RestTemplate.exchange()
将对URL中的所有无效字符进行编码,但不会对+
进行编码,因为+
是有效的URL字符。但是如何在任何URL的查询参数中传递+
?
答案 0 :(得分:1)
如果您传递给RestTemplate的URI已将编码集设置为true,那么它将不会对您传递的URI执行编码,否则它将执行。
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);
}
}
因此,如果您需要在其中传递带有+
的查询参数,那么RestTemplate不会对+
进行编码,但是其他所有无效的URL字符都为+
,这是一个有效的URL字符。因此,您必须首先对参数(URLEncoder.encode("abc+123=", "UTF-8")
)进行编码,然后将编码后的参数传递给RestTemplate,以声明URI已经使用builder.build(true).toUri();
进行了编码,其中true
告诉RestTemplate URI是格式已编码,因此不再需要再次编码,因此+
将作为%2B
传递。
builder.build(true).toUri();
输出: http://example.com/endpoint?param1=abc%2B123%3D ,因为编码将执行一次。builder.build().toUri();
输出: http://example.com/endpoint?param1=abc%252B123%253D ,因为编码将进行两次。