如何使用 Spring RestTemplate POST 表单数据？

Question

我想将以下（工作）curl 片段转换为 RestTemplate 调用：

curl -i -X POST -d "email=first.last@example.com" https://app.example.com/hr/email

如何正确传递电子邮件参数？ 以下代码导致 404 Not Found 响应：

String url = "https://app.example.com/hr/email";

Map<String, String> params = new HashMap<String, String>();
params.put("email", "first.last@example.com");

RestTemplate restTemplate = new RestTemplate();
ResponseEntity<String> response = restTemplate.postForEntity( url, params, String.class );

我尝试在 PostMan 中制定正确的调用，并且可以通过将 email 参数指定为正文中的“form-data”参数来使其正常工作。 在 RestTemplate 中实现此功能的正确方法是什么？

Answer 1

POST 方法应该与 HTTP 请求对象一起发送。 并且请求可能包含 HTTP 标头或 HTTP 正文或两者。

因此，让我们创建一个 HTTP 实体并在正文中发送标头和参数。

HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);

MultiValueMap<String, String> map= new LinkedMultiValueMap<String, String>();
map.add("email", "first.last@example.com");

HttpEntity<MultiValueMap<String, String>> request = new HttpEntity<MultiValueMap<String, String>>(map, headers);

ResponseEntity<String> response = restTemplate.postForEntity( url, request , String.class );

http://docs.spring.io/spring/docs/current/javadoc-api/org/springframework/web/client/RestTemplate.html#postForObject-java.lang.String-java.lang.Object-java.lang。类-java.lang.Object...-

Answer 2

如何在一个请求中发布混合数据：文件、字符串[]、字符串。

您可以只使用您需要的东西。

private String doPOST(File file, String[] array, String name) {
    RestTemplate restTemplate = new RestTemplate(true);

    //add file
    LinkedMultiValueMap<String, Object> params = new LinkedMultiValueMap<>();
    params.add("file", new FileSystemResource(file));

    //add array
    UriComponentsBuilder builder = UriComponentsBuilder.fromHttpUrl("https://my_url");
    for (String item : array) {
        builder.queryParam("array", item);
    }

    //add some String
    builder.queryParam("name", name);

    //another staff
    String result = "";
    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.MULTIPART_FORM_DATA);

    HttpEntity<LinkedMultiValueMap<String, Object>> requestEntity =
            new HttpEntity<>(params, headers);

    ResponseEntity<String> responseEntity = restTemplate.exchange(
            builder.build().encode().toUri(),
            HttpMethod.POST,
            requestEntity,
            String.class);

    HttpStatus statusCode = responseEntity.getStatusCode();
    if (statusCode == HttpStatus.ACCEPTED) {
        result = responseEntity.getBody();
    }
    return result;
}

POST 请求的正文和下一个结构中将包含 File：

POST https://my_url?array=your_value1&array=your_value2&name=bob 

Answer 3

这是使用 spring 的 RestTemplate 进行 POST 休息调用的完整程序。

import java.util.HashMap;
import java.util.Map;

import org.springframework.http.HttpEntity;
import org.springframework.http.ResponseEntity;
import org.springframework.util.LinkedMultiValueMap;
import org.springframework.util.MultiValueMap;
import org.springframework.web.client.RestTemplate;

import com.ituple.common.dto.ServiceResponse;

   public class PostRequestMain {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        MultiValueMap<String, String> headers = new LinkedMultiValueMap<String, String>();
        Map map = new HashMap<String, String>();
        map.put("Content-Type", "application/json");

        headers.setAll(map);

        Map req_payload = new HashMap();
        req_payload.put("name", "piyush");

        HttpEntity<?> request = new HttpEntity<>(req_payload, headers);
        String url = "http://localhost:8080/xxx/xxx/";

        ResponseEntity<?> response = new RestTemplate().postForEntity(url, request, String.class);
        ServiceResponse entityResponse = (ServiceResponse) response.getBody();
        System.out.println(entityResponse.getData());
    }

}

Answer 4

客户端.java

@PostMapping(value = "/employee", consumes = "application/json")
public Employee createProducts(@RequestBody Employee product) {
    HttpHeaders headers = new HttpHeaders();
    headers.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));
    HttpEntity<Employee> entity = new HttpEntity<Employee>(product,headers);

    ResponseEntity<Employee> response = restTemplate.exchange(
            "http://hello-server/rest/employee", HttpMethod.POST, entity, Employee.class);

    return response.getBody();
}

服务器端.java

private static List<Employee> list = new ArrayList<>();

@PostMapping(path="rest/employee", consumes = "application/json")
public Employee createEmployee(@RequestBody Employee employee)

{
    list.add(employee);
    return employee;
}
static
{
    list.add(new Employee(1, "albert", "Associate", "mphasis"));
    list.add(new Employee(2, "sachin", "software engineer", "mphasis"));
    list.add(new Employee(3, "dhilip", "Lead engineer", "IBM"));
}

雇员.java

public class Employee {

private Integer id;
private String name;
private String Designation;
private String company;
 // generate getter setter and toString()
}

1.发布请求

Answer 5

您的url String需要为您传递的地图使用变量标记，例如：

String url = "https://app.example.com/hr/email?{email}";

或者你可以显式地将查询参数编码到字符串中以开始，而不必传递地图，例如：

String url = "https://app.example.com/hr/email?email=first.last@example.com";

另请参见https://stackoverflow.com/a/47045624/1357094