如何使用Xpath正確傳遞XML中的對象

我將嘗試描述我的問題。我有一個結構的XML文檔（不看俄文本;它是確定）：如何使用Xpath正確傳遞XML中的對象

<Books> 
    <Book ganre="fantasy"> 
    <bookId>FD46</bookId> 
    <bookName>Меч предназначения</bookName> 
    <bookAuthor>Анджей Сапковский</bookAuthor> 
    <bookYear>1994</bookYear> 
    <bookAvailable>false</bookAvailable> 
    </Book> 
    <Book ganre="fantasy"> 
    <bookId>0RD7</bookId> 
    <bookName>Башня ласточки</bookName> 
    <bookAuthor>Анджей Сапковский</bookAuthor> 
    <bookYear>1997</bookYear> 
    <bookAvailable>false</bookAvailable> 
    </Book> 
    <Book ganre="action"> 
    <bookId>709F</bookId> 
    <bookName>Автостопом по галактике</bookName> 
    <bookAuthor>Дуглас Адамс</bookAuthor> 
    <bookYear>1979</bookYear> 
    <bookAvailable>false</bookAvailable> 
    </Book> 
    </Books>

試圖通過bookID與此XPath

FileInputStream fileInputStream = new FileInputStream("Test/Books.xml"); 
      DocumentBuilderFactory documentBuilderFactory = DocumentBuilderFactory.newInstance(); 
      DocumentBuilder documentBuilder = documentBuilderFactory.newDocumentBuilder(); 

      Document document = documentBuilder.parse(fileInputStream); 

      XPathFactory xPathFactory = XPathFactory.newInstance(); 
      XPath xPath = xPathFactory.newXPath(); 

      Node node = (Node) xPath 
        .evaluate("//Book[bookId/text()='" + bookID + "']", document.getDocumentElement(), XPathConstants.NODE);

bookID的幫助找到元素 - 這是用戶輸入（例如，像這樣）

Scanner sc = new Scanner(System.in, "cp866"); 
String bookID; 
bookID = sc.nextLine();

所以想法是從XML節點從這個對象，我們目前的ID在一個字符串01發現返回所以我可以放入另一個xml。

像

String takenBookName = new XMLDocument(xml).xpath("/Books/Book/bookName/text").get(0); //it will doesnt work ;)

來源

2015-09-28 TheOriginalNickname

請加最後一個元素（這樣的人可以複製/粘貼） –

@DmytroPastovenskyi解決 – TheOriginalNickname

我剛剛與http://www.freeformatter.com/xpath-tester.html#ad-output檢查。您的請求在那裏工作。你可以再次檢查XML和你的xpath嗎？ –

必須爲你工作。

//Book[bookId/text()='709F']/bookName/text()

來源

2015-09-28 12:20:43

如何使用Xpath正確傳遞XML中的對象

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