将异常转换为JSON

一尘不染

将异常转换为JSON

json

在Java 7中是否可以将Exception对象转换为Json？

例：

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

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2020-07-27

共1个答案

一尘不染

从理论上讲，您还可以遍历堆栈跟踪中的元素并生成如下内容：

{ "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()"
        }
    }
}

您可以像这样遍历异常：

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() + "()");
    }
}

2020-07-27