type=Not Acceptable, status=406 用于生成 XML 的 Spring Rest 错误 [英] type=Not Acceptable, status=406 Error In Spring Rest For Producing XML

问题描述

我正在研究 Spring Rest 和我的 Spring Rest 应用程序，如果我尝试生成 json 一切正常.我可以在浏览器上看到它.没有错误.

I'm working on Spring Rest and in my Spring Rest app, if I try to produce json everything is OK. I can see it on browser. There is no error.

但是如果我想生成 XML，我会使用生成 = "application/xml" 或生成=MediaType.TEXT_XML_VALUE 和我收到此错误:

But if I want to produce XML, I use produces = "application/xml" or produces=MediaType.TEXT_XML_VALUE and I getting this error:

Whitelabel Error Page

This application has no explicit mapping for /error, so you are seeing this as a fallback.

Sun Oct 23 18:30:51 EEST 2016
There was an unexpected error (type=Not Acceptable, status=406).
Could not find acceptable representation

我的其余代码是:

package getExample;

import java.util.ArrayList;
import java.util.List;

import org.springframework.http.MediaType;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.bind.annotation.RestController;

import pojo.Address;
import pojo.Person;

@RestController
public class GetExampleController {

    @RequestMapping(value = "getExample",method=RequestMethod.GET,produces=MediaType.TEXT_XML_VALUE)
    public List<Person> getExample1(@RequestParam(value = "personId", defaultValue = "0") String id) {
        List<Person> personList = new ArrayList<>();
        Person person1 = new Person("1", "ilkay", "günel",
                new Address("Cennet Mah.", "K.Çekmece", "İstanbul", "TÜRKİYE"));
        personList.add(person1);

        Person person2 = new Person("2", "alican", "akkuş",
                new Address("Cennet Mah.", "K.Çekmece", "İstanbul", "TÜRKİYE"));
        personList.add(person2);

        Person person3 = new Person("3", "mustafa", "demir",
                new Address("Cennet Mah.", "K.Çekmece", "İstanbul", "TÜRKİYE"));
        personList.add(person3);

        if (id.equals("0")) {
            return personList;
        }
        else {
            return personList.subList(Integer.parseInt(id)-1, Integer.parseInt(id));
        }
    }
}

错误是什么?为什么我可以得到 XML 输出?我该如何解决这个问题?

What is the error? Why can I get XML output? How can I solve this?

推荐答案

需要添加jackson-dataformat-xml的依赖:

<dependency>
    <groupId>com.fasterxml.jackson.dataformat</groupId>
    <artifactId>jackson-dataformat-xml</artifactId>
</dependency>

否则，您可以使用 JAXB 注释来注释您的 bean.

Otherwise, you may annotate your bean with JAXB annotations.

如果您有 Jackson XML 扩展名 (jackson-dataformat-xml)类路径，它将用于呈现 XML 响应和非常相同的我们用于 JSON 的示例会起作用.

If you have the Jackson XML extension (jackson-dataformat-xml) on the classpath, it will be used to render XML responses and the very same example as we used for JSON would work.

...

如果 Jackson 的 XML 扩展不可用，JAXB(默认提供)在 JDK 中)将被使用，并有额外的要求[你的班级] 注释为@XmlRootElement...

If Jackson’s XML extension is not available, JAXB (provided by default in the JDK) will be used, with the additional requirement to have [your class] annotated as @XmlRootElement...

...

要让服务器呈现 XML 而不是 JSON，您可能必须发送Accept: text/xml 标头(或使用浏览器).

To get the server to render XML instead of JSON you might have to send an Accept: text/xml header (or use a browser).

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