如何调用following方法JsonParser:

  /**
   * Converts a JSON document to XML.
   * @param io input
   * @param options parser options
   * @return parser
   * @throws IOException I/O exception
   */
  private static IOContent toXML(final IO io, final JsonParserOptions options) throws IOException {
    final JsonConverter conv = JsonConverter.get(options);
    final IOContent xml = new IOContent(conv.convert(io).serialize().finish());
    xml.name(io.name());
    return xml;
  }

但我肯定不会在IDE中看到此方法:

java - JsonParser中的toXML在哪里,为什么该方法不可用?-LMLPHP

method is in JavaDocs:
Method Detail

    toXML

    public static IOContent toXML(IO io,
                  JsonParserOptions options)
                           throws java.io.IOException

    Converts a JSON document to XML.

    Parameters:
        io - input
        options - parser options
    Returns:
        parser
    Throws:
        java.io.IOException - I/O exception

生成文件正在使用:
compile group: 'org.basex', name: 'basex', version: '9.2.4'
这是我在存储库中看到的最新版本:
maven { url "https://mvnrepository.com/" }

我一直努力组装项目,并从生成的.class中的JAR中提取了BaseX文件,但没有进一步寻找该方法是否存在。

也许我只是invoking方法不正确?

最佳答案

哎呀:

package basex;

import java.io.IOException;
import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Paths;
import java.util.logging.Logger;
import org.basex.build.json.JsonParser;
import org.basex.build.xml.SAXWrapper;
import org.basex.core.MainOptions;
import org.basex.io.IOFile;

public class JsonToXmlTransformer {

    private static final Logger log = Logger.getLogger(JsonToXmlTransformer.class.getName());

    public JsonToXmlTransformer() {
    }

    private void baseXparseJsonFile(String fileName) throws IOException {
        org.basex.build.json.JsonParser jsonParser = new org.basex.build.json.JsonParser(new IOFile(fileName), new MainOptions());

        SAXWrapper foo = org.basex.build.json.JsonParser.xmlParser(new IOFile(fileName));
        foo.parse();
        String bar = foo.toString();
        log.info(bar);
    }

    public void transform(String fileName) throws IOException {
        String content = new String(Files.readAllBytes(Paths.get(fileName)), StandardCharsets.UTF_8);
        org.json.JSONObject json = new org.json.JSONObject(content);
        log.info(org.json.XML.toString(json));
    }
}

包装错误...

09-10 07:10
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