Question

我正在使用带有MySql数据库的Java spring MVC，而且我正在尝试使用REST api。
问题在于日期字段，这是我的模型：

import com.fasterxml.jackson.annotation.JsonAutoDetect;
import javax.persistence.*;
import java.util.Date;

@Entity
@Table(name = "patient_mesure")
@JsonAutoDetect
public class PatientMesure {
    @Id
    @Column(name = "id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;
    @Column(name = "mesure_patient")
    private float mesure_patient;
    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "mesure", nullable = false)
    private Mesure mesure;
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "patient", nullable = false)
    private Personne patient;
    @Column(name = "date")
    private Date date;
    @Column(name = "inserted", nullable = false)
    private Date inserted;
    @Column(name = "updated", nullable = false)
    private Date updated;

    // getters and setters
}

这是我的Rest Controller：

import com.eheio.spring.models.PatientMesure;
import com.eheio.spring.models.Personne;
import com.eheio.spring.services.PatientMesureService;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.web.bind.annotation.*;

import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpSession;
import java.util.List;

@RestController
public class PatientMesureRestController {

    private PatientMesureService patientMesureService;

    @Autowired
    public void setPatientMesureService(PatientMesureService patientMesureService) {
        this.patientMesureService = patientMesureService;
    }

    @GetMapping(value = "/getAllMesures")
    public List<PatientMesure> listAllUsers(HttpServletRequest httpServletRequest) {
        HttpSession httpSession = httpServletRequest.getSession(true);
        Personne personne = (Personne) httpSession.getAttribute("personne");
        return patientMesureService.findMesuresByPatient(personne.getId());
    }
}

我得到的结果：

{ id: 1, date: 1527801252000, inserted: 1527801252000, mesure_patient: 50, updated: 1527801252000}

如何获取日期字段的正确表单？像2018-05-31 21:14:12

这样的东西

Answer 1

设置自定义JsonSerialize并在那里格式化，SimpleDateFormat这样：

首先创建JsonDateSerializer类：

    year   X10.   X30.    X50.  X80.   DOY    NDVI
   <dbl>  <dbl>  <dbl>   <dbl> <dbl> <dbl>   <dbl>
 1 2013. -0.844 -0.459 -0.0144 0.583  247. -0.844 
 2 2013. -0.844 -0.459 -0.0144 0.583   96. -0.447 
 3 2013. -0.844 -0.459 -0.0144 0.583  202. -0.0144
 4 2013. -0.844 -0.459 -0.0144 0.583   59.  0.584 
 5 2014. -0.811 -0.403 -0.0136 0.623  128. -0.818 
 6 2014. -0.811 -0.403 -0.0136 0.623   37. -0.410 
 7 2014. -0.811 -0.403 -0.0136 0.623  187. -0.0136
 8 2014. -0.811 -0.403 -0.0136 0.623  278.  0.620 
 9 2015. -0.890 -0.494 -0.0332 0.646  280. -0.887 
10 2015. -0.890 -0.494 -0.0332 0.646  330. -0.488

然后在你的getter方法中：

import java.io.IOException;
import java.text.SimpleDateFormat;
import java.util.Date;
import org.codehaus.jackson.JsonGenerator;
import org.codehaus.jackson.JsonProcessingException;
import org.codehaus.jackson.map.JsonSerializer;
import org.codehaus.jackson.map.SerializerProvider;
import org.springframework.stereotype.Component;

@Component
public class JsonDateSerializer extends JsonSerializer<Date>{
    private static final SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss.S");

    @Override
    public void serialize(Date date, JsonGenerator gen, SerializerProvider provider)
    throws IOException, JsonProcessingException {
        String formattedDate = dateFormat.format(date);
        gen.writeString(formattedDate);
    }
}

另外，您使用的是Java 8，DateTimeFormatter是线程安全的（正如@Philippe Marschall指出的那样）：

@JsonSerialize(using=JsonDateSerializer.class)
public Date getDate() {
    return date;
}

Answer 2

您可以使用如下

import com.fasterxml.jackson.annotation.JsonFormat;


    @JsonFormat(shape = JsonFormat.Shape.STRING,  pattern = "yyyy-MM-dd HH:mm:ss.SS a")
    @Column(name = "date")
    private Date date;

    @JsonFormat(shape = JsonFormat.Shape.STRING,  pattern = "yyyy-MM-dd HH:mm:ss.SS a")
    @Column(name = "inserted", nullable = false)
    private Date inserted;

    @JsonFormat(shape = JsonFormat.Shape.STRING,  pattern = "yyyy-MM-dd HH:mm:ss.SS a")
    @Column(name = "updated", nullable = false)
    private Date updated;

如何使用Java和MySql在Json中获取正确的日期格式

2 个答案: