Xml处理——快速定位Xml节点和获取数据

来源：互联网发布：尤克里里淘宝网编辑：程序博客网时间：2024/05/20 00:38

场景一：快速定位到某一个具体的节点，并获取节点数据

public static Node getNode(Document doc, String path) {if (doc != null) {if (doc.selectSingleNode(path) != null) {return doc.selectSingleNode(path);}}return null;}

场景二：获取某个列表类型的节点的数据

public static List<Node> getNodeList(Document doc, String path) {if (doc != null) {if (doc.selectNodes(path) != null) {return doc.selectNodes(path);}}return null;}

场景三：获取指定节点的数据，并将数据注入到指定的JavaBean中（前提是JavaBean已经进行了注解的处理）

调用示例：

getXmlNode("xml格式的string类型",“指定节点的路径”,"想要转换成的javaBean类","节点替换前的名称","节点替换后的名称")

getXmlNode(xml, "//item/degreeInfo",Student.class,"degreeInfo","personBaseInfo");

public static Object getXmlNode(String xml, String path, Class<?> cls, String... jss) {Document doc;String strXml = null;try {doc = DocumentHelper.parseText(xml);if (doc != null) {if (doc.selectSingleNode(xpath) != null) {strXml = doc.selectSingleNode(path).asXML();}}if (jss != null && jss.length >= 2 && StringUtils.hasLength(strXml)) {return XmlBase.xmlToBean(cls, strXml.replace(jss[0], jss[1]));} else if (StringUtils.hasLength(strXml)) {return XmlBase.xmlToBean(cls, strXml);}} catch (DocumentException e) {e.printStackTrace();}return null;}

场景四：获取xml某列表节点数据，并注入到相应的JavaBean中（前提是JavaBean已经进行了注解的处理）

getXmlNode("xml格式的string类型",“指定节点的路径”,JavaBean列表对象,"想要转换成的javaBean类","节点替换前的名称","节点替换后的名称")

getXmlNode(xml, "//item/degreeInfo",list,Student.class,"degreeInfo","personBaseInfo");

public static void getXmlNode(String xml, String xpath,List list, Class<?> cls, String... jss) {Document doc;String strXml = null;try {doc = DocumentHelper.parseText(xml);if (doc != null) {if (doc.selectNodes(xpath) != null) {List<Node> lst= doc.selectNodes(xpath);for(Node l:lst){strXml=l.asXML();if (jss != null && jss.length >= 2 && StringUtils.hasLength(strXml)) {list.add(XmlBase.xmlToBean(cls, strXml.replace(jss[0], jss[1])));}else{list.add(XmlBase.xmlToBean(cls, strXml));}}}}} catch (DocumentException e) {e.printStackTrace();}}

1 0