無法將整個 JSON 文件讀入 ArrayList（ArrayList 反序列化）

Question

正如標題所說，我無法將整個 JSON 文件反序列化為 ArrayList，更具體地說，我的代碼僅讀取文件中的第一項並忽略 Z65E8800B5C6800AAD896F888B2A6。

這是我的寫作方法：

private static void writeJSON(MyClass myClassObject) throws JsonGenerationException, JsonMappingException, IOException {
    ObjectMapper mapper = new ObjectMapper();
    PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("myclass.json", true)));
    mapper.writeValue(out, myClassObject);
}

每當我想序列化 object 時，我都會在我的主代碼中調用此方法，它工作正常。

這是我的閱讀方法：

private static void readFromFile() throws IOException {
    String jsonString = FileUtils.readFileToString(new File("myclass.json"), StandardCharsets.UTF_8);
    ObjectMapper mapper = new ObjectMapper();
    mapper.configure(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY, true);
    ArrayList<MyClass> myClassArray = mapper.readValue(jsonString, new TypeReference<ArrayList<MyClass>>() {});
}

在添加這部分之前

mapper.configure(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY, true);

和 MyClass class 中的默認構造函數，我遇到了一些異常，更具體地說是“無法從 START_OBJECT 令牌中反序列化 java.util.ArrayList 的實例”

但是，現在它可以正常工作，沒有任何錯誤或異常，但是通過打印 myClassArray ArrayList 我意識到它只包含第一個 object （及其值）。 試圖找出問題，我注意到如果我打印jsonString值，我會得到整個 JSON 文件（作為字符串）。

所以我猜這個問題與這部分代碼有關

ArrayList<MyClass> myClassArray = mapper.readValue(jsonString, new TypeReference<ArrayList<MyClass>>() {});

但我無法弄清楚。

MyClass class 看起來像這樣：

public class MyClass { 

private String name;
private String country;
private double features[];
private Date timestamp = new Date();


public City() {     //The empty constructor is used for the needs of the readFromFile method in Main class.
    
}

public City(String name, String country) {
    super();
    this.name = name;
    this.country = country;
}

public City(String name,String country,double[] features) {
    this.name = name;
    this.country = country;
    this.features = Arrays.copyOf(features, features.length);
}

public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public String getCountry() {
    return country;
}

public void setCountry(String country) {
    this.country = country;
}

public double[] getFeatures() {
    return Arrays.copyOf(features, features.length);
}

public void setFeatures(double[] features) {
    this.features = features;
}

public Date getTimestamp() {
    return timestamp;
}

public void setTimestamp(Date timestamp) {
    this.timestamp = timestamp;
}

和 JSON 文件現在（忽略零和數字，整個事情是半完成的）：

{"name":"John","country":"gr","features":[0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.7123809523809526,0.75,0.0],"timestamp":1637924593676}{"name":"Scott","country":"gb","features":[0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.6598639455782312,0.2,0.15601209271073035],"timestamp":1637924610010}
{"name":"Michael","country":"it","features":[0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.6877551020408165,0.75,0.06856164464370273],"timestamp":1638458784201}

Answer 1

您的 json 文件不包含有效的 json 數組。 它應該看起來像：

[ 
 { ... },
 { ... }
]

不是

{ ... }
{ ... }

我認為您將 json 寫為單個對象，而不是數組。

Answer 2

您的 JSON 文件無效。 您只有幾個 JSON 對象，但您需要將它們捆綁在一個 JSON 數組中，如下所示：

[
  {
    "name": "John",
    "country": "gr",
    "features": [
      0.0,
      0.0,
      0.0,
      0.0,
      0.0,
      0.0,
      0.0,
      0.7123809523809526,
      0.75,
      0.0
    ],
    "timestamp": 1637924593676
  },
  {
    "name": "Scott",
    "country": "gb",
    "features": [
      0.0,
      0.0,
      0.0,
      0.0,
      0.0,
      0.0,
      0.0,
      0.6598639455782312,
      0.2,
      0.15601209271073035
    ],
    "timestamp": 1637924610010
  },
  {
    "name": "Michael",
    "country": "it",
    "features": [
      0.0,
      0.0,
      0.0,
      0.0,
      0.0,
      0.0,
      0.0,
      0.6877551020408165,
      0.75,
      0.06856164464370273
    ],
    "timestamp": 1638458784201
  }
]

---

為了修正你的寫作方法，你需要為它提供一個MyClass對象的列表，而不是一次一個（你不需要 append 在使用這個解決方案寫作時）：

public class Main {
    public static void main(String[]args) throws IOException {
        List<MyClass> objectsToSerialize = new ArrayList<>();
        objectsToSerialize.add(new MyClass("Name1", "Country1", new double[] { 1.0 }));
        objectsToSerialize.add(new MyClass("Name2", "Country2", new double[] { 2.0 }));
        objectsToSerialize.add(new MyClass("Name3", "Country3", new double[] { 3.0 }));

        ObjectMapper mapper = new ObjectMapper();
        PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("myclass.json")));
        mapper.writeValue(out, objectsToSerialize);
    }
}

Answer 3

我假設您使用的映射器是 Jackson 的Object Mapper 。 在您的代碼行中：

ArrayList<MyClass> myClassArray = mapper.readValue(jsonString, new TypeReference<ArrayList<MyClass>>() {});

沒有真正的錯誤。 相反，您糾正先前錯誤的方法是有缺陷的：

mapper.configure(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY, true);

您只是繞過了最初的錯誤，問題是您的 JSON 格式不正確，因此無法將其轉換為集合。 檢查您的原始文件，或將字符串放入在線 JSON 格式化程序以驗證它是否有效。 這是一個潛在的鏈接： https://jsonlint.com/ 。