在Firebase中制作POJO时,可以使用ServerValue.TIMESTAMP吗？

Question

当您创建一个简单的旧Java对象,该对象旨在从Firebase序列化和反序列化时,有没有办法使用该ServerValue.TIMESTAMP值？

例如,假设我想要一个对象,其中一个属性是最后一次编辑,并且您想要使用该ServerValue.TIMESTAMP值.

在POJO类中,您可能具有:

private String timeLastChanged;

要么

private Map<String, String> timeLastChanged;

在第一个例子中String,我遇到了设置问题timeLastChange = ServerValue.TIMESTAMP;,因为ServerValue.TIMESTAMP是Map.

在第二个例子中,Map<String, String>我得到了一个"未能去抖"的错误,因为它无法正确地将数据库中存储的long反序列化为一个Map<String, String>.这有什么工作吗？

Answer 1

2016年12月27日更新

正如许多人提到的那样,为@Exclude切换了@JsonIgnore .

我终于想出了一个灵活的解决方案来处理Dates和ServerValue.TIMESTAMP.这是Ivan V,Ossama和puf的例子.

我无法想出一个办法来对付之间的转换long和HashMap<String, String>,但如果你窝在一个更通用的属性HashMap<String, Object>就可以进入该数据库可作为单一的长值("日期","1443765561874")或为ServerValue.TIMESTAMP哈希映射("日期",{".sv","servertime"}).然后当你把它拉出来时,它总是一个HashMap("date","some long number").然后,您可以使用@JsonIgnore @Exclude批注在POJO类中创建辅助方法(意味着Firebase将忽略它,而不将其视为用于序列化到数据库或从数据库序列化的方法),以便从返回的HashMap轻松获取long值.在您的应用中使用.

下面是POJO类的完整示例:

import com.google.firebase.database.Exclude;
import com.firebase.client.ServerValue;

import java.util.HashMap;
import java.util.Map;

public class ExampleObject {
    private String name;
    private String owner;
    private HashMap<String, Object> dateCreated;
    private HashMap<String, Object> dateLastChanged;

    /**
     * Required public constructor
     */
    public ExampleObject() {
    }

    public ExampleObject(String name, String owner, HashMap<String,Object> dateCreated) {
        this.name = name;
        this.owner = owner;
        this.dateCreated = dateCreated;

        //Date last changed will always be set to ServerValue.TIMESTAMP
        HashMap<String, Object> dateLastChangedObj = new HashMap<String, Object>();
        dateLastChangedObj.put("date", ServerValue.TIMESTAMP);
        this.dateLastChanged = dateLastChangedObj;
    }

    public String getName() {
        return name;
    }

    public String getOwner() {
        return owner;
    }

    public HashMap<String, Object> getDateLastChanged() {
        return dateLastChanged;
    }

    public HashMap<String, Object> getDateCreated() {
      //If there is a dateCreated object already, then return that
        if (dateCreated != null) {
            return dateCreated;
        }
        //Otherwise make a new object set to ServerValue.TIMESTAMP
        HashMap<String, Object> dateCreatedObj = new HashMap<String, Object>();
        dateCreatedObj.put("date", ServerValue.TIMESTAMP);
        return dateCreatedObj;
    }

// Use the method described in https://stackoverflow.com/questions/25500138/android-chat-crashes-on-datasnapshot-getvalue-for-timestamp/25512747#25512747
// to get the long values from the date object.
    @Exclude
    public long getDateLastChangedLong() {

        return (long)dateLastChanged.get("date");
    }

    @Exclude
    public long getDateCreatedLong() {
        return (long)dateCreated.get("date");
    }

}

Answer 2

我想稍微改善一下Lyla的答案.首先,我想摆脱公共方法public HashMap<String, Object> getDateLastChanged() public HashMap<String, Object> getDateCreated().为此,您可以dateCreated使用@JsonProperty注释标记属性.另一种可能的方式这样做是要改变性质检测这样的:@JsonAutoDetect(fieldVisibility = Visibility.ANY, getterVisibility = Visibility.NONE, setterVisibility = Visibility.NONE)
第二,我不明白为什么我们需要把ServerValue.TIMESTAMP进HashMap,而我们只需将其保存为属性.所以我的最终代码是:

import com.fasterxml.jackson.annotation.JsonIgnore;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.firebase.client.ServerValue;

public class ShoppingList {
    private String listName;
    private String owner;
    @JsonProperty
    private Object created;

    public ShoppingList() {
    }

    public ShoppingList(String listName, String owner) {
        this.listName = listName;
        this.owner = owner;
        this.created = ServerValue.TIMESTAMP;
    }

    public String getListName() {
        return listName;
    }

    public String getOwner() {
        return owner;
    }

    @JsonIgnore
    public Long getCreatedTimestamp() {
        if (created instanceof Long) {
            return (Long) created;
        }
        else {
            return null;
        }
    }

    @Override
    public String toString() {
        return listName + " by " + owner;
    }

}

Answer 3

这些解决方案对我来说似乎有点困难,因为我不知道@JsonIgnore在做什么.我试图以一种简单的方式做到这一点,看起来很有效.

private Object dateLastChanged;

public Object getDateLastChanged() {
    return dateLastChanged;
}

为了获得人类可读的东西,我只需将返回值dateLastChanged Object转换为以下方法,方法是将其转换为Long.

static String convertTime(Long unixtime) {
    Date dateObject = new Date(unixtime);
    SimpleDateFormat dateFormatter = new SimpleDateFormat("dd-MM-yy hh:mmaa");
    return dateFormatter.format(dateObject);
}

欢迎就我的解决方案发表意见^^

Answer 4

您可以使用Firebase本机,而不是使用@JsonIgnore @Exclude.例如,我在一个类似的项目中工作,我的模型是这样的.

package com.limte.app.borrador_1.mModel;

import com.google.firebase.database.Exclude;
import com.google.firebase.database.ServerValue;

/**
 * Created by Familia on 20/12/2016.
 */

public class ChatItem {
    String chatName;
    Long creationDate;


    public ChatItem() {
    }

    public String getChatName() {
        return chatName;
    }

    public void setChatName(String chatName) {
        this.chatName = chatName;
    }

    public java.util.Map<String, String> getCreationDate() {
        return ServerValue.TIMESTAMP;
    }

    @Exclude
    public Long getCreationDateLong() {
        return creationDate;
    }
    public void setCreationDate(Long creationDate) {
        this.creationDate = creationDate;
    }

}

这段代码有效!在Firebase中,结果是: 结果