Question

我是JPQL的新手，所以我正在使用Spring Boot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。

我尝试像这样做@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)")

但它只返回该员工的所有rfcnumber，我希望它只提取以字母I开头的rfc编号，我试着通过在网上搜索来做CONCAT，同样的事情。我是新手，所以我觉得它会更简单，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;
}

}

这是员工的模型：

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

我认为您应该使用与SQL中相同的内容：

喜欢＆＃39; I％＆＃39;

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

百分比字符（％） - 匹配任何字符的零个或多个。块引用

请尝试以下方法：

   @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。

<强>示例：

LIKE：代码％

stackoverflow问题中还包含其他示例。

Answer 2

@Query（＆＃34;从Incident x中选择x，其中x.recipient.login =：login和（x.rfcnumber，如I％或x.rfcnumber = null）＆＃34; +＆＃34;和x.status NOT IN（8,6,18,7,24））＆＃34;

尝试此查询

将sql翻译为JPQL

2 个答案:

硬编码字符

参数