Question

我在春季靴子里有休息服务。我有一个帖子方法。在post方法中，我接受json。这是工作。但是当客户端向我发送userId字段的字符（短）时，我采取json解析错误。在异常之前我可以控制它吗？我想自定义错误消息，但异常消息太复杂了。

<%= f.submit foo.persisted? ? 'Update Foo' : 'Create Foo' %>

错误讯息是： JSON解析错误：无法从字符串“a”反序列化类型@RequestMapping(value = "/customers", method = RequestMethod.POST) public ResponseEntity<?> insertCustomer(@Valid @RequestBody InsertParams params) /* */} public class InsertParams() { private Short userId; }的值：不是有效的短值

Answer 1

我是这样做的。

@ExceptionHandler(HttpMessageNotReadableException.class)
public ResponseEntity<ExceptionResponse> 
handleHttpMessageNotReadable(HttpMessageNotReadableException ex) {

    String fieldName = "";

    String message = "";

    if (!(ex.getCause() instanceof JsonMappingException)) {
        return new ResponseEntity<ExceptionResponse>(HttpStatus.BAD_REQUEST);
    }

    JsonMappingException e = (JsonMappingException) ex.getCause();

    for (JsonMappingException.Reference reference : e.getPath()) {
        fieldName = reference.getFieldName();
    }

    if (fieldName.toLowerCase().contains("date"))
        message = fieldName + " format exception. Should be yyyy-MM-dd";
    else
        message = fieldName + " can not take characters.";

    ExceptionResponse er = new ExceptionResponse();
    er.setResult("Failure");
    er.setStatus(HttpStatus.BAD_REQUEST.value());
    er.setDescription(message);

    return new ResponseEntity<ExceptionResponse>(er, HttpStatus.BAD_REQUEST);

休息服务号验证

1 个答案: