Question

我真的很讨厌问题。我一直在寻找谷歌和stackoverflow的决心，试图解决它，但我尝试了很多解决方案，它仍然无法正常工作。

我想使用DocumentBuilder从XML文件中读取XML数据。

这是我的代码：

public void LoadFromFile(){

    try{
        //File fXmlFile = new File("fishlist.xml");
        DocumentBuilderFactory dbFactory = 
        DocumentBuilderFactory.newInstance();
        DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
        Document doc = dBuilder.parse("file:///C:/fishlist.xml"); // <--- there is exception run
        doc.getDocumentElement().normalize();


    }catch(Exception ex){
        ex.printStackTrace();
    }
}

当我尝试运行此方法时，它会显示异常：

W/System.err: java.io.FileNotFoundException: /C:/fishlist.xml: open failed: 
ENOENT (No such file or directory)
W/System.err:     at libcore.io.IoBridge.open(IoBridge.java:456)
W/System.err:     at java.io.FileInputStream.<init>(FileInputStream.java:76)
W/System.err:     at             
libcore.net.url.FileURLConnection.connect(FileURLConnection.java:123)
W/System.err:     at org.apache.harmony.xml.parsers.DocumentBuilderImpl.parse(DocumentBuilderImpl.jav
a:117)
W/System.err:     at 
javax.xml.parsers.DocumentBuilder.parse(DocumentBuilder.java:155)

W/System.err: Caused by: android.system.ErrnoException: open failed: ENOENT (No such file or directory)
W/System.err:     at libcore.io.Posix.open(Native Method)
W/System.err:     at libcore.io.BlockGuardOs.open(BlockGuardOs.java:186)
W/System.err:     at libcore.io.IoBridge.open(IoBridge.java:442)

是的。我有这个方向的文件。请帮忙。

Answer 1

要阅读的数据

<?xml version="1.0"?>
<company>
    <staff id="1001">
        <firstname>yong</firstname>
        <lastname>mook kim</lastname>
        <nickname>mkyong</nickname>
        <salary>100000</salary>
    </staff>
    <staff id="2001">
        <firstname>low</firstname>
        <lastname>yin fong</lastname>
        <nickname>fong fong</nickname>
        <salary>200000</salary>
    </staff>
</company>

尝试使用此代码的java代码

try {

    File fXmlFile = new File("/Users/mkyong/staff.xml");
    DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
    DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
    Document doc = dBuilder.parse(fXmlFile);

    //optional, but recommended
    //read this - http://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());

    NodeList nList = doc.getElementsByTagName("staff");

    System.out.println("----------------------------");

    for (int temp = 0; temp < nList.getLength(); temp++) {

        Node nNode = nList.item(temp);

        System.out.println("\nCurrent Element :" + nNode.getNodeName());

        if (nNode.getNodeType() == Node.ELEMENT_NODE) {

            Element eElement = (Element) nNode;

            System.out.println("Staff id : " + eElement.getAttribute("id"));
            System.out.println("First Name : " + eElement.getElementsByTagName("firstname").item(0).getTextContent());
            System.out.println("Last Name : " + eElement.getElementsByTagName("lastname").item(0).getTextContent());
            System.out.println("Nick Name : " + eElement.getElementsByTagName("nickname").item(0).getTextContent());
            System.out.println("Salary : " + eElement.getElementsByTagName("salary").item(0).getTextContent());

        }
    }
    } catch (Exception e) {
    e.printStackTrace();
    }

根元素：公司

Current Element :staff
Staff id : 1001
First Name : yong
Last Name : mook kim
Nick Name : mkyong
Salary : 100000

Current Element :staff
Staff id : 2001
First Name : low
Last Name : yin fong
Nick Name : fong fong
Salary : 200000

Answer 2

过去我的一个潜在解决方法是为文件创建一个File对象。

File xmlFile = new File("C:/fishlist.xml");

然后，tou可以做以下两件事之一：

为File创建FileInputStream并将其传递到DocumentBuilder的解析方法
继续像往常一样发送uri：

Document doc = dBuilder.parse(xmlFile.toURI().toString);

使用DocumentBuilder.parse（）方法读取XML文件。文件丢失了

2 个答案:

根元素：公司