在JPA查询过程中无法解析实体属性

Question

我正在尝试查询hibernate表中的RunEntity。查询中的第一个where子句搜索testName=传递值testName的RunEntities。在stacktrace中，它提到在RunEntity中找不到testname类型的匹配项，但是RunEntity对象显式地有一个名为testname的字符串，带有setter和getter以及@Column符号。

表格设置

CREATE TABLE RunEntity (ID INTEGER IDENTITY,TestNumber INTEGER NOT NULL, TestName varchar(50) NOT NULL, ENVIRONMENT VARCHAR(50) NOT NULL, Source VARCHAR(50), Date TIMESTAMP, RESULTFILES BLOB);

查询

@Query("SELECT r FROM RunEntity r WHERE r.testName = :testname AND r.testNumber = :testnumber AND r.environment = :environment AND r.source = :source")
public List<RunEntity> findByNameNumberEnvironmentSource(
        @Param("testname") String testname,
        @Param("testnumber") int testnumber,
        @Param("environment") String environment,
        @Param("source") String source);

实体

package com.web_application;

 import java.sql.Timestamp;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.Table;
import javax.persistence.Lob;

@Entity
@Table(name = "TESTRUNS")
public class RunEntity {

private int ID;
private int testNumber;
private String testName;
private String environment;
private String source;
private String passOrFail;
private Timestamp date;
private byte[] resultFiles;


@Id
@Column(name = "ID")
@GeneratedValue
public int getID()
{
    return this.ID;
}
public void setID(int ID){this.ID = ID;}

@Column(name="TestNumber")
public int getTestNumber()
{
    return this.testNumber;
}
public void setTestNumber(int testNum){this.testNumber = testNum;}

@Column(name="TestName")
public String testName()
{
    return this.testName;
}
public void setTestName(String testName){this.testName = testName;}

@Column(name="Environment")
public String getEnvironment()
{
    return this.environment;
}
public void setEnvironment(String enviro){this.environment = enviro;}

@Column(name="Source")
public String getSource()
{
    return this.source;
}
public void setSource(String src){this.source = src;}

@Column(name="PassOrFail")
public String getPassOrFail()
{
    return this.passOrFail;
}
public void setPassOrFail(String pOrF){this.passOrFail = pOrF;}

@Column(name="Date")
public Timestamp getDate()
{
    return this.date;
}
public void setDate(Timestamp dates){this.date = dates;}

@Lob
@Column(name="ResultFiles")
public byte[] getResultFiles()
{
    return this.resultFiles;
}
public void setResultFiles(byte[] file){this.resultFiles = file;}

}

stacktrace的一部分

Caused by: org.hibernate.QueryException: could not resolve property: testname of: com.web_application.RunEntity [SELECT r FROM com.web_application.RunEntity r WHERE r.testname = :testname AND r.testNumber = :testnumber AND r.environment = :environment AND r.source = :source]
at org.hibernate.QueryException.generateQueryException(QueryException.java:137)
at org.hibernate.QueryException.wrapWithQueryString(QueryException.java:120)
at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:234)
at org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:158)
at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:126)
at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:88)
at org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:190)
at org.hibernate.internal.AbstractSessionImpl.getHQLQueryPlan(AbstractSessionImpl.java:301)
at org.hibernate.internal.AbstractSessionImpl.createQuery(AbstractSessionImpl.java:236)
at org.hibernate.internal.SessionImpl.createQuery(SessionImpl.java:1800)
at org.hibernate.jpa.spi.AbstractEntityManagerImpl.createQuery(AbstractEntityManagerImpl.java:328)
... 66 more

Answer 1

将您的testName（）getter更正为getTestName（）。您使用的是属性访问，必须遵守JavaBeans约定。

Answer 2

改变这个

 @Column(name="TestName")
public String testName()
{
    return this.testName;
}

到

@Column(name="TestName")
public String getTestName()
{
    return this.testName;
}

属性访问命名约定很重要。例如，尝试使用IDE（Eclipse Getter Setter，而不是手动执行）