获得Java异常的堆栈信息

来源:互联网 发布:js实现动态树形结构 编辑:程序博客网 时间:2024/05/17 03:47
public static String getExceptionStack(Exception e) {        StackTraceElement[] stackTraceElements = e.getStackTrace();        String result = e.toString() + "\n";        for (int index = stackTraceElements.length - 1; index >= 0; --index) {                result += "at [" + stackTraceElements[index].getClassName() + ",";                result += stackTraceElements[index].getFileName() + ",";                result += stackTraceElements[index].getMethodName() + ",";                result += stackTraceElements[index].getLineNumber() + "]\n";        }        return result;}

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