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))
  }
}

如果您正在寻找一种不使用外部库的方法来实现它，下面的代码将帮助您。

public static Map<String, String> splitQuery(URL url) throws UnsupportedEncodingException {
    Map<String, String> query_pairs = new LinkedHashMap<String, String>();
    String query = url.getQuery();
    String[] pairs = query.split("&");
    for (String pair : pairs) {
        int idx = pair.indexOf("=");
        query_pairs.put(URLDecoder.decode(pair.substring(0, idx), "UTF-8"), URLDecoder.decode(pair.substring(idx + 1), "UTF-8"));
    }
    return query_pairs;
}

您可以使用< Map >.get(“client_id”)访问返回的Map，在您的问题中给出的URL将返回“SS”。

添加了UPDATE url -解码

由于这个答案仍然很受欢迎，我对上面的方法做了一个改进版本，它可以处理具有相同键的多个参数和没有值的参数。

public static Map<String, List<String>> splitQuery(URL url) throws UnsupportedEncodingException {
  final Map<String, List<String>> query_pairs = new LinkedHashMap<String, List<String>>();
  final String[] pairs = url.getQuery().split("&");
  for (String pair : pairs) {
    final int idx = pair.indexOf("=");
    final String key = idx > 0 ? URLDecoder.decode(pair.substring(0, idx), "UTF-8") : pair;
    if (!query_pairs.containsKey(key)) {
      query_pairs.put(key, new LinkedList<String>());
    }
    final String value = idx > 0 && pair.length() > idx + 1 ? URLDecoder.decode(pair.substring(idx + 1), "UTF-8") : null;
    query_pairs.get(key).add(value);
  }
  return query_pairs;
}

更新Java8版本

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, mapping(Map.Entry::getValue, toList())));
}

public SimpleImmutableEntry<String, String> splitQueryParameter(String it) {
    final int idx = it.indexOf("=");
    final String key = idx > 0 ? it.substring(0, idx) : it;
    final String value = idx > 0 && it.length() > idx + 1 ? it.substring(idx + 1) : null;
    return new SimpleImmutableEntry<>(
        URLDecoder.decode(key, StandardCharsets.UTF_8),
        URLDecoder.decode(value, StandardCharsets.UTF_8)
    );
}

使用URL运行上述方法

https://stackoverflow.com?param1=value1&param2=&param3=value3&param3

返回这个Map:

{param1=["value1"], param2=[null], param3=["value3", null]}

org.apache.http.client.utils.URLEncodedUtils

是否有一个知名的库可以帮你做到这一点

import org.apache.hc.client5.http.utils.URLEncodedUtils

String url = "http://www.example.com/something.html?one=1&two=2&three=3&three=3a";

List<NameValuePair> params = URLEncodedUtils.parse(new URI(url), Charset.forName("UTF-8"));

for (NameValuePair param : params) {
  System.out.println(param.getName() + " : " + param.getValue());
}

输出

one : 1
two : 2
three : 3
three : 3a

如果你正在使用Spring框架:

public static void main(String[] args) {
    String uri = "http://my.test.com/test?param1=ab&param2=cd&param2=ef";
    MultiValueMap<String, String> parameters =
            UriComponentsBuilder.fromUriString(uri).build().getQueryParams();
    List<String> param1 = parameters.get("param1");
    List<String> param2 = parameters.get("param2");
    System.out.println("param1: " + param1.get(0));
    System.out.println("param2: " + param2.get(0) + "," + param2.get(1));
}

你会得到:

param1: ab
param2: cd,ef

如果你正在使用servlet doGet，试试这个

request.getParameterMap()

返回此请求参数的java.util.Map。 返回: 一个不可变的java.util.Map，包含参数名作为键，参数值作为映射值。参数映射中的键类型为String。参数映射中的值类型为字符串数组。 (Java文档)

Eclipse Jersey REST框架通过UriComponent支持这一点。例子:

import org.glassfish.jersey.uri.UriComponent;

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";
MultivaluedMap<String, String> params = UriComponent.decodeQuery(URI.create(uri), true);
for (String key : params.keySet()) {
  System.out.println(key + ": " + params.getFirst(key));
}