RestTemplate.exchange()
方法会对URL中所有无效字符进行编码,但不会将+
编码,因为它是一个合法的URL字符。但是如果想在任何URL的查询参数中传递+
,该怎么办呢?
RestTemplate.exchange()
方法会对URL中所有无效字符进行编码,但不会将+
编码,因为它是一个合法的URL字符。但是如果想在任何URL的查询参数中传递+
,该怎么办呢?
如果您向RestTemplate传递的URI中将encoded设置为true,则它不会对您传递的URI进行编码,否则它将执行编码。
import java.io.UnsupportedEncodingException;
import java.net.URI;
import java.net.URLEncoder;
import java.util.Collections;
import org.springframework.http.HttpEntity;
import org.springframework.http.HttpHeaders;
import org.springframework.http.HttpMethod;
import org.springframework.http.ResponseEntity;
import org.springframework.http.client.BufferingClientHttpRequestFactory;
import org.springframework.http.client.SimpleClientHttpRequestFactory;
import org.springframework.web.client.RestTemplate;
import org.springframework.web.util.UriComponentsBuilder;
class Scratch {
public static void main(String[] args) {
RestTemplate rest = new RestTemplate(
new BufferingClientHttpRequestFactory(new SimpleClientHttpRequestFactory()));
HttpHeaders headers = new HttpHeaders();
headers.add("Content-Type", "application/json");
headers.add("Accept", "application/json");
HttpEntity<String> requestEntity = new HttpEntity<>(headers);
UriComponentsBuilder builder = null;
try {
builder = UriComponentsBuilder.fromUriString("http://example.com/endpoint")
.queryParam("param1", URLEncoder.encode("abc+123=", "UTF-8"));
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}
URI uri = builder.build(true).toUri();
ResponseEntity responseEntity = rest.exchange(uri, HttpMethod.GET, requestEntity, String.class);
}
}
如果您需要传递带有+
的查询参数,则RestTemplate不会对+
进行编码,但将所有其他无效的URL字符都编码为+
,因为+
是URL中有效的字符。因此,您必须先对参数进行编码(URLEncoder.encode("abc+123=", "UTF-8")
),然后将编码后的参数传递给RestTemplate,指定URI已经使用builder.build(true).toUri();
进行了编码,其中true
告诉RestTemplate URI已经被编码过了,所以不要再次编码,因此+
将被传递为%2B
。
builder.build(true).toUri();
输出: http://example.com/endpoint?param1=abc%2B123%3D 因为编码只会执行一次。builder.build().toUri();
输出: http://example.com/endpoint?param1=abc%252B123%253D 因为编码将执行两次。