"JSON parse error: Cannot deserialize value of type `int` from String \"true\": not a valid `int` value; nested exception is com.fasterxml.jackson.databind.exc.InvalidFormatException: Cannot deserialize value of type `int` from String \"true\": not a valid `int` value\n at [Source: (PushbackInputStream); line: 1, column: 13] (through reference chain: com.neusoft.ebip.dashboard.pojo.dto.role.RoleDTO[\"isEnable\"])"
时间: 2023-06-15 12:03:02 浏览: 4726
这个错误是因为你试图将一个字符串类型的"true"转换成一个整数类型的值。在JSON中,"true"是一个布尔类型的关键字,而不是整数类型。检查一下你的代码,看看是否在将JSON字符串转换成Java对象时出现了类型不匹配的问题。你可以使用Jackson等库来进行JSON转换,并且在Java对象中将"isEnable"字段声明为布尔类型,以避免这种错误。
相关问题
JSON parse error: Cannot deserialize value of type `java.sql.Date` from String "9.1":
JSON parse error: Cannot deserialize value of type `java.sql.Date` from String "9.1" means that there is an issue with parsing a JSON string into a `java.sql.Date` object. The error occurs because the value "9.1" cannot be converted to a `java.sql.Date` object.
To fix this error, you need to ensure that the JSON string contains a valid date format that can be parsed into a `java.sql.Date` object. The format should be in accordance with the expected format for dates in your application.
If the JSON string contains a date in a different format, you can either modify the JSON string to match the expected format or use a custom deserializer to handle the conversion.
JSON parse error: Cannot deserialize value of type `java.util.Date` from String "2023-01-01": not a valid representation (error: Failed to parse Date value '2023-01-01': Unparseable date: "2023-01-01"); nested exception is com.fasterxml.jackson.databind.exc.InvalidFormatException: Cannot deserialize value of type `java.util.Date` from String "2023-01-01": not a valid representation (error: Failed to parse Date value '2023-01-01': Unparseable date: "2023-01-01")
这个错误是因为在将 JSON 字符串转换为 Java 对象时,日期格式不匹配导致的。可以通过在对应的属性上使用 `@JsonFormat` 注解指定日期格式来解决这个问题。
例如,如果你的日期格式为 "yyyy-MM-dd",可以在 Java 对象的对应日期类型属性上添加注解:
```
@JsonFormat(pattern="yyyy-MM-dd")
private Date date;
```
这样,在将 JSON 字符串转换为 Java 对象时,就会按照指定的日期格式进行解析,避免出现日期格式不匹配的错误。
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