理解Java中的xml和url [英] understand xml and urls in Java

问题描述

我正在尝试使用此代码在.Net环境中重新创建类似的应用程序，但我对逻辑不太清楚。因此，我想寻求一些帮助，解释下面代码中发生了什么：



有一个contentCreator类，它会产生一个问题&我不太确定，它在这里做什么返回api.create（publication.findLink（issue），Utils.renderXml（document））; ：

I am trying to use this code to recreate similar application in .Net environment, but I am little unclear of the logic. Hence I would like to seek some assistance, in explaining, what is going on in the code below:

There is a contentCreator class, which creates a issue & I am little unsure, what its doing here return api.create(publication.findLink("issue"), Utils.renderXml(document));:

//ContentCreator

	public static ApiObject createIssue(NotessaApi api, ApiObject publication, Issue issue) throws IOException {
		Document document = Utils.createEmptyDocument();
		Node node = Utils.createNode(document, "issue");
		
		Utils.addXml(document, node, "title", issue.getTitle());
		Utils.addXml(document, node, "approval_status", issue.getApprovalStatus());
		Utils.addXml(document, node, "display_title", issue.getDisplayTitle());
		Utils.addXml(document, node, "display_date", issue.getDisplayDate());
		List<Property> properties = issue.getProperties();
		if(!properties.isEmpty()) {
			Node container = Utils.createNode(document, node, "properties");
			for(Property property : properties) {
				Element element = Utils.createNode(document, container, "property");
				element.setAttribute("name", property.getName());
				Utils.addXml(document, element, "value", property.getValue());
			}
		}
		
		return api.create(publication.findLink("issue"), Utils.renderXml(document));
	}

是否创建问题xml并将其转发到API Object类的findlink url。



API Object类创建以下两种方法：

Is it creating the issue xml and forwarding it to the findlink url from the API Object class.

API Object class creates the following two methods:

import java.io.IOException;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;

import javax.xml.namespace.QName;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathExpressionException;
import javax.xml.xpath.XPathFactory;
	
public URL findLink(String rel) {
		String result = xpath("//link[contains(@rel,'" + rel + "')]/@href");
		try {
			return new URL(result);
		} catch(MalformedURLException e) {
			throw new RuntimeException(e);
		} 
	}

	public URL getSelfLink() throws MalformedURLException {
		String uri = xpath("//@uri");
		return new URL(uri); 
	}

任何进一步的建议，将非常感谢。非常感谢。

Any further advice, would be most appreciated. Many thanks.

推荐答案

return语句调出 api.create（publication.findLink（issue），Utils.renderXml （document））; 并返回该方法返回的任何内容。

The return statement makes a call out to api.create(publication.findLink("issue"), Utils.renderXml(document)); and returns whatever is returned by that method.

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