我正在使用Spring MVC 3.2 @RequestMapping和@ResponseBody作为REST服务。示例端点如下所示:
@RequestMapping(value = "query", method = RequestMethod.GET)
@ResponseBody
public Locations searchHandler(@RequestParam String q, HttpServletRequest request, HttpServletResponse response) {
...
发送错误的非现有端点请求或缺少GET参数q将显示Tomcat 7错误报告:
<html>
<head>
<title>Apache Tomcat/7.0.50 - Error report</title>
<style>
<!--H1 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:22px;} H2 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:16px;} H3 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:14px;} BODY {font-family:Tahoma,Arial,sans-serif;color:black;background-color:white;} B {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;} P {font-family:Tahoma,Arial,sans-serif;background:white;color:black;font-size:12px;}A {color : black;}A.name {color : black;}HR {color : #525D76;}-->
</style>
</head>
<body>
<h1>HTTP Status 404 - </h1>
<HR size="1" noshade="noshade">
<p>
<b>type</b> Status report
</p>
<p>
<b>message</b>
<u></u>
</p>
<p>
<b>description</b>
<u>The requested resource is not available.</u>
</p>
<HR size="1" noshade="noshade">
<h3>Apache Tomcat/7.0.50</h3>
</body>
</html>
如何禁用此错误页面。我只是想将错误消息作为内容而没有任何HTML或其他信息。
答案 0 :(得分:3)
好的,我找到了解决方案。我实现了一个异常处理程序,如上面的链接所述:
@ControllerAdvice
public class ErrorController {
/**
* .
* @param request .
* @param response .
* @throws Exception .
*/
@ExceptionHandler(Exception.class)
public void handleConflict(HttpServletRequest request, HttpServletResponse response, Exception e) throws Exception {
// If the exception is annotated with @ResponseStatus rethrow it and let
// the framework handle it - like the OrderNotFoundException example
// at the start of this post.
// AnnotationUtils is a Spring Framework utility class.
if (AnnotationUtils.findAnnotation(e.getClass(), ResponseStatus.class) != null) {
throw e;
}
response.setStatus(400);
response.getWriter().println(e.getMessage());
}
}
使用response.getWriter()...而不是使用response.sendError()非常重要。