将异常转换为 JSON

json exception java

2022-09-03 14:03:42

在Java 7中，是否有可能将对象转换为Json？Exception

例：

try {      
    //something
} catch(Exception ex) {     
    Gson gson = new Gson();
    System.out.println(gson.toJson(ex));
}

答案 1

好吧，可以做这样的事情，尽管您不想转换异常对象本身，而是转换它其中的消息，使用您设计的格式，如下所示：

// […]
} catch (Exception ex) {
    Gson gson = new Gson();
    Map<String, String> exc_map = new HashMap<String, String>();
    exc_map.put("message", ex.toString());
    exc_map.put("stacktrace", getStackTrace(ex));
    System.out.println(gson.toJson(exc_map));
}

与定义 as 建议答案：getStackTrace()

public static String getStackTrace(final Throwable throwable) {
     final StringWriter sw = new StringWriter();
     final PrintWriter pw = new PrintWriter(sw, true);
     throwable.printStackTrace(pw);
     return sw.getBuffer().toString();
}

答案 2

从理论上讲，您还可以迭代堆栈跟踪中的元素，并生成如下所示的内容：

{ "NullPointerException" :
    { "Exception in thread \"main\" java.lang.NullPointerException",
        { 
          "Book.java:16" : "com.example.myproject.Book.getTitle",
          "Author.java:25" : "at com.example.myproject.Author.getBookTitles",
          "Bootstrap.java:14" : "at com.example.myproject.Bootstrap.main()"
        }
    },
  "Caused By" :
    { "Exception in thread \"main\" java.lang.NullPointerException",
        { 
          "Book.java:16" : "com.example.myproject.Book.getTitle",
          "Author.java:25" : "at com.example.myproject.Author.getBookTitles",
          "Bootstrap.java:14" : "at com.example.myproject.Bootstrap.main()"
        }
    }
}

您可以像这样迭代异常：

catch (Exception cause) {
    StackTraceElement elements[] = cause.getStackTrace();
    for (int i = 0, n = elements.length; i < n; i++) {       
        System.err.println(elements[i].getFileName()
            + ":" + elements[i].getLineNumber() 
            + ">> "
            + elements[i].getMethodName() + "()");
    }
}