How to use xpath in Java to find the value or node attribute in xml and replace it with another value?

Question

Here is my code:

// locate the node(s) XPath xpath = XPathFactory.newInstance().newXPath(); NodeList nodes = (NodeList)xpath.evaluate("//root[text()='input[1]']", doc, XPathConstants.NODESET); // make the change for (int idx = 0; idx < nodes.getLength(); idx++) { nodes.item(idx).setTextContent(input[3]); } // save the result Transformer xformer = TransformerFactory.newInstance().newTransformer(); xformer.transform(new DOMSource(doc), new StreamResult(new File(outputFile))); 

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Input [1] is what I'm looking for in XML, and input [3] is what it replaces.

Here is the XML:

 <root> <accounts> <account name="Bill Gates"> <position>CEO</position> <phoneNumber>123-485-1854</phoneNumber> <notes>Runs the company</notes> </account> <account name="Service Account"/> <account name="Burt Mackland"> <position>CFO</position> <phoneNumber>345-415-4813</phoneNumber> <notes>Great CFO</notes> </account> <account name="Joe Smith"> <position>Accountant</position> <reportsTo>Burt Mackland</reportsTo> <phoneNumber>135-118-7815</phoneNumber> <notes>Must have ID checked at all Gates</notes> </account> </accounts> <departments> <deparment name="Finance"> <employeeCount>2</employeeCount> </deparment> <department name="Human Resources"> <employeeCount>0</employeeCount> </department> <department name="Executive"> <employeeCount>2</employeeCount> </department> </departments> </root> 

The user may not know what is in the XML. Therefore, I cannot encode Xpath in code. Please, anyone would appreciate it!

Answer 1

As you want to search for text in any element (not just <root> ), change the XPath expression from

 //root[text()='TEXT-TO-BE-FOUND'] 

For

 //*[text()='TEXT-TO-BE-FOUND'] 

To find all attributes with the same value, you must use the following XPath expression:

 //*/@*[.='TEXT-TO-BE-FOUND'] 

Also, since you cannot make hardcode, make a TEXT-TO-BE-FOUND variable. (Note: TEXT-TO-BE-FOUND must be escaped , it may not contain ' , as this will affect the XPath expression.)

The working code is below. It will replace all elements and all attributes with an input value:

 import java.io.File; import java.util.Scanner; import javax.xml.parsers.*; import javax.xml.transform.*; import javax.xml.transform.dom.DOMSource; import javax.xml.transform.stream.StreamResult; import javax.xml.xpath.*; import org.w3c.dom.*; public class XmlChange { public static void main(String argv[]) throws Exception { Scanner keyboard = new Scanner(System.in); System.out.print("Type like textToFind,textToReplace: "); // type, for example CEO,Chief Executive Officer String next = keyboard.nextLine(); String[] input = next.split(","); String textToFind = input[0].replace("'", "\\'"); //"CEO"; String textToReplace = input[1].replace("'", "\\'"); // "Chief Executive Officer"; String filepath = "root.xml"; String fileToBeSaved = "root2.xml"; DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance(); DocumentBuilder docBuilder = docFactory.newDocumentBuilder(); Document doc = docBuilder.parse(filepath); XPath xpath = XPathFactory.newInstance().newXPath(); // change ELEMENTS String xPathExpression = "//*[text()='" + textToFind + "']"; NodeList nodes = (NodeList) xpath.evaluate(xPathExpression, doc, XPathConstants.NODESET); for (int idx = 0; idx < nodes.getLength(); idx++) { nodes.item(idx).setTextContent(textToReplace); } // change ATTRIBUTES String xPathExpressionAttr = "//*/@*[.='" + textToFind + "']"; NodeList nodesAttr = (NodeList) xpath.evaluate(xPathExpressionAttr, doc, XPathConstants.NODESET); for(int i=0; i<nodesAttr.getLength(); i++) { nodesAttr.item(i).setTextContent(textToReplace); } System.out.println("Everything replaced."); // save xml file back TransformerFactory transformerFactory = TransformerFactory.newInstance(); Transformer transformer = transformerFactory.newTransformer(); DOMSource source = new DOMSource(doc); StreamResult result = new StreamResult(new File(fileToBeSaved)); transformer.transform(source, result); } }