[英]How to parse this XML SOAP response to a POJO?
我在将XML SOAP返回到相应的POJO类时遇到问题。 XML返回如下所示:
<env:Envelope xmlns:env="http://schemas.xmlsoap.org/soap/envelope/">
<env:Header></env:Header>
<env:Body>
<ns2:teste xmlns:ns2="http://teste.com.br/">
<retorno>
<codigoRetorno>000</codigoRetorno>
<descricao>Consulta Realizada com Sucesso</descricao>
<item>
<a>teste</a>
<b>teste</b>
<c>teste</c>
</item>
<item>
<a>teste</a>
<b>teste</b>
<c>teste</c>
</item>
</retorno>
</ns2:teste >
</env:Body>
</env:Envelope>
我尝试使用Jackson XMLmapper,但我不能让它在反序列化期间将'RETURN'节点视为ROOT元素。 它将'Envelope'节点视为ROOT节点。
我只需要提取返回节点并转换为我的pojo类。
另一个问题是'item'节点应该是集合的一部分,但是没有父节点对这些元素进行分组。
有没有人知道解析这种类型的xml的解析器?
您可以通过以下方式合并流式XML分析器(StAX)和XmlMapper
:
import java.io.StringReader;
import javax.xml.stream.XMLInputFactory;
import javax.xml.stream.XMLStreamReader;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;
public class Deser {
// @formatter:off
private static final String JSON = " <env:Envelope xmlns:env=\"http://schemas.xmlsoap.org/soap/envelope/\">\n" +
" <env:Header></env:Header>\n" +
" <env:Body>\n" +
" <ns2:teste xmlns:ns2=\"http://teste.com.br/\">\n" +
" <retorno>\n" +
" <codigoRetorno>000</codigoRetorno>\n" +
" <descricao>Consulta Realizada com Sucesso</descricao>\n" +
" <item>\n" +
" <a>teste</a>\n" +
" <b>teste</b>\n" +
" <c>teste</c>\n" +
" </item>\n" +
" <item>\n" +
" <a>teste</a>\n" +
" <b>teste</b>\n" +
" <c>teste</c>\n" +
" </item>\n" +
" </retorno>\n" +
" </ns2:teste >\n" +
" </env:Body>\n" +
"</env:Envelope>";
// @formatter:on
private static final String TARGET_ELEMENT = "retorno";
public static void main(String[] args) throws Exception {
XmlMapper xmlMapper = new XmlMapper();
xmlMapper.disable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES);
XMLInputFactory f = XMLInputFactory.newFactory();
XMLStreamReader sr = f.createXMLStreamReader(new StringReader(JSON));
while (sr.hasNext()) {
int type = sr.next();
if (type == XMLStreamReader.START_ELEMENT && TARGET_ELEMENT.equals(sr.getLocalName())) {
Retorno r = xmlMapper.readValue(sr, Retorno.class);
System.out.println(r.getDescricao());
}
}
}
}
class Retorno {
private int codigoRetorno;
private String descricao;
public int getCodigoRetorno() {
return codigoRetorno;
}
public void setCodigoRetorno(int codigoRetorno) {
this.codigoRetorno = codigoRetorno;
}
public String getDescricao() {
return descricao;
}
public void setDescricao(String descricao) {
this.descricao = descricao;
}
}
这会产生:
Consulta Realizada com Sucesso
根据需要调整代码,这只是为了证明如何让它做你需要的!
我发现最干净的解决方案是使用JSOUP:
private <T> T parseResponse(HttpEntity entity, Class<T> typeTarget) throws Exception {
try {
String xmlSoapResponse = EntityUtils.toString(entity, StandardCharsets.UTF_8);
String xmlRetorno = extractXmlElement(xmlSoapResponse, "retorno");
XmlMapper xmlMapper = new XmlMapper();
xmlMapper.disable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES);
xmlMapper.registerModule(new JavaTimeModule());
return xmlMapper.readValue(xmlRetorno.toString(), typeTarget);
} catch (Exception e) {
throw new Exception("Fail during parser", e);
}
}
private String extractXmlElement(String xmlString, String nodeTagNameElement) {
Document document = Jsoup.parse(xmlString, "", Parser.xmlParser());
document.outputSettings().prettyPrint(false);
Elements retorno = document.getElementsByTag(nodeTagNameElement);
return retorno.toString();
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.