在这里回答，因为这是一个流行的线程。这是一个干净的Kotlin解决方案，使用推荐的UrlQuerySanitizer api。请参阅官方文档。我添加了一个字符串构建器来连接和显示参数。

    var myURL: String? = null
    // if the url is sent from a different activity where you set it to a value
    if (intent.hasExtra("my_value")) {
        myURL = intent.extras.getString("my_value")
    } else {
        myURL = intent.dataString
    }

    val sanitizer = UrlQuerySanitizer(myURL)
    // We don't want to manually define every expected query *key*, so we set this to true
    sanitizer.allowUnregisteredParamaters = true
    val parameterNamesToValues: List<UrlQuerySanitizer.ParameterValuePair> = sanitizer.parameterList
    val parameterIterator: Iterator<UrlQuerySanitizer.ParameterValuePair> = parameterNamesToValues.iterator()

    // Helper simply so we can display all values on screen
    val stringBuilder = StringBuilder()

    while (parameterIterator.hasNext()) {
        val parameterValuePair: UrlQuerySanitizer.ParameterValuePair = parameterIterator.next()
        val parameterName: String = parameterValuePair.mParameter
        val parameterValue: String = parameterValuePair.mValue

        // Append string to display all key value pairs
        stringBuilder.append("Key: $parameterName\nValue: $parameterValue\n\n")
    }

    // Set a textView's text to display the string
    val paramListString = stringBuilder.toString()
    val textView: TextView = findViewById(R.id.activity_title) as TextView
    textView.text = "Paramlist is \n\n$paramListString"

    // to check if the url has specific keys
    if (sanitizer.hasParameter("type")) {
        val type = sanitizer.getValue("type")
        println("sanitizer has type param $type")
    }

仅供参考，这是我最终得到的结果(基于URLEncodedUtils，并返回一个Map)。

特点:

它接受url的查询字符串部分(你可以使用request.getQueryString()) 空查询字符串将产生空Map 没有值的参数(?test)将被映射为空List<String>

代码:

public static Map<String, List<String>> getParameterMapOfLists(String queryString) {
    Map<String, List<String>> mapOfLists = new HashMap<String, List<String>>();
    if (queryString == null || queryString.length() == 0) {
        return mapOfLists;
    }
    List<NameValuePair> list = URLEncodedUtils.parse(URI.create("http://localhost/?" + queryString), "UTF-8");
    for (NameValuePair pair : list) {
        List<String> values = mapOfLists.get(pair.getName());
        if (values == null) {
            values = new ArrayList<String>();
            mapOfLists.put(pair.getName(), values);
        }
        if (pair.getValue() != null) {
            values.add(pair.getValue());
        }
    }

    return mapOfLists;
}

兼容性帮助器(值存储在String数组中，就像在ServletRequest.getParameterMap()中一样):

public static Map<String, String[]> getParameterMap(String queryString) {
    Map<String, List<String>> mapOfLists = getParameterMapOfLists(queryString);

    Map<String, String[]> mapOfArrays = new HashMap<String, String[]>();
    for (String key : mapOfLists.keySet()) {
        mapOfArrays.put(key, mapOfLists.get(key).toArray(new String[] {}));
    }

    return mapOfArrays;
}

在Android上，Apache库提供了一个Query解析器:

http://developer.android.com/reference/org/apache/http/client/utils/URLEncodedUtils.html和http://hc.apache.org/httpcomponents-client-ga/httpclient/apidocs/org/apache/http/client/utils/URLEncodedUtils.html

Apache AXIS2具有QueryStringParser.java的自包含实现。如果不使用Axis2，只需从这里下载源代码和测试用例

http://svn.apache.org/repos/asf/axis/axis2/java/core/trunk/modules/kernel/src/org/apache/axis2/transport/http/util/QueryStringParser.java

http://svn.apache.org/repos/asf/axis/axis2/java/core/trunk/modules/kernel/test/org/apache/axis2/transport/http/util/QueryStringParserTest.java

解析查询字符串比看起来要复杂一些，这取决于您希望有多宽容。

首先，查询字符串是ascii字节。每次读入一个字节，然后将它们转换成字符。如果角色是?或者&，然后它表示参数名的开始。如果字符为=，则它表示一个参数值的开始。如果字符为%，则表示已编码字节的开始。这就是棘手的地方。

When you read in a % char you have to read the next two bytes and interpret them as hex digits. That means the next two bytes will be 0-9, a-f or A-F. Glue these two hex digits together to get your byte value. But remember, bytes are not characters. You have to know what encoding was used to encode the characters. The character é does not encode the same in UTF-8 as it does in ISO-8859-1. In general it's impossible to know what encoding was used for a given character set. I always use UTF-8 because my web site is configured to always serve everything using UTF-8 but in practice you can't be certain. Some user-agents will tell you the character encoding in the request; you can try to read that if you have a full HTTP request. If you just have a url in isolation, good luck.

不管怎样，假设您正在使用UTF-8或其他一些多字节字符编码，现在您已经解码了一个已编码的字节，您必须将其放在一边，直到捕获下一个字节。您需要所有已编码的字节放在一起，因为您不能一次正确地对一个字节进行url解码。把所有在一起的字节放在一边，然后立刻解码，重新构建你的角色。

另外，如果你想要宽容一些，并解释用户代理破坏url，它会变得更有趣。例如，一些webmail客户端会对内容进行双重编码。或者使用两个?&=字符(例如:http://yoursite.com/blah??p1==v1&&p2==v2)。如果您想尝试优雅地处理这个问题，就需要向解析器添加更多的逻辑。

这对我有用。 我不知道为什么每个人都想要一个地图，列表> 我所需要的只是一个简单的名称值Map。

为了简单起见，我使用URI.getQuery()中的构建;

public static Map<String, String> getUrlParameters(URI uri)
    throws UnsupportedEncodingException {
    Map<String, String> params = new HashMap<String, String>();
    for (String param : uri.getQuery().split("&")) {
        String pair[] = param.split("=");
        String key = URLDecoder.decode(pair[0], "UTF-8");
        String value = "";
        if (pair.length > 1) {
            value = URLDecoder.decode(pair[1], "UTF-8");
        }
        params.put(new String(key), new String(value));
    }
    return params;
}