JSON parse error: Cannot deserialize value of type `com.baolong.javaweb.pojo.Dept` from Array value (token `JsonToken.START_ARRAY`); nested exception is com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot deserialize value of type `com.baolong.javaweb.pojo.Dept` from Array value (token `JsonToken.START_ARRAY`)

这个错误通常意味着在 JSON 解析过程中，期望得到的是一个对象，但实际上得到的是一个数组。这可能是因为你的 JSON 字符串格式不正确或者你的 Java 对象定义与 JSON 字符串不匹配导致的。你可以检查你的 JSON 字符串格式是否正确，或者检查你的 Java 对象定义是否正确。如果你的 JSON 字符串确实是一个数组，你需要将其转换为一个包含多个对象的 JSON 数组，然后使用相应的 Java 对象列表来解析它。

相关问题

JSON parse error: Cannot deserialize value of type java.time.LocalDateTime

This error occurs when trying to parse a JSON string that contains a value for a LocalDateTime object, but the parsing library is unable to convert it into a LocalDateTime object.

To resolve this error, you can try using a JSON parsing library that supports LocalDateTime objects, such as Jackson or Gson. Alternatively, you can modify the JSON string to use a format that can be easily converted to a LocalDateTime object, such as ISO-8601 format.

Here's an example of parsing a JSON string containing a LocalDateTime object using Jackson:

ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(new JavaTimeModule()); // register module to support LocalDateTime objects
String json = "{\"dateTime\":\"2021-10-25T10:15:30\"}";
MyClass myObj = mapper.readValue(json, MyClass.class);

In this example, the `JavaTimeModule` is registered to support LocalDateTime objects, and the JSON string is parsed into a `MyClass` object.

If the JSON string cannot be modified, you may need to manually parse the string and convert it to a LocalDateTime object using a DateTimeFormatter. Here's an example:

String json = "{\"dateTime\":\"2021-10-25T10:15:30\"}";
JsonObject jsonObject = JsonParser.parseString(json).getAsJsonObject();
String dateTimeString = jsonObject.get("dateTime").getAsString();
DateTimeFormatter formatter = DateTimeFormatter.ISO_LOCAL_DATE_TIME;
LocalDateTime dateTime = LocalDateTime.parse(dateTimeString, formatter);

In this example, the JSON string is manually parsed using `JsonParser`, the `dateTime` value is retrieved from the `JsonObject`, and a `DateTimeFormatter` is used to convert the string into a `LocalDateTime` object.

JSON parse error: Cannot deserialize value of type `java.util.Date

在Java中，当我们从JSON字符串中反序列化日期时，可能会遇到“JSON parse error: Cannot deserialize value of type `java.util.Date`”的错误。这是因为默认情况下，Jackson JSON库期望日期格式为“yyyy-MM-dd HH:mm:ss”，而我们提供的日期格式不匹配。为了解决这个问题，我们可以使用@JsonFormat注解来指定日期格式，例如：

@JsonFormat(shape = JsonFormat.Shape.STRING, pattern="yyyy-MM-dd'T'HH:mm:ss.SSS'Z'", timezone = "GMT")
private Date date;

在这个例子中，我们使用了ISO 8601日期格式，并指定了时区为GMT。这样，当我们从JSON字符串中反序列化日期时，就不会再出现“JSON parse error: Cannot deserialize value of type `java.util.Date`”的错误了。