Question

我的工具中有一个方法，它将XML String作为条目：

"<file type=\"JNH\"> " + 
"<field code=\"AAAAA\" name=\"NAME1\">string1</field> " + 
"<field code=\"BBBBB\" name=\"NAME2\">string2</field> " +
"<desc code=\"CCCC\" name=\"DESC\"> " +  
    "<txt>bla.</txt> " + 
    "<txt>bla.</txt> " + 
"</desc> " + 
"</file> "

主要是格式，实际上我的工具需要XML字符串输入，但我想在我的应用程序中自动构建这个XML字符串。我的意思是代替myService(String xml)，我想要myService(String string1, String string2, String[] txt)。

这样做的最佳解决方案是什么？使用.replaceAll()方法执行XML模板？（文本行数可以更改）

非常感谢。

Answer 1

我的示例代码在这里创建了xml字符串。我希望这可以帮到你。

import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.StringWriter;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerException;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;
import org.w3c.dom.Document;
import org.w3c.dom.Element;


public class CreateXMLClass {

    public CreateXMLClass() {
        String xmlString = createXML("value1", "value2", "value3");
        System.out.println(xmlString);
    }

    // creates xml and returns the content
    private String createXML(String v1, String v2, String v3) {
        String data="";
        TransformerFactory transformerFactory;
        Transformer transformer;
        DOMSource source;
        StreamResult result;
        StringWriter writer;

        try{
            DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
            DocumentBuilder docBuilder = docFactory.newDocumentBuilder();

            //root elements
            Document doc = docBuilder.newDocument();
            Element soapEnvelope = doc.createElement("soapenv:Envelope");
            soapEnvelope.setAttribute("xmlns:xsi", "http://www.w3.org/2001/XMLSchema-instance");
            soapEnvelope.setAttribute("xmlns:xsd", "http://www.w3.org/2001/XMLSchema");
            soapEnvelope.setAttribute("xmlns:soapenv", "http://schemas.xmlsoap.org/soap/envelope/");
            soapEnvelope.setAttribute("xmlns:ns0", "https://ws.mycompany.com/");
            doc.appendChild(soapEnvelope);



            //elements
            Element header = doc.createElement("soapenv:Header");
            soapEnvelope.appendChild(header);

            Element soapBody = doc.createElement("soapenv:Body");
            soapEnvelope.appendChild(soapBody);

            Element ns0 = doc.createElement("ns0:SampleApplication");
            ns0.setAttribute("xmlns:ns0", "http://ws.mycompany.com");
            soapBody.appendChild(ns0);


            Element element01 = doc.createElement("Sample1");
            element01.appendChild(doc.createTextNode(v1));
            ns0.appendChild(element01);

            Element element02 = doc.createElement("Sample2");
            element02.appendChild(doc.createTextNode(v2));
            ns0.appendChild(element02);

            Element element03 = doc.createElement("Sample3");
            element03.appendChild(doc.createTextNode(v2));
            ns0.appendChild(element03);



            writer = new StringWriter();
            result = new StreamResult(writer);

            transformerFactory = TransformerFactory.newInstance();
            transformer = transformerFactory.newTransformer();
            source = new DOMSource(doc);
            transformer.transform(source, result);

            data =  writer.getBuffer().toString();

        } catch(ParserConfigurationException pce){
            pce.printStackTrace();
        } catch(TransformerException tfe){
            tfe.printStackTrace();
        }

        return data;
    }

    public static void main(String args[]) {
        new CreateXMLClass();
        System.exit(0);
    }
}

Answer 2

不要使用字符串创建XML - 它很容易输入错误（忘记关闭标记，忘记转义引号）。使用javax.xml包中的标准API。它看起来像这样：

import java.io.StringWriter;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;

import org.w3c.dom.Document;
import org.w3c.dom.Element;

public class Test {
    public String getXMLString(String string1, String string2, String[] txts) throws Exception {
        // Create document
        DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder docBuilder = docFactory.newDocumentBuilder();
        Document doc = docBuilder.newDocument();

        // Root element
        Element rootElement = doc.createElement("file");
        rootElement.setAttribute("type", "JNH");
        doc.appendChild(rootElement);

        // Field 1
        Element field1Element = doc.createElement("field");
        field1Element.setAttribute("code", "AAAAA");
        field1Element.setAttribute("name", "NAME1");
        field1Element.setTextContent(string1);
        rootElement.appendChild(field1Element);

        // likewise for Field 2

        // Desc
        Element descElement = doc.createElement("desc");
        descElement.setAttribute("code", "CCCC");
        descElement.setAttribute("name", "DESC");
        descElement.setTextContent(string1);
        rootElement.appendChild(descElement);
        // Children
        for (String txt : txts) {
            Element txtElement = doc.createElement("desc");
            txtElement.setTextContent(txt);
            descElement.appendChild(txtElement);
        }

        // Save document as string
        TransformerFactory tf = TransformerFactory.newInstance();
        Transformer transformer = tf.newTransformer();
        StringWriter writer = new StringWriter();
        transformer.transform(new DOMSource(doc), new StreamResult(writer));
        return writer.getBuffer().toString();
    }
}

用Java构建XML模板

2 个答案: