【问题标题】:How to get specific elements from an file whose structure is XML-like in Java如何从Java中结构类似于XML的文件中获取特定元素
【发布时间】:2019-11-06 08:21:47
【问题描述】:

我有一个 .sic 文件,其结构类似于 XML,但不完全。我有一个Channel2 部分,我想在其中阅读一些元素。部分是这样的:

.
.
.
<SI name = "Channel2" type = "list">
           <SI name = "SecsPortConfig" type = "list">
              <SI name = "PortType" type = "string">'XXX'</SI>
              <SI name = "Protocol" type = "string">'XXX'</SI>
              <SI name = "Serial" type = "list">
                 <SI name = "Port" type = "int">'XXX'</SI>
                 <SI name = "Speed" type = "int">'XXXX'</SI>
              </SI>
              <SI name = "Socket" type = "list">
                 <SI name = "ConnectionMode" type = "string">'XXX'</SI>
                 <SI name = "LocalHost" type = "string">'XXX.XXX.XXX.XXX'</SI>
                 <SI name = "LocalPort" type = "int">'XXX'</SI>
                 <SI name = "RemoteHost" type = "string">'XXX.XXX.XXX'</SI>
                 <SI name = "RemotePort" type = "int">'XXX'</SI>
              </SI>
              <SI name = "HSMS" type = "list">
                 <SI name = "T5" type = "int">'XXX'</SI>
                 <SI name = "T6" type = "int">'XXX'</SI>
                 <SI name = "T7" type = "int">'XXX'</SI>
                 <SI name = "T8" type = "int">'XXX'</SI>
                 <SI name = "LinkTestTime" type = "int">'XXX'</SI>
              </SI>
              <SI name = "SECSI" type = "list">
                 <SI name = "T1" type = "int">'XXX'</SI>
                 <SI name = "T2" type = "int">'XXX'</SI>
                 <SI name = "T4" type = "int">'XXX'</SI>
                 <SI name = "RTY" type = "int">'XXX'</SI>
                 <SI name = "IsHost" type = "bool">'XXX'</SI>
                 <SI name = "IsMaster" type = "bool">'XXX'</SI>
                 <SI name = "InterleaveBlocks" type = "bool">'XXX'</SI>
              </SI>
              <SI name = "SECSII" type = "list">
                 <SI name = "DeviceID" type = "int">'XXX'</SI>
                 <SI name = "T3" type = "int">'XXX'</SI>
                 <SI name = "MultipleOpen" type = "bool">'XXX'</SI>
                 <SI name = "AutoDeviceID" type = "bool">'XXX'</SI>
              </SI>
              <SI name = "Log" type = "list">
                 <SI name = "LogCharError" type = "bool">'XXX'</SI>
                 <SI name = "LogCharEvent" type = "bool">'XXX'</SI>
                 <SI name = "LogCharReceive" type = "bool">'XXX'</SI>
                 <SI name = "LogCharSend" type = "bool">'XXX'</SI>
                 <SI name = "LogSecsIHsmsError" type = "bool">'XXX'</SI>
                 <SI name = "LogSecsIHsmsEvent" type = "bool">'XXX'</SI>
                 <SI name = "LogSecsIHsmsReceive" type = "bool">'XXX'</SI>
                 <SI name = "LogSecsIHsmsSend" type = "bool">'XXX'</SI>
                 <SI name = "LogSecsIIError" type = "bool">'XXX'</SI>
                 <SI name = "LogSecsIIEvent" type = "bool">'XXX'</SI>
                 <SI name = "LogSecsIIReceive" type = "bool">'XXX'</SI>
                 <SI name = "LogSecsIISend" type = "bool">'XXX'</SI>
              </SI>
           </SI>
           <SI name = "UseSeparateSECSLogFile" type = "bool">'XXX'</SI>
           <SI name = "Connected" type = "bool">'XXX'</SI>
           <SI name = "MessageFilters" type = "list">
              <SI name = "DeviceIDList" type = "list"/>
              <SI name = "StreamFunctionList" type = "list"/>
           </SI>
           <SI name = "SafeMessageFilters" type = "list">
              <SI name = "DeviceIDList" type = "list"/>
              <SI name = "StreamFunctionList" type = "list"/>
           </SI>
        </SI>
        .
        .
        .

如果它是一个 xml 文件,我可以解析它并读出元素,但是我该如何处理这种文件呢? 我想提取元素RemoteHostRemotePort。 我现在使用 BufferedReader 进行了尝试,并从文件中获取了 Section Channel2,并将此部分插入到字符串中,但是如何提取我想要的元素的特定值?我可能可以使用子字符串和其他一些字符串方法来做到这一点,但没有更简单的方法吗? 到目前为止,这是我的代码:

    File file = new File("C:\\Users\\but\\Desktop\\ExternalswPassThroughSrv.sic");

    int counter = 0;

    BufferedReader br = new BufferedReader(new FileReader(file));

    String cl;
    String finalString = "";
    while ((cl = br.readLine()) != null) {
        if (cl.contains("Channel2")) {
            counter = 63;
        }
        if(counter != 0){
            //System.out.println(cl);
            finalString += cl + "\n";
            counter--;
        }
    }
    System.out.println(finalString);

【问题讨论】:

  • "如何处理这种文件?" - 就像所有其他 xml 文件一样。您是否研究过解析此文件以及如何使用 xpath 语句查找元素?
  • 也许我是瞎子……但是,为什么这不是 XML 文件?
  • 您向我们展示的部分似乎是格式良好的 XML。我很困惑这个问题是关于什么的。
  • 它以 xml 格式形成,但它们不像普通 xml 那样被标记。使用 DOM 解析器时应该如何获取元素?我可以解析它,但像 getElementsByTagName 这样的方法无法读取元素。一个普通的 xml 文件如下所示: 'LokeshGuptaIndia . 。我要读取的文件总是以 SI 名​​称开头,所以我不知道如何访问元素。

标签: java xml file substring


【解决方案1】:

由于我们不知道整个文件是如何形成的:
即使不是完整的 XML 文档,您也可以从文件的其余部分中提取 XML-Fragment,并通过添加根元素将其转换为格式良好的 XML-Document。

之后,您可以将其解析为 Document 并使用 XPath 提取所需信息。

这是一些可以为您工作的示例 Java 代码(为了清楚起见,我没有包含 xml)

import org.w3c.dom.Document;
import org.xml.sax.InputSource;
import org.xml.sax.SAXException;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.transform.TransformerException;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathExpressionException;
import javax.xml.xpath.XPathFactory;
import java.io.IOException;
import java.io.StringReader;

public class ConvertXml {
    public static void main(String[] args) throws ParserConfigurationException, IOException, SAXException, TransformerException, XPathExpressionException {
        // Your XML-like content
        String xmlString = "xml here";

        // transform xml-Fragment into well-formed xml with root element
        String xmlStringWellformed = "<content>" + xmlString + "</content>";

        // parse well-formed xml
        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        DocumentBuilder builder = factory.newDocumentBuilder();
        Document document = builder.parse(new InputSource(new StringReader(xmlStringWellformed)));

        // build xpath expression
        String xPathRemoteHost = "//SI[@name='Channel2']/SI[@name='SecsPortConfig']/SI[@name='Socket']/SI[@name='RemoteHost']/text()";
        String xPathRemotePort = "//SI[@name='Channel2']/SI[@name='SecsPortConfig']/SI[@name='Socket']/SI[@name='RemotePort']/text()";
        XPath xPath = XPathFactory.newInstance().newXPath();

        // Use XPath for extraction
        String remoteHost = (String) xPath.compile(xPathRemoteHost).evaluate(document, XPathConstants.STRING);
        String remotePort = (String) xPath.compile(xPathRemotePort).evaluate(document, XPathConstants.STRING);

        System.out.println("RemoteHost: " + remoteHost);
        System.out.println("RemotePort: " + remotePort);
    }
}

来源: Baeldung - Intro to XPath with Java

【讨论】:

  • 感谢您的回答!我切换到 python 并解决了这个问题,但我必须用 jython 在 java 中实现 python 代码。有了这个,我可以在不使用 python 的情况下做到这一点。我现在试了一下,效果很好:)
【解决方案2】:
Document _myDoc = null;

LSInput input  = implLS.createLSInput();

input.setStringData(requestXML);

_myDoc = parser.parse(input);

SI = ((NodeList)_myDoc.getElementsByTagName("MessageFilters")).item(0).getFirstChild().getNodeValue();

您可以使用 getElementsByTagName 获取 XML 元素的节点值。但为此,您需要有不同的元素名称。这不是答案。这不是答案。只是一个提示。试试这个。

【讨论】:

    猜你喜欢
    • 1970-01-01
    • 2019-12-28
    • 2014-10-26
    • 1970-01-01
    • 2014-08-31
    • 1970-01-01
    • 1970-01-01
    相关资源
    最近更新 更多