JDK1.6的单向加密:
BASE64Encoder base64Encoder = new BASE64Encoder();
String newStr =java.util.Base64.getEncoder().encodeToString(srcStr.getBytes());
JDK1.8的单向加密:
String newStr = Base64.getEncoder().encodeToString(md5Bytes);
BASE64Encoder base64Encoder = new BASE64Encoder();
String newStr =java.util.Base64.getEncoder().encodeToString(srcStr.getBytes());
JDK1.8的单向加密:
String newStr = Base64.getEncoder().encodeToString(md5Bytes);
2018-12-25
String otpCode = RandomStringUtils.randomNumeric(6);
用apache的commons工具包也可以获取随机数
用apache的commons工具包也可以获取随机数
2018-12-24
@麻了一一一
你好,的确,对应没有走到controller层面内的错误补获不到,可以采用controlleradvice注解另外定义一个类,然后实现一个方法,类似于这样
@ControllerAdvice
public class Test {
@ExceptionHandler(value = Exception.class)
@ResponseBody
public CommonReturnType error(HttpServletRequest request, HttpServletResponse response, Exception e)
}
你好,的确,对应没有走到controller层面内的错误补获不到,可以采用controlleradvice注解另外定义一个类,然后实现一个方法,类似于这样
@ControllerAdvice
public class Test {
@ExceptionHandler(value = Exception.class)
@ResponseBody
public CommonReturnType error(HttpServletRequest request, HttpServletResponse response, Exception e)
}
2018-12-22
public CommonReturnType getUser(@RequestParam(name = "id")Integer id)运行没有问题。
但是通常写:@RequestParam(value = "id")Integer id
但是通常写:@RequestParam(value = "id")Integer id
2018-12-21