java 如何从查询字符串中删除查询参数

how to remove a query parameter from a query string

提问by Marcus Junius Brutus

I am using UriBuilderto remove a parameter from a URI:

我正在使用UriBuilder从 URI 中删除参数：

public static URI removeParameterFromURI(URI uri, String param) {
    UriBuilder uriBuilder = UriBuilder.fromUri(uri);
    return uriBuilder.replaceQueryParam(param, "").build();
}

public static String removeParameterFromURIString(String uriString, String param) {
    try {
        URI uri = removeParameterFromURI(new URI(uriString), param);
        return uri.toString();
    } catch (URISyntaxException e) {
        throw new RuntimeException(e);
    }
}

The above sort of works and modifies:

上述类型的工作和修改：

http://a.b.c/d/e/f?foo=1&bar=2&zar=3

http://abc/d/e/f?foo=1&bar=2&zar=3

… into:

… 进入：

http://a.b.c/d/e/f?bar=&foo=1&zar=3

http://abc/d/e/f?bar=&foo=1&zar=3

But it has the following issues:

但它存在以下问题：

1. It messes up the order of the parameters. I know that the order is not relevant but it still bothers me.
2. it doesn't fully remove the parameter, it just sets its value to the empty string. I would prefer is the parameter is completely removed from the query string.

1. 它弄乱了参数的顺序。我知道订单不相关，但它仍然困扰着我。
2. 它并没有完全删除参数，它只是将其值设置为空字符串。我更喜欢从查询字符串中完全删除参数。

Is there some standard or commonly used library that can achieve the above neatly without having to parse and hack the query string myself?

是否有一些标准或常用的库可以巧妙地实现上述功能，而无需自己解析和破解查询字符串？

回答by Flips

Using httpclient URIBuilderwould be much cleaner if you can.

如果可以的话，使用httpclient URIBuilder会更干净。

public String removeQueryParameter(String url, String parameterName) throws URISyntaxException {
    URIBuilder uriBuilder = new URIBuilder(url);
    List<NameValuePair> queryParameters = uriBuilder.getQueryParams();
    for (Iterator<NameValuePair> queryParameterItr = queryParameters.iterator(); queryParameterItr.hasNext();) {
        NameValuePair queryParameter = queryParameterItr.next();
        if (queryParameter.getName().equals(parameterName)) {
            queryParameterItr.remove();
        }
    }
    uriBuilder.setParameters(queryParameters);
    return uriBuilder.build().toString();
}

回答by TTKatrina

In Android, without import any library. I write a util method inspired by this answerReplace query parameters in Uri.Builder in Android?(Replace query parameters in Uri.Builder in Android?)

在 Android 中，无需导入任何库。我写了一个受这个答案启发的 util 方法Replace query parameters in Uri.Builder in Android?（在 Android 中替换 Uri.Builder 中的查询参数？）

Hope can help you. Code below:

希望能帮到你。代码如下：

public static Uri removeUriParameter(Uri uri, String key) {
    final Set<String> params = uri.getQueryParameterNames();
    final Uri.Builder newUri = uri.buildUpon().clearQuery();
    for (String param : params) {
        if (!param.equals(key)) {
            newUri.appendQueryParameter(param, uri.getQueryParameter(param));
        }
    }
    return newUri.build();
}

回答by VolkerK

Using streams and URIBuilder from httpclientit would look like this

使用httpclient 中的流和URIBuilder看起来像这样

public String removeQueryParameter(String url, String parameterName) throws URISyntaxException {
    URIBuilder uriBuilder = new URIBuilder(url);
    List<NameValuePair> queryParameters = uriBuilder.getQueryParams()
              .stream()
              .filter(p -> !p.getName().equals(parameterName))
              .collect(Collectors.toList());
    if (queryParameters.isEmpty()) {
        uriBuilder.removeQuery();
    } else {
        uriBuilder.setParameters(queryParameters);
    }
    return uriBuilder.build().toString();
}

回答by Grisha

To fully remove the parameter, you can use

要完全删除参数，您可以使用

public static URI removeParameterFromURI(URI uri, String param) {
    UriBuilder uriBuilder = UriBuilder.fromUri(uri);
    return uriBuilder.replaceQueryParam(param, (Object[]) null).build();
}

回答by Daniel F

If you are on Android and want to remove allquery parameters, you can use

如果您使用的是 Android 并且想要删除所有查询参数，则可以使用

Uri uriWithoutQuery = Uri.parse(urlWithQuery).buildUpon().clearQuery().build();

Uri uriWithoutQuery = Uri.parse(urlWithQuery).buildUpon().clearQuery().build();

回答by linkaipeng

public static String removeQueryParameter(String url, List<String> removeNames) {
    try {
        Map<String, String> queryMap = new HashMap<>();
        Uri uri = Uri.parse(url);
        Set<String> queryParameterNames = uri.getQueryParameterNames();
        for (String queryParameterName : queryParameterNames) {
            if (TextUtils.isEmpty(queryParameterName)
                    ||TextUtils.isEmpty(uri.getQueryParameter(queryParameterName))
                    || removeNames.contains(queryParameterName)) {
                continue;
            }
            queryMap.put(queryParameterName, uri.getQueryParameter(queryParameterName));
        }
        // remove all params
        Uri.Builder uriBuilder = uri.buildUpon().clearQuery();

        for (String name : queryMap.keySet()) {
            uriBuilder.appendQueryParameter(name, queryMap.get(name));
        }
        return uriBuilder.build().toString();
    } catch (Exception e) {
        return url;
    }
}

回答by Sedrick

I am not sure if there is some library to help, but I would just split the string on "?" and take the second half and split it on "&". Then I would rebuild the string accordingly.

