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:
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...