相关疑难解决方法(0)

我需要从字符串缓冲区或字符串解析xml数据..i代码如下.这里就行了document doc =db.parse(eventXml )......抛出异常 - plesae找到下面的代码和异常.请帮助我

码

eventXml = strBuffer.toString();
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();

Document doc = db.parse(eventXml );

例外

 java.net.MalformedURLException: no protocol: <?xml version="1.0" encoding="UTF-8
    " standalone="yes"?>%0A<EventInfo xmlns="http://www.telenet.be/oms/event">%0A
     <TelenetEventInfo className="be.telenet.oms.events.OmsAsapJsrpNotifyEvent">%0A
           <SimpleFields>%0A            <SimpleField>%0A                <FieldName>C
    ompletion_Date_Time</FieldName>%0A                <FieldValue>04/08/2009 08:34:0
    1</FieldValue>%0A            </SimpleField>%0A            <SimpleField>%0A
              <FieldName>Originator</FieldName>%0A                <FieldValue>System
    </FieldValue>%0A            </SimpleField>%0A            <SimpleField>%0A
             <FieldName>Status</FieldName>%0A                <FieldValue>S</FieldVal
    ue>%0A            </SimpleField>%0A            <SimpleField>%0A                <
    FieldName>Workorder_ID</FieldName>%0A                <FieldValue>I00054132231-09
    8</FieldValue>%0A            </SimpleField>%0A        </SimpleFields>%0A
    <ArrayData>%0A            <ArrayNames>%0A                <ArrayName>Parameters</
    ArrayName>%0A                <ArrayFieldEntry>%0A                    <ArraySubFi
    eld>[0].Parameter_Name</ArraySubField>%0A                    <ArraySubFieldValue
    >WARNING_TEXT</ArraySubFieldValue>%0A                </ArrayFieldEntry>%0A
              <ArrayFieldEntry>%0A                    <ArraySubField>[0].Parameter_v
    alue</ArraySubField>%0A                    <ArraySubFieldValue>UnknownKeyExcepti
    on-Retrieving a webURL%0A                    for …

解析下面的XML时,首先url-malformed-exception解析代码而不是给出xml字符串我使用了这段代码

Document doc=dBuilder.parse(newInputSource(newByteArrayInputStream(xmlResponse.getBytes("utf-8"))));

根据这个链接

java.net.MalformedURLException:没有协议

现在我得到的节点值为null.我怎样才能克服这一点.在for循环的代码中,我已经提到过节点的空值即将来临

我使用以下代码:

try {
    DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
    DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
    Document doc = dBuilder.parse(new InputSource(new ByteArrayInputStream(xmlResponse.getBytes("utf-8"))));
    //read this - https://stackoverflow.com/questions/13786607/normalization-in-dom-parsing-with-java-how-does-it-work
    doc.getDocumentElement().normalize();
    System.out.println("Root element :" + doc.getDocumentElement().getNodeName());
    XPath xPath =  XPathFactory.newInstance().newXPath()
    String expression = "/GetMatchingProductForIdResponse/GetMatchingProductForIdResult/Products/Product"
    System.out.println(expression)
    NodeList nodeList = (NodeList) xPath.compile(expression).evaluate(doc, XPathConstants.NODESET)
    System.out.println("the size will be of the node list ${nodeList.getLength()}");
    for (int i = 0; i < nodeList.getLength(); i++) {
            System.out.println(nodeList.item(i).getNodeValue()+"the value coming will be "); // here i am …

我尝试了所有不同类型的组合，但我只是没有得到它想要的：

java.net.MalformedURLException: no protocol: <?xml version="1.0" encoding="UTF-8" standalone="yes"?><ds><ds>...
    at java.net.URL.<init>(URL.java:593)
    at java.net.URL.<init>(URL.java:490)
    at java.net.URL.<init>(URL.java:439)
    at com.sun.org.apache.xerces.internal.impl.XMLEntityManager.setupCurrentEntity(XMLEntityManager.java:620)
    at com.sun.org.apache.xerces.internal.impl.XMLVersionDetector.determineDocVersion(XMLVersionDetector.java:148)
    at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:805)
    at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:770)
    at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(XMLParser.java:141)
    at com.sun.org.apache.xerces.internal.parsers.DOMParser.parse(DOMParser.java:243)
    at com.sun.org.apache.xerces.internal.jaxp.DocumentBuilderImpl.parse(DocumentBuilderImpl.java:339)
    at javax.xml.parsers.DocumentBuilder.parse(DocumentBuilder.java:177)

我查看了其他问题，但我不明白我的 XML 有什么问题。

这是完整的 XML（删除了一些纯文本）：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<ds>
    <ds>
        <cue>ABGB : §§ 786 , 810 , 812 </cue>Die Kosten ... <cue>Anmerkung : </cue>
        ... <cue>Bestätigung von </cue>7 
        <Relation bewertung="1">Ob 56/10a </Relation>= Zak 
        <Relation bewertung="1">2010/773 , 440 </Relation>. 
    </ds>
</ds>

生产代码：

DocumentBuilderFactory icFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder icBuilder;

Document parse …