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[英]How to handle "org.springframework.http.converter.HttpMessageNotReadableException: JSON parse error" in Spring?
[英]Handle DefaultHandlerExceptionResolver : org.springframework.http.converter.HttpMessageNotReadableException: JSON parse error: Unexpected character
我收到以下錯誤消息
WARN 13712 --- [nio-8080-exec-9] .w.s.m.s.DefaultHandlerExceptionResolver : Resolved exception caused by handler execution: org.springframework.http.converter.HttpMessageNotReadableException:
JSON parse error: Unexpected character (',' (code 44)): expected a value; nested exception is com.fasterxml.jackson.core.JsonParseException: Unexpected character (',' (code 44)): expected a value at [Source: java.io.PushbackInputStream@1db33478; line: 2, column: 16]
我正在使用這樣的 spring 框架。
@GetMapping("/access")
public HashMap<String, Object> getAccess() {
HashMap<String, Object> response = new HashMap<String, Object>();
ArrayList<Access> list = new ArrayList<Access>();
BdConector bd = BdConector.getInstance();
list = bd.getAllAccess();
if (list != null) {
response.put("result", "200");
} else {
response.put("result", "500");
}
response.put("items", list);
return response;
}
我想在發生解析器異常時記錄 json body-content。
嘗試這樣的事情......
@ExceptionHandler(com.fasterxml.jackson.core.JsonParseException.class)
public ResponseEntity<Object> handleException(com.fasterxml.jackson.core.JsonParseException ex) {
LOG.error("Error parse json " + ex.getMessage());
...
}
不太確定如何提取無效 json 的原始字符串,也許它在異常的“getMessage()”中,如我所示。 無論如何,您可以調試“ex”歸檔並找到如何正確獲取它。
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