是否有广泛使用的Java库可以执行dojo.objectToQuery()之类的操作?例如。 (假设使用HttpCore的HttpParams对象,但任何键值映射都可以):
HttpParams params = new BasicHttpParams()
.setParameter("foo", "bar")
.setParameter("thud", "grunt");
UnknownLibrary.toQueryString(params);
应该产生“foo = bar& thud = grunt”。
我知道写起来并不难,但似乎应该已经写好了。我找不到它。
答案 0 :(得分:11)
为什么重新发明轮子? Apache HttpClient有URLEncodedUtils:
List<BasicNameValuePair> params = Arrays.asList(new BasicNameValuePair("\\%^ &=@#:", "\\%^ &=@#:"));
String query = URLEncodedUtils.format(params, "UTF-8");
答案 1 :(得分:5)
你真的不需要一个库。我在HttpCore中找不到一个,所以我写了这样的东西,
public String toString() {
StringBuilder sb = new StringBuilder(baseUrl);
char separator;
if (baseUrl.indexOf('?') > 0)
separator = '&';
else
separator = '?';
for (Parameter p : parameters) {
sb.append(separator);
try {
sb.append(URLEncoder.encode(p.name, "UTF-8"));
if (p.value != null) {
sb.append('=');
sb.append(URLEncoder.encode(p.value, "UTF-8"));
}
} catch (UnsupportedEncodingException e) {
// Not really possible, throw unchecked exception
throw new IllegalStateException("No UTF-8");
}
separator = '&';
}
return sb.toString();
}
答案 2 :(得分:5)
我最终写了自己的。它可以被称为
URIUtils.withQuery(uri, "param1", "value1", "param2", "value2");
这不是那么糟糕。
/**
* Concatenates <code>uri</code> with a query string generated from
* <code>params</code>.
*
* @param uri the base URI
* @param params a <code>Map</code> of key/value pairs
* @return a new <code>URI</code>
*/
public static URI withQuery(URI uri, Map<String, String> params) {
StringBuilder query = new StringBuilder();
char separator = '?';
for (Entry<String, String> param : params.entrySet()) {
query.append(separator);
separator = '&';
try {
query.append(URLEncoder.encode(param.getKey(), "UTF-8"));
if (!StringUtils.isEmpty(param.getValue())) {
query.append('=');
query.append(URLEncoder.encode(param.getValue(), "UTF-8"));
}
} catch (UnsupportedEncodingException e) {
throw new RuntimeException(e);
}
}
return URI.create(uri.toString() + query.toString());
}
/**
* Concatenates <code>uri</code> with a query string generated from
* <code>params</code>. The members of <code>params</code> will be
* interpreted as {key1, val1, key2, val2}. Empty values can be given
* as <code>""</code> or <code>null</code>.
*
* @param uri the base URI
* @param params the key/value pairs in sequence
* @return a new <code>URI</code>
*/
public static URI withQuery(URI uri, String... params) {
Map<String, String> map = new LinkedHashMap<String, String>();
for (int i = 0; i < params.length; i += 2) {
String key = params[i];
String val = i + 1 < params.length ? params[i + 1] : "";
map.put(key, val);
}
return withQuery(uri, map);
}
答案 3 :(得分:1)
如果你有JAX-RS,那就是javax.ws.rs.core.UriBuilder;除此之外,您应该首先了解URLEncoder和URLDecoder,它们至少会编码参数列表的各个部分。
答案 4 :(得分:0)
有很多库可以帮助你构建URI(不要重新发明轮子)。这是一个让你入门的人:
start_urls
另请参阅: GIST > URI Builder Tests