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

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
    }
}

有一个新版本的Apache HTTP客户端- org.apache.httpcomponents。client5 -其中URLEncodedUtils现在已弃用。应该改为使用URIBuilder:

import org.apache.hc.core5.http.NameValuePair;
import org.apache.hc.core5.net.URIBuilder;

private static Map<String, String> getQueryParameters(final String url) throws URISyntaxException {
    return new URIBuilder(new URI(url), StandardCharsets.UTF_8).getQueryParams()
                                                               .stream()
                                                               .collect(Collectors.toMap(NameValuePair::getName,
                                                                                         nameValuePair -> URLDecoder.decode(nameValuePair.getValue(), StandardCharsets.UTF_8)));
}

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。

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

一个kotlin版本

由马提亚提供的答案

fun decomposeQueryString(query: String, charset: Charset): Map<String, String?> {
   return if (query.split("?").size <= 1)
       emptyMap()
   else {
       query.split("?")[1]
            .split("&")
            .map { it.split(Pattern.compile("="), 2) }
            .associate {
                Pair(
                        URLDecoder.decode(it[0], charset.name()),
                        if (it.size > 1) URLDecoder.decode(it[1], charset.name()) else null
                )
            }
     }
}

这需要问号'?’。