我不确定是否有一些库可以提供帮助，但我只是将字符串拆分为“？” 并将下半部分拆分为“&”。然后我会相应地重建字符串。

    public static void main(String[] args) {
        // TODO code application logic here
        System.out.println("original: http://a.b.c/d/e/f?foo=1&bar=2&zar=3");
        System.out.println("new     : " + fixString("http://a.b.c/d/e/f?foo=1&bar=2&zar=3"));
    }

    static String fixString(String original)
    {
        String[] processing = original.split("\?");
        String[] processing2ndHalf = processing[1].split("&");

        return processing[0] + "?" + processing2ndHalf[1] + "&" + processing2ndHalf[0] + "&" + processing2ndHalf[2];
    }

Output:

输出：

To remove a paramater just remove it from the return string.

要删除参数，只需将其从返回字符串中删除。

回答by Marcus Junius Brutus

Based on the suggestion by JB Nizzet, this is what I ended up doing (I added some extra logic to be able to assert whether I expect the parameter to be present, and if so, how many times):

根据JB Nizzet的建议，这就是我最终要做的（我添加了一些额外的逻辑，以便能够断言我是否希望参数存在，如果是，则出现多少次）：

public static URI removeParameterFromURI(URI uri, String parameter, boolean assertAtLeastOneIsFound, Integer assertHowManyAreExpected) {
    Assert.assertFalse("it makes no sense to expect 0 or less", (assertHowManyAreExpected!=null) && (assertHowManyAreExpected<=0) );
    Assert.assertFalse("it makes no sense to not assert that at least one is found and at the same time assert a definite expected number", (!assertAtLeastOneIsFound) && (assertHowManyAreExpected!=null) );
    String queryString = uri.getQuery();
    if (queryString==null)
        return uri;
    Map<String, List<String>> params = parseQuery(queryString);
    Map<String, List<String>> paramsModified = new LinkedHashMap<>();
    boolean found = false;
    for (String key: params.keySet()) {
        if (!key.equals(parameter))
            Assert.assertNull(paramsModified.put(key, params.get(key)));
        else {
            found = true;
            if (assertHowManyAreExpected!=null) {
                Assert.assertEquals((long) assertHowManyAreExpected, params.get(key).size());
            }
        }
    }
    if (assertAtLeastOneIsFound)
        Assert.assertTrue(found);
    UriBuilder uriBuilder = UriBuilder.fromUri(uri)
        .replaceQuery("");
    for (String key: paramsModified.keySet()) {
        List<String> values = paramsModified.get(key);
        uriBuilder = uriBuilder.queryParam(key, (Object[]) values.toArray(new String[values.size()]));
    }
    return uriBuilder.build();
}

public static String removeParameterFromURI(String uri, String parameter, boolean assertAtLeastOneIsFound, Integer assertHowManyAreExpected) {
    try {
        return removeParameterFromURI(new URI(uri), parameter, assertAtLeastOneIsFound, assertHowManyAreExpected).toString();
    } catch (URISyntaxException e) {
        throw new RuntimeException(e);
    }
}

private static Map<String, List<String>> parseQuery(String queryString) {
    try {
        final Map<String, List<String>> query_pairs = new LinkedHashMap<String, List<String>>();
        final String[] pairs = queryString.split("&");
        for (String pair : pairs) {
            final int idx = pair.indexOf("=");
            final String key = idx > 0 ? URLDecoder.decode(pair.substring(0, idx), StandardCharsets.UTF_8.name()) : pair;
            if (!query_pairs.containsKey(key)) {
                query_pairs.put(key, new ArrayList<String>());
            }
            final String value = idx > 0 && pair.length() > idx + 1 ? URLDecoder.decode(pair.substring(idx + 1), StandardCharsets.UTF_8.name()) : null;
            query_pairs.get(key).add(value);
        }
        return query_pairs;
    } catch (UnsupportedEncodingException e) {
        throw new RuntimeException(e);
    }
}

回答by sendon1982

You can use simpler method from Collection based on @Flips solution:

您可以使用基于@Flips 解决方案的 Collection 中更简单的方法：

public String removeQueryParameter(String url, String parameterName) throws URISyntaxException {
    URIBuilder uriBuilder = new URIBuilder(url);
    List<NameValuePair> queryParameters = uriBuilder.getQueryParams();

    queryParameters.removeIf(param -> 
         param.getName().equals(parameterName));

    uriBuilder.setParameters(queryParameters);

    return uriBuilder.build().toString();
}