spring 如何在jaxb生成的xml字符串中转义双引号

wfveoks0 于 2022-12-10 发布在 Spring

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基本上，我需要构建一个xml文件，然后将xml请求提交给restTemplate post操作，如下所示：

final RestTemplate restTemplate1 = new RestTemplate();
        List<HttpMessageConverter<?>> messageConverters = new ArrayList<>();
        MappingJackson2HttpMessageConverter converter = new MappingJackson2HttpMessageConverter();
        converter.setSupportedMediaTypes(Collections.singletonList(MediaType.APPLICATION_XML));
        messageConverters.add(converter);
        restTemplate1.setMessageConverters(messageConverters);
    HttpHeaders headers = new HttpHeaders();
    headers.setBasicAuth("okmAdmin", "admin");
    headers.setContentType(MediaType.APPLICATION_XML);

    JAXBContext context = null;
    context = JAXBContext.newInstance(Folder.class);
    Marshaller mar = null;
    mar = context.createMarshaller();
    mar.setProperty("jaxb.encoding", "Unicode");
    mar.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.FALSE);

    //Print XML String to Console
    StringWriter sw = new StringWriter();

    Folder folder = new Folder();
    folder.setPath("/okm:root/test4");

    //Write XML to StringWriter
    mar.marshal(folder, sw);

    String xmlContent = sw.toString();

    System.out.println(xmlContent);

    HttpEntity<String> request = new HttpEntity<String>(xmlContent, headers);

    ResponseEntity<Folder> response = restTemplate1.postForEntity("http://localhost:8080/OpenKM/services/rest/folder/create", request, Folder.class);

生成的xml如下：

<?xml version="1.0" encoding="Unicode" standalone="yes"?><folder><path>/okm:root/test4</path><permissions>0</permissions><subscribed>false</subscribed><hasChildren>false</hasChildren><style>0</style></folder>

但是，当resttemplate提交xml时，我得到了以下错误：

2022-12-05 16:50:21.121 ERROR 15764 --- [nio-8082-exec-1] o.a.c.c.C.[.[.[/].[dispatcherServlet]    : Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Request processing failed; nested exception is org.springframework.web.client.HttpClientErrorException$BadRequest: 400 : "JAXBException occurred : Unexpected character '"' (code 34) in prolog; expected '<'<EOL> at [row,col {unknown-source}]: [1,1]. "] with root cause

根据错误消息，我需要转义双引号。
我怎样才能摆脱它呢？

spring

来源：https://stackoverflow.com/questions/74688774/how-to-escape-double-quote-in-xml-string-generated-by-jaxb

2条答案

按热度按时间

w51jfk4q1#

尝试使用StringEscapeUtils.escapeXml（），如下所示：

HttpEntity<String> request = new HttpEntity<String>(StringEscapeUtils.escapeXml(xmlContent), headers);

赞(0）回复(0）举报 2022-12-10

x6yk4ghg2#

我设法通过更改以下代码来解决此问题：删除以下内容：

MappingJackson2HttpMessageConverter converter = new MappingJackson2HttpMessageConverter();

添加了以下内容：

messageConverters.add(new StringHttpMessageConverter());

我没有将对象转换为Folder，而是使用String：

ResponseEntity<String> response = restTemplate1.postForEntity("http://localhost:8080/OpenKM/services/rest/folder/create", request, String.class);

完整的代码如下：

final RestTemplate restTemplate1 = new RestTemplate();
        List<HttpMessageConverter<?>> messageConverters = new ArrayList<>();
        messageConverters.add(new StringHttpMessageConverter());

    restTemplate1.setMessageConverters(messageConverters);

        HttpHeaders headers = new HttpHeaders();
        headers.setBasicAuth("okmAdmin", "admin");
        headers.setContentType(MediaType.APPLICATION_XML);

        JAXBContext context = null;
        context = JAXBContext.newInstance(Folder.class);
        Marshaller mar = null;
        mar = context.createMarshaller();

        mar.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.FALSE);

        //Print XML String to Console
        StringWriter sw = new StringWriter();

        Folder folder = new Folder();
        folder.setPath("/okm:root/test23");

        //Write XML to StringWriter
        mar.marshal(folder, sw);

        HttpEntity<String> request = new HttpEntity<String>(xmlString, headers);

        ResponseEntity<String> response = restTemplate1.postForEntity("http://localhost:8080/OpenKM/services/rest/folder/create", request, String.class);

        System.out.println(response);

    }

我补充这一点的情况下，如果别人有同样的问题。
谢谢

赞(0）回复(0）举报 2022-12-10

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spring 如何在jaxb生成的xml字符串中转义双引号

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