Question

我尝试使用以下代码调用WEB API并获取JSON输出。

我已经使用POSTMAN对此进行了测试，并且运行正常。标头中有两个值，POST请求的主体中有一些值。

但是当我尝试使用 Apache HTTP Client访问相同的WEB API时，我得到以下输出和错误：

Response Code : 200
Result:{"errorCode":3000,"errorMessage":"Invalid request parameters"}

代码：

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;

import java.util.logging.Level;
import java.util.logging.Logger;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.HttpClient;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.HttpClientBuilder;
import org.apache.http.message.BasicNameValuePair;


public class WhiteSourceAPI {

public static void main(String[] args) {

    try {
        String url = "https://example.com/api";

        HttpClient client = HttpClientBuilder.create().build();
        HttpPost post = new HttpPost(url);

        //header
        post.setHeader("Content-Type", "application/json");
        post.setHeader("Accept-Charset", "UTF-8");

        List<NameValuePair> urlParameters = new ArrayList<>();
        urlParameters.add(new BasicNameValuePair("requestType", "xxxx"));
        urlParameters.add(new BasicNameValuePair("projectToken", "xxxx-xxxx-xxxx-xxxx"));

        post.setEntity(new UrlEncodedFormEntity(urlParameters));

        HttpResponse response = client.execute(post);
        System.out.println("Response Code : " + response.getStatusLine().getStatusCode());

        BufferedReader rd = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));

        StringBuilder result = new StringBuilder();
        String line = "";
        while ((line = rd.readLine()) != null) {
            result.append(line);
        }
        System.out.println("Result:" +result);
    } catch (IOException ex) {
        Logger.getLogger(WhiteSourceAPI.class.getName()).log(Level.SEVERE, null, ex);
    }
}
}

Answer 1

正如@VitalyZ所提到的，我不应该使用UrlEncodedFormEntity作为正文，而是使用原始JSON。

toSort = "<div class='touchpoint_filterattributes' id='BODIVID_1571' style = 'background:#b7b7b7;'><div class='color_box_label' style = 'color:#000000;'>Customer Experience Team</div></div><br/><div class='touchpoint_filterattributes' id='BODIVID_1567' style = 'background:#b7b7b7;'><div class='color_box_label' style = 'color:#000000;'>1234</div></div><br/><div class='touchpoint_filterattributes' id='BODIVID_1569' style = 'background:#b7b7b7;'><div class='color_box_label' style = 'color:#000000;'>Claims</div></div><br/><div class='touchpoint_filterattributes' id='BODIVID_1266' style = 'background:#C9C9C9;'><div class='color_box_label' style = 'color:#000000;'>Claims</div></div><br/><div class='touchpoint_filterattributes' id='BODIVID_1570' style = 'background:#b7b7b7;'><div class='color_box_label' style = 'color:#000000;'>Corporate Communication</div></div><br/><div class='touchpoint_filterattributes' id='BODIVID_1260' style='background:#C9C9C9;'><div class='color_box_label' style = 'color:#000000;'>Claims</div></div><br/>";

a = toSort.split("<br/>");
obj = {}
for(key in a){
	id = a[key].substr(53,4);
  obj[id] = a[key];
}
newhtml = "";
for(key in obj){
	newhtml += obj[key]+"<br/>";
}
console.log(newhtml);

访问使用POSTMAN运行良好的Web API时出错

1 个答案: