
        this.description = description;
    }
}
扩展BaseEntity-Entry.java的实体示例:

package com.agileplayers.applicationname.core.domain;

import javax.persistence.Entity;
import javax.persistence.ManyToOne;
import java.util.Date;

@Entity
public class Entry  extends BaseEntity{

    private String type;

    @ManyToOne
    private Account account;

    public Entry() {
    }

    public String getType() {
        return type;
    }

    public void setType(String type) {
        this.type = type;
    }

    public Account getAccount() {
        return account;
    }

    public void setAccount(Account account) {
        this.account = account;
    }
}

我已经解决了这个问题。应该有spring.jpa.properties.hibernate.default\u schema=schemaName,而不是spring.datasource.schema=schemaName

在本例中,生成表所需的一组必要属性如下:

spring.datasource.url = jdbc:postgresql://localhost:5432/postgres
spring.jpa.properties.hibernate.default_schema = schemaName
spring.datasource.username = userName
spring.datasource.password = userPassword
spring.jpa.hibernate.ddl-auto = create-drop
spring.jpa.show-sql = true

我已经解决了这个问题。应该有spring.jpa.properties.hibernate.default\u schema=schemaName,而不是spring.datasource.schema=schemaName

在本例中,生成表所需的一组必要属性如下:

spring.datasource.url = jdbc:postgresql://localhost:5432/postgres
spring.jpa.properties.hibernate.default_schema = schemaName
spring.datasource.username = userName
spring.datasource.password = userPassword
spring.jpa.hibernate.ddl-auto = create-drop
spring.jpa.show-sql = true

默认情况下,postgres使用公共模式名。spring.jpa.properties.hibernate.default_schema-对我来说就是这样!!10xBy default postgres使用一个公共模式名。spring.jpa.properties.hibernate.default_schema-我就是这样!!10倍