将 sql 翻译成 JPQL

Question

我是 JPQL 的新手，所以我正在开发 Spring 启动应用程序，我有这个 SQL 查询部分：

select TOP 10 RFC_NUMBER, RECIPIENT_ID from [50004].SD_REQUEST S
INNER JOIN [50004].AM_EMPLOYEE E
--ON S.RECIPIENT_ID = E.EMPLOYEE_ID
WHERE E.AVAILABLE_FIELD_5  ='j.doe'  
        AND SD_REQUEST.STATUS_ID NOT IN (8,6,18,7,24)
        AND SD_REQUEST.RFC_NUMBER like 'I%'

到 JPQL。

我试过像这样做一个@Query：

   @Query("select x from Incident x  Left join x.recipient recip where recip.login=:login and (x.rfcnumber like :I_% or :rfcnumber = null )"
+ " and x.status NOT IN (8,6,18,7,24)")

但它只是 returns 该员工的所有 rfcnumber，我希望它只提取以字母 I 开头的 rfc 编号，我试着通过在当时的网络上搜索来做 CONCAT，同样的事情。我对此很陌生，所以我认为它会简单得多，我认为这只是语法问题。

非常感谢。

编辑（添加模型）：

import java.io.Serializable;
import java.sql.Date;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.Table;


@Entity
@Table(name="SD_REQUEST")
public class Incident implements Serializable {


private static final long serialVersionUID = -8235081865121541091L;


@Id
@Column(name="REQUEST_ID")
private Integer inid;

@ManyToOne
@JoinColumn(name = "SUBMITTED_BY")
private Employee sender;


@Column(name="RFC_NUMBER")
private String rfcnumber;

@Column(name="CREATION_DATE_UT")
private Date date;

@Column(name="DESCRIPTION")
private String description;

@Column(name="COMMENT")
private String comment;

@Column(name="STATUS_ID")
private Integer status;

@ManyToOne
@JoinColumn(name = "RECIPIENT_ID")
private Employee recipient;

public Incident()
{

}

public Incident(int inid,String rfcnumber,Date date,String description,String comment,Integer status)
{
    this.inid=inid; 
    this.rfcnumber= rfcnumber;
    this.date=date;
    this.description=description;
    this.comment=comment;
    this.status=status;

}

public int getInid() {
    return inid;
}

public void setInid(int inid) {
    this.inid = inid;
}

public String getRfcnumber() {
    return rfcnumber;
}

public void setRfcnumber(String rfcnumber) {
    this.rfcnumber = rfcnumber;
}

public Date getDate() {
    return date;
}

public void setDate(Date date) {
    this.date = date;
}

public String getDescription() {
    return description;
}

public void setDescription(String description) {
    this.description = description;
}

public String getComment() {
    return comment;
}

public void setComment(String comment) {
    this.comment = comment;
}

public Integer getStatus() {
    return status;
}

public void setStatus(Integer status) {
    this.status = status;
}

public Employee getSender() {
    return sender;
}

public void setSender(Employee sender) {
    this.sender = sender;
}

public Employee getRecipient() {
    return recipient;
}

public void setRecipient(Employee recipient) {
    this.recipient = recipient;
}

public void setInid(Integer inid) {
    this.inid = inid;
}

}

这是 Employee 的模型：

import javax.persistence.*;
import javax.persistence.Table;

import com.fasterxml.jackson.annotation.JsonIgnore;
import com.fasterxml.jackson.databind.annotation.JsonDeserialize;

@Entity
@JsonDeserialize(as =Employee.class)
@Table(name = "AM_EMPLOYEE")
public class Employee implements Serializable {

    private static final long serialVersionUID = 5071617893593927440L;

@Id
@Column(name = "EMPLOYEE_ID" )
private Integer id;


@Column(name = "LAST_NAME")
private String lastName;


@Column(name = "AVAILABLE_FIELD_5")
private String login;

@OneToMany(mappedBy="sender")
@JsonIgnore
private List<Incident> myCreatedIncidents;  
@OneToMany(mappedBy="recipient")
@JsonIgnore
private List<Incident> myOtherIncidents;

@Column(name = "PASSWD")
private String password;

public Employee() {
//super();
}

public Employee (String login,String password)
{

}
public Employee(Integer id, String lastName,String login, String password) {
    this.id = id;
    this.lastName = lastName;
    this.login = login;
    this.password = password;

}

/**
 * @return the id
 */
public Integer getId() {
    return id;
}

/**
 * @param id
 *            the id to set
 */
public void setId(Integer id) {
    this.id = id;
}

/**
 * @return the lastName
 */
public String getLastName() {
    return lastName;
}

/**
 * @param lastName
 *            the lastName to set
 */
public void setLastName(String lastName) {
    this.lastName = lastName;
}

/**
 * @return the login
 */
public String getLogin() {
    return login;
}

/**
 * @param login
 *            the login to set
 */
public void setLogin(String login) {
    this.login = login;
}

/**
 * @return the password
 */
public String getPassword() {
    return password;
}

/**
 * @param password
 *            the password to set
 */
public void setPassword(String password) {
    this.password = password;
}

public List<Incident> getMyCreatedIncidents() {
    return myCreatedIncidents;
}

public void setMyCreatedIncidents(List<Incident> myCreatedIncidents) {
    this.myCreatedIncidents = myCreatedIncidents;
}

public List<Incident> getMyOtherIncidents() {
    return myOtherIncidents;
}

public void setMyOtherIncidents(List<Incident> myOtherIncidents) {
    this.myOtherIncidents = myOtherIncidents;
}

}

Answer 1

硬编码字符

我认为你应该使用与 SQL 相同的方法：

like 'I%'

具体根据@http://www.objectdb.com/java/jpa/query/jpql/string#LIKE_-_String_Pattern_Matching_with_Wildcards_文章：

The percent character (%) - which matches zero or more of any character. Blockquote

所以请尝试以下操作：

   @Query("select x from Incident x  Left join x.recipient recip where recip.login=:login and (x.rfcnumber like 'I%' or :rfcnumber = null )"
+ " and x.status NOT IN (8,6,18,7,24)"

)

参数

如果您使用参数，请参阅解决方案@Parameter in like clause JPQL。

示例：

LIKE :code%

Whosebug 问题中还包含其他示例。

Answer 2

@查询("select x from Incident x where x.recipient.login=:login and (x.rfcnumber like I% or x.rfcnumber = null )" + " 和 x.status 不在 (8,6,18,7,24))"

试试这个查询

将 sql 翻译成 JPQL

translate sql to JPQL

jpa

jpql

spring-boot

硬编码字符

参数