如何在Java中使用XQuery在XML文件的新行中顯示多個元素？

Question

這是我的XML文件

<bookstore> 
    <book category="COOKING"> 
    <title lang="en">Everyday Italian</title> 
    <author>Giada De Laurentiis</author> 
    <year>2005</year> 
    <price>30.00</price> 
    </book> 
    <book category="CHILDREN"> 
    <title lang="en">Harry Potter</title> 
    <author>J K. Rowling</author> 
    <year>2005</year> 
    <price>29.99</price> 
    </book> 
    <book category="WEB"> 
    <title lang="en">Learning XML</title> 
    <author>Erik T. Ray</author> 
    <year>2003</year> 
    <price>39.95</price> 
</book> 
</bookstore> 

，這是我的Java代碼：

import javax.xml.namespace.QName; 
import java.util.Properties; 

import com.ddtek.xquery3.XQConnection; 
import com.ddtek.xquery3.XQException; 
import com.ddtek.xquery3.XQExpression; 
import com.ddtek.xquery3.XQItemType; 
import com.ddtek.xquery3.XQSequence; 
import com.ddtek.xquery3.xqj.DDXQDataSource; 

public class XQueryTester3 { 

// Filename for XML document to query 
private String filename; 

// Data Source for querying 
private DDXQDataSource dataSource; 

// Connection for querying 
private XQConnection conn; 

public XQueryTester3(String filename) { 
this.filename = "untitled1.xml"; 
} 

public void init() throws XQException { 
dataSource = new DDXQDataSource(); 
conn = dataSource.getConnection(); 
} 

public String query(String queryString) throws XQException { 
XQExpression expression = conn.createExpression(); 
expression.bindString(new QName("docName"), filename, 
    conn.createAtomicType(XQItemType.XQBASETYPE_STRING)); 
XQSequence results = expression.executeQuery(queryString); 
return results.getSequenceAsString(new Properties()); 
} 

public static void main(String[] args) { 

     try { 

      XQueryTester3 tester = new XQueryTester3("untitled1.xml"); 
      tester.init(); 

      final String sep = System.getProperty("line.separator"); 
      String queryString = 
      "declare variable $docName as xs:string external;" + sep + 
          "  for $book in doc($docName)/bookstore/book " + 
          " where $book/year =2003 " + 
          " return " + 
          "$book/author/text()"; 
      System.out.println(tester.query(queryString)); 
      } catch (Exception e) { 
      e.printStackTrace(System.err); 
      System.err.println(e.getMessage()); 
      } 
} 
} 

輸出將是：

Erik T. Ray 

我最大的問題是我想顯示作者&這本書的標題，所以我修改了代碼

"declare variable $docName as xs:string external;" + sep + 
          "  for $book in doc($docName)/bookstore/book " + 
          " where $book/year =2003 " + 
          " return " + 
          "$book/author/text()"; 
      System.out.println(tester.query(queryString)); 

是這樣：

"declare variable $docName as xs:string external;" + sep + 
          "  for $book in doc($docName)/bookstore/book " + 
          " where $book/year =2003 " + 
          " return " + 
          "$book/author/text() " + 
          "$book/title/text()"; 
      System.out.println(tester.query(queryString)); 

和我得到一個錯誤：

Unexpected token "$" beyond end of query 

但是，如果我改變這樣的代碼（添加HTML標籤和花括號{}）：

"declare variable $docName as xs:string external;" + sep + 
          "  for $book in doc($docName)/bookstore/book " + 
          " where $book/year =2003 " + 
          " return " + 
          "<i>{$book/author/text()} " + 
          "{$book/title/text()}</i>"; 
      System.out.println(tester.query(queryString)); 

我可以得到的輸出：

<i>Erik T. RayLearning XML</i> 

的問題是，我不希望在我的輸出HTML標記，我想作者的名字&書是在一個新的生產線的冠軍。

誰能幫我上面提到的這個問題？

如何在輸出中沒有HTML的新行中顯示多個元素？

輸出應該是這樣的：

Erik T. Ray 

Learning XML 

Answer 1

你可以嘗試使用concat()：

"declare variable $docName as xs:string external;" + sep + 
          "  for $book in doc($docName)/bookstore/book " + 
          " where $book/year =2003 " + 
          " return " + 
          "concat($book/author/text(),'
',$book/title/text())"; 
      System.out.println(tester.query(queryString)); 

Answer 2

我已經找到了我的問題的解決方案：我使用的XPath，XQuery的不是。

幸運的是我碰到了this link

這說明了一切。