如何将查询参数附加到现有URL？

Question

我想将键值对作为查询参数附加到现有URL。虽然我可以通过检查URL是否有查询部分或片段部分来执行此操作，并通过跳过一堆if子句来执行追加，但我想知道如果通过Apache执行此操作是否有干净的方法Commons库或类似的东西。

http://example.com将为http://example.com?name=John

http://example.com#fragment将为http://example.com?name=John#fragment

http://example.com?email=john.doe@email.com将为http://example.com?email=john.doe@email.com&name=John

http://example.com?email=john.doe@email.com#fragment将为http://example.com?email=john.doe@email.com&name=John#fragment

我之前已多次运行此方案，并且我希望在不破坏URL的情况下执行此操作。

Answer 1

有很多库可以帮助你构建URI（不要重新发明轮子）。这里有三个让你入门：

Java EE 7

import javax.ws.rs.core.UriBuilder;
...
return UriBuilder.fromUri(url).queryParam(key, value).build();

org.apache.httpcomponents:httpclient:4.5.2

import org.apache.http.client.utils.URIBuilder;
...
return new URIBuilder(url).addParameter(key, value).build();

org.springframework:spring-web:4.2.5.RELEASE

import org.springframework.web.util.UriComponentsBuilder;
...
return UriComponentsBuilder.fromUriString(url).queryParam(key, value).build().toUri();

另请参阅： GIST > URI Builder Tests

Answer 2

这可以通过使用java.net.URI类使用现有部分构建新实例来完成，这应该确保它符合URI语法。

查询部分将为null或现有字符串，因此您可以决定使用＆amp;添加另一个参数。或者开始一个新的查询。

public class StackOverflow26177749 {

    public static URI appendUri(String uri, String appendQuery) throws URISyntaxException {
        URI oldUri = new URI(uri);

        String newQuery = oldUri.getQuery();
        if (newQuery == null) {
            newQuery = appendQuery;
        } else {
            newQuery += "&" + appendQuery;  
        }

        URI newUri = new URI(oldUri.getScheme(), oldUri.getAuthority(),
                oldUri.getPath(), newQuery, oldUri.getFragment());

        return newUri;
    }

    public static void main(String[] args) throws Exception {
        System.out.println(appendUri("http://example.com", "name=John"));
        System.out.println(appendUri("http://example.com#fragment", "name=John"));
        System.out.println(appendUri("http://example.com?email=john.doe@email.com", "name=John"));
        System.out.println(appendUri("http://example.com?email=john.doe@email.com#fragment", "name=John"));
    }
}

输出

http://example.com?name=John
http://example.com?name=John#fragment
http://example.com?email=john.doe@email.com&name=John
http://example.com?email=john.doe@email.com&name=John#fragment

Answer 3

使用URI类。

使用您现有的URI创建一个新的String，以及＃34;将其分解为＆＃34;部分，并实例化另一个以组装修改后的URL：

URI u = new URI("http://example.com?email=john@email.com&name=John#fragment");

// Modify the query: append your new parameter
StringBuilder sb = new StringBuilder(u.getQuery() == null ? "" : u.getQuery());
if (sb.length() > 0)
    sb.append('&');
sb.append(URLEncoder.encode("paramName", "UTF-8"));
sb.append('=');
sb.append(URLEncoder.encode("paramValue", "UTF-8"));

// Build the new url with the modified query:
URI u2 = new URI(u.getScheme(), u.getAuthority(), u.getPath(),
    sb.toString(), u.getFragment());

Answer 4

我建议对接受HashMap作为参数的Adam答案进行改进

/**
 * Append parameters to given url
 * @param url
 * @param parameters
 * @return new String url with given parameters
 * @throws URISyntaxException
 */
public static String appendToUrl(String url, HashMap<String, String> parameters) throws URISyntaxException
{
    URI uri = new URI(url);
    String query = uri.getQuery();

    StringBuilder builder = new StringBuilder();

    if (query != null)
        builder.append(query);

    for (Map.Entry<String, String> entry: parameters.entrySet())
    {
        String keyValueParam = entry.getKey() + "=" + entry.getValue();
        if (!builder.toString().isEmpty())
            builder.append("&");

        builder.append(keyValueParam);
    }

    URI newUri = new URI(uri.getScheme(), uri.getAuthority(), uri.getPath(), builder.toString(), uri.getFragment());
    return newUri.toString();
}

Answer 5

Kotlin＆clean，因此您无需在代码审查之前进行重构：

private fun addQueryParameters(url: String?): String? {
        val uri = URI(url)

        val queryParams = StringBuilder(uri.query.orEmpty())
        if (queryParams.isNotEmpty())
            queryParams.append('&')

        queryParams.append(URLEncoder.encode("$QUERY_PARAM=$param", Xml.Encoding.UTF_8.name))
        return URI(uri.scheme, uri.authority, uri.path, queryParams.toString(), uri.fragment).toString()
    }

Answer 6

亚当答案的更新也考虑了胰蛋白酶的答案。不必在循环中实例化一个字符串。

public static URI appendUri(String uri, Map<String, String> parameters) throws URISyntaxException {
    URI oldUri = new URI(uri);
    StringBuilder queries = new StringBuilder();

    for(Map.Entry<String, String> query: parameters.entrySet()) {
        queries.append( "&" + query.getKey()+"="+query.getValue());
    }

    String newQuery = oldUri.getQuery();
    if (newQuery == null) {
        newQuery = queries.substring(1);
    } else {
        newQuery += queries.toString();
    }

    URI newUri = new URI(oldUri.getScheme(), oldUri.getAuthority(),
            oldUri.getPath(), newQuery, oldUri.getFragment());

    return newUri;
}

Answer 7

对于android，使用： https://developer.android.com/reference/android/net/Uri#buildUpon()

URI oldUri = new URI(uri);
Uri.Builder builder = oldUri.buildUpon();
 builder.appendQueryParameter("newParameter", "dummyvalue");
 Uri newUri =  builder.build();