一种现成的URI查询部分解码解决方案(包括解码和多参数值)

评论

我对https://stackoverflow.com/a/13592567/1211082中@Pr0gr4mm3r提供的代码不满意。基于流的解决方案不做URLDecoding，可变版本的笨拙。

因此，我阐述了一个解决方案

Can decompose a URI query part into a Map<String, List<Optional<String>>> Can handle multiple values for the same parameter name Can represent parameters without a value properly (Optional.empty() instead of null) Decodes parameter names and values correctly via URLdecode Is based on Java 8 Streams Is directly usable (see code including imports below) Allows for proper error handling (here via turning a checked exception UnsupportedEncodingExceptioninto a runtime exception RuntimeUnsupportedEncodingException that allows interplay with stream. (Wrapping regular function into functions throwing checked exceptions is a pain. And Scala Try is not available in the Java language default.)

Java代码

import java.io.UnsupportedEncodingException;
import java.net.URLDecoder;
import java.util.*;
import static java.util.stream.Collectors.*;

public class URIParameterDecode {
    /**
     * Decode parameters in query part of a URI into a map from parameter name to its parameter values.
     * For parameters that occur multiple times each value is collected.
     * Proper decoding of the parameters is performed.
     * 
     * Example
     *   <pre>a=1&b=2&c=&a=4</pre>
     * is converted into
     *   <pre>{a=[Optional[1], Optional[4]], b=[Optional[2]], c=[Optional.empty]}</pre>
     * @param query the query part of an URI 
     * @return map of parameters names into a list of their values.
     *         
     */
    public static Map<String, List<Optional<String>>> splitQuery(String query) {
        if (query == null || query.isEmpty()) {
            return Collections.emptyMap();
        }

        return Arrays.stream(query.split("&"))
                    .map(p -> splitQueryParameter(p))
                    .collect(groupingBy(e -> e.get0(), // group by parameter name
                            mapping(e -> e.get1(), toList())));// keep parameter values and assemble into list
    }

    public static Pair<String, Optional<String>> splitQueryParameter(String parameter) {
        final String enc = "UTF-8";
        List<String> keyValue = Arrays.stream(parameter.split("="))
                .map(e -> {
                    try {
                        return URLDecoder.decode(e, enc);
                    } catch (UnsupportedEncodingException ex) {
                        throw new RuntimeUnsupportedEncodingException(ex);
                    }
                }).collect(toList());

        if (keyValue.size() == 2) {
            return new Pair(keyValue.get(0), Optional.of(keyValue.get(1)));
        } else {
            return new Pair(keyValue.get(0), Optional.empty());
        }
    }

    /** Runtime exception (instead of checked exception) to denote unsupported enconding */
    public static class RuntimeUnsupportedEncodingException extends RuntimeException {
        public RuntimeUnsupportedEncodingException(Throwable cause) {
            super(cause);
        }
    }

    /**
     * A simple pair of two elements
     * @param <U> first element
     * @param <V> second element
     */
    public static class Pair<U, V> {
        U a;
        V b;

        public Pair(U u, V v) {
            this.a = u;
            this.b = v;
        }

        public U get0() {
            return a;
        }

        public V get1() {
            return b;
        }
    }
}

Scala代码

．.． 为了完整起见，我忍不住要用Scala提供简洁美观的解决方案

import java.net.URLDecoder

object Decode {
  def main(args: Array[String]): Unit = {
    val input = "a=1&b=2&c=&a=4";
    println(separate(input))
  }

  def separate(input: String) : Map[String, List[Option[String]]] = {
    case class Parameter(key: String, value: Option[String])

    def separateParameter(parameter: String) : Parameter =
      parameter.split("=")
               .map(e => URLDecoder.decode(e, "UTF-8")) match {
      case Array(key, value) =>  Parameter(key, Some(value))
      case Array(key) => Parameter(key, None)
    }

    input.split("&").toList
      .map(p => separateParameter(p))
      .groupBy(p => p.key)
      .mapValues(vs => vs.map(p => p.value))
  }
}

