我得到了这样的URI:

https://google.com.ua/oauth/authorize?client_id=SS&response_type=code&scope=N_FULL&access_type=offline&redirect_uri=http://localhost/Callback

我需要一个包含已解析元素的集合:

NAME               VALUE
------------------------
client_id          SS
response_type      code
scope              N_FULL
access_type        offline
redirect_uri       http://localhost/Callback

确切地说,我需要一个与c# /等价的Java。净HttpUtility。ParseQueryString方法。


当前回答

如果你碰巧在类路径上有cxf-core,并且你知道你没有重复的查询参数,你可能想要使用UrlUtils.parseQueryString。

其他回答

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

Hutool框架通过HttpUtil来支持这一点。例子:

import cn.hutool.http.HttpUtil;

    String url ="https://google.com.ua/oauth/authorize?client_id=SS&response_type=code&scope=N_FULL&access_type=offline&redirect_uri=http://localhost/Callback";
    Map<String, List<String>> stringListMap = HttpUtil.decodeParams(url, "UTF-8");
    System.out.println("decodeParams:" + stringListMap);

你会得到:

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

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 (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")
    }

我有一个Kotlin版本,看看这是如何在谷歌的顶部结果。

@Throws(UnsupportedEncodingException::class)
fun splitQuery(url: URL): Map<String, List<String>> {

    val queryPairs = LinkedHashMap<String, ArrayList<String>>()

    url.query.split("&".toRegex())
            .dropLastWhile { it.isEmpty() }
            .map { it.split('=') }
            .map { it.getOrEmpty(0).decodeToUTF8() to it.getOrEmpty(1).decodeToUTF8() }
            .forEach { (key, value) ->

                if (!queryPairs.containsKey(key)) {
                    queryPairs[key] = arrayListOf(value)
                } else {

                    if(!queryPairs[key]!!.contains(value)) {
                        queryPairs[key]!!.add(value)
                    }
                }
            }

    return queryPairs
}

还有扩展方法

fun List<String>.getOrEmpty(index: Int) : String {
    return getOrElse(index) {""}
}

fun String.decodeToUTF8(): String { 
    URLDecoder.decode(this, "UTF-8")
}