【问题标题】:Use CORS in a http request在 http 请求中使用 CORS
【发布时间】:2018-03-05 14:39:07
【问题描述】:

今天,我有以下问题。我想向 rest api 发出 http 请求,但我只收到 405 错误。我搜索并发现,api使用cors。 那么如何使用 cors 发出 http 请求呢?

这是我的 http 客户端:

package community.opencode.minetools4j.util.http;

import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NonNull;
import org.apache.commons.io.IOUtils;

import java.io.DataOutputStream;
import java.io.IOException;
import java.io.UnsupportedEncodingException;
import java.net.HttpURLConnection;
import java.net.URL;
import java.net.URLEncoder;
import java.util.HashMap;
import java.util.Map;

/**
 * This is an utility class.
 * It simplify making HTTP requests
 */
public class HttpRequest {

    /**
     * Performs a new HTTP request
     *
     * @param requestBuilder See {@link RequestBuilder}
     * @return See {@link RequestResponse}
     * @throws IOException Will be thrown if the request can't be executed
     */
    public static RequestResponse performRequest(RequestBuilder requestBuilder) throws IOException {
        URL newUrl = new URL(requestBuilder.getPath());

        HttpURLConnection connection = (HttpURLConnection) newUrl.openConnection();
        connection.setRequestMethod(requestBuilder.getMethod().getType());
        requestBuilder.getHeaders().forEach(connection::setRequestProperty);
        connection.setConnectTimeout(5000);
        connection.setReadTimeout(5000);
        connection.setDoOutput(true);

        DataOutputStream out = new DataOutputStream(connection.getOutputStream());
        out.writeBytes(getParametersString(requestBuilder.getParams()));

        int status = connection.getResponseCode();
        String result = IOUtils.toString(connection.getInputStream(), "UTF-8");
        String contentType = connection.getHeaderField("Content-Type");
        connection.getInputStream().close();
        connection.disconnect();
        return new RequestResponse(status, result, contentType);
    }

    /**
     * Makes from a map a valid HTTP string
     * Example: "myField=Hey!&myPassword=abc"
     *
     * @param params The parameters
     * @return The formed String
     * @throws UnsupportedEncodingException Should never be thrown
     */
    private static String getParametersString(Map<String, String> params) throws UnsupportedEncodingException {
        StringBuilder builder = new StringBuilder();
        for (Map.Entry<String, String> entry : params.entrySet()) {
            builder.append(URLEncoder.encode(entry.getKey(), "UTF-8"));
            builder.append("=");
            builder.append(URLEncoder.encode(entry.getValue(), "UTF-8"));
            builder.append("&");
        }
        String result = builder.toString();
        return result.length() > 0 ? result.substring(0, result.length() - 1) : result;
    }

    /**
     * A simple builder class for {@link #performRequest(RequestBuilder)}
     */
    @Data
    public static class RequestBuilder {

        private final String path;
        private final HttpRequestMethod method;

        private final Map<String, String> params = new HashMap<>();
        private final Map<String, String> headers = new HashMap<>();

        public RequestBuilder addParam(@NonNull String name, @NonNull String value) {
            this.params.put(name, value);
            return this;
        }

        public RequestBuilder addHeader(@NonNull String name, @NonNull String value) {
            this.headers.put(name, value);
            return this;
        }
    }

    /**
     * A simplified http response.
     * Including status, message and the returned content type
     */
    @Data
    @AllArgsConstructor
    public static class RequestResponse {

        private int status;
        private String resultMessage;
        private String contentType;
    }
}

package community.opencode.minetools4j.util.http;

import lombok.Getter;

/**
 * Simple enum for {@link HttpRequest}.
 * It's a better way than just write manually the method name
 */
public enum HttpRequestMethod {
    POST("POST"),
    GET("GET"),
    PUT("PUT"),
    HEAD("HEAD"),
    OPTIONS("OPTIONS"),
    DELETE("DELETE"),
    TRACE("TRACE");

    @Getter
    private String type;

    HttpRequestMethod(String type) {
        this.type = type;
    }
}

这是我发出http请求的方法:

HttpRequest.RequestResponse requestResponse = HttpRequest.performRequest(new HttpRequest.RequestBuilder(API_URI + "/ping/" + host + "/" + port, HttpRequestMethod.GET));

感谢您的关注。

真诚地, 斯凯莱格

对不起,我的英语不好。我希望你能理解我;)

【问题讨论】:

  • 第一个代码块中有两个类。抱歉,这是我的第一个问题 :)
  • 405 表示“方法不允许”。您的问题与 CORS 无关,它只影响来自浏览器的 AJAX 请求。服务器可能需要一个帖子,而不是一个获取。

标签: java http cors httprequest


【解决方案1】:

什么是 HTTP 代码 405

状态 405 表示您发送的请求不允许您使用的方法。

Source

如何解决

您需要检查 API 文档,看看您是否有以下错误之一:

  • 缺少 Auth 标头
  • 缺少内容
  • 网址错误
  • 方法错误

简化代码

使用 unirest 可以简化您的请求代码,并涵盖边缘情况

【讨论】:

    【解决方案2】:

    如果您遇到此类错误:

    对预检请求的响应未通过访问控制检查:请求的资源上不存在“Access-Control-Allow-Origin”标头。因此,Origin 'http://localhost:4200' 不允许访问。响应的 HTTP 状态代码为 405。

    您必须创建一个 servlet 过滤器并在 doFilter 方法上添加以下代码。

        public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse, FilterChain chain)
            throws IOException, ServletException {
        HttpServletRequest request = (HttpServletRequest) servletRequest;
    
        HttpServletResponse resp = (HttpServletResponse) servletResponse;
        resp.addHeader("Access-Control-Allow-Origin", "*");
        resp.addHeader("Access-Control-Allow-Methods", "GET, POST, PUT, DELETE,HEAD");
        resp.addHeader("Access-Control-Allow-Headers", "Origin, X-Requested-With, Content-Type, Accept");
        resp.addHeader("Accept-Encoding", "multipart/form-data");
    
        if (request.getMethod().equals("OPTIONS")) {
            resp.setStatus(200);
            return;
        }
        chain.doFilter(request, servletResponse);
      }
    

    【讨论】:

    • 这在浏览器中是准确的,但在浏览器外部的客户端应用程序中,它可能指向使用了错误的方法
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