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 将会 http://example.com?name=John

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

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

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

我之前已多次运行此场景,并且我希望在不破坏URL的情况下执行此操作.

Answer 1

有很多库可以帮助您构建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的:4.5.2

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

org.springframework:弹簧网:4.2.5.RELEASE

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

另请参见: GIST> URI Builder测试

Answer 2

这可以通过使用java.net.URI类使用现有部件构造新实例来完成,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&附加另一个参数,或者启动新查询.

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"));
    }
}

产量

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

对于 Android，请使用： 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();

Answer 4

使用该URI课程.

URI用你现有的东西创建一个新的String"分解"部分,然后实例化另一个来组装修改后的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 5

我建议改进 Adam 的答案，接受 HashMap 作为参数

/**
 * 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();
}