另外，我建议使用基于正则表达式的URLParser实现

import java.util.regex.Matcher;
import java.util.regex.Pattern;

class URLParser {
    private final String query;
    
    public URLParser(String query) {
        this.query = query;
    }
    
    public String get(String name) {
        String regex = "(?:^|\\?|&)" + name + "=(.*?)(?:&|$)";
        Pattern pattern = Pattern.compile(regex);
        Matcher matcher = pattern.matcher(this.query);

        if (matcher.find()) {
            return matcher.group(1);
        }
        
        return "";
    }
}

这个类很容易使用。它只需要初始化时的URL或查询字符串，并根据给定的键解析值。

class Main {
    public static void main(String[] args) {
        URLParser parser = new URLParser("https://www.google.com/search?q=java+parse+url+params&oq=java+parse+url+params&aqs=chrome..69i57j0i10.18908j0j7&sourceid=chrome&ie=UTF-8");
        System.out.println(parser.get("q"));  // java+parse+url+params
        System.out.println(parser.get("sourceid"));  // chrome
        System.out.println(parser.get("ie"));  // UTF-8
    }
}

用谷歌番石榴，分成两行:

import java.util.Map;
import com.google.common.base.Splitter;

public class Parser {
    public static void main(String... args) {
        String uri = "https://google.com.ua/oauth/authorize?client_id=SS&response_type=code&scope=N_FULL&access_type=offline&redirect_uri=http://localhost/Callback";
        String query = uri.split("\\?")[1];
        final Map<String, String> map = Splitter.on('&').trimResults().withKeyValueSeparator('=').split(query);
        System.out.println(map);
    }
}

这让你

{client_id=SS, response_type=code, scope=N_FULL, access_type=offline, redirect_uri=http://localhost/Callback}

如果您正在使用Java 8，并且愿意编写一些可重用的方法，那么可以在一行中完成。

private Map<String, List<String>> parse(final String query) {
    return Arrays.asList(query.split("&")).stream().map(p -> p.split("=")).collect(Collectors.toMap(s -> decode(index(s, 0)), s -> Arrays.asList(decode(index(s, 1))), this::mergeLists));
}

private <T> List<T> mergeLists(final List<T> l1, final List<T> l2) {
    List<T> list = new ArrayList<>();
    list.addAll(l1);
    list.addAll(l2);
    return list;
}

private static <T> T index(final T[] array, final int index) {
    return index >= array.length ? null : array[index];
}

private static String decode(final String encoded) {
    try {
        return encoded == null ? null : URLDecoder.decode(encoded, "UTF-8");
    } catch(final UnsupportedEncodingException e) {
        throw new RuntimeException("Impossible: UTF-8 is a required encoding", e);
    }
}

但这是一条很残酷的线。

只是Java 8版本的更新

public Map<String, List<String>> splitQuery(URL url) {
    if (Strings.isNullOrEmpty(url.getQuery())) {
        return Collections.emptyMap();
    }
    return Arrays.stream(url.getQuery().split("&"))
            .map(this::splitQueryParameter)
            .collect(Collectors.groupingBy(SimpleImmutableEntry::getKey, LinkedHashMap::new, **Collectors**.mapping(Map.Entry::getValue, **Collectors**.toList())));
}

mapping和toList()方法必须用于顶部答案中没有提到的collector。否则它会在IDE中抛出编译错误

在我看来，这是最好的方法:

static Map<String, String> decomposeQueryString(String query, Charset charset) {
    return Arrays.stream(query.split("&"))
        .map(pair -> pair.split("=", 2))
        .collect(Collectors.toMap(
            pair -> URLDecoder.decode(pair[0], charset),
            pair -> pair.length > 1 ? URLDecoder.decode(pair[1], charset) : null)
        );
}

前提条件是查询语法不允许重复参数。