如何在Java中使用Elasticsearch Rest api？ [英] How to use Elasticsearch Rest api in java?

问题描述

我正在使用Apache Http客户端使用ElasticSearch Rest Api，但是我总是将HTTP错误代码作为200.请帮助

Java代码

  import java.io.BufferedReader; 
 import java.io.File; 
 import java.io.IOException; 
 import java.io.InputStreamReader; 
 import java.net.MalformedURLException; 
 import java.util.Scanner; 
 
 import org.apache.http.HttpResponse; 
 import org.apache.http.client.methods.HttpPost; 
 import org.apache.http.entity.StringEntity; 
 import org.apache.http.impl.client.DefaultHttpClient; 
 
 public class ApacheHttpClientPost {
 
 public static void main（String [] args）{
 String path =C：\\Tools\\ ElasticSearchApi\\javadoc.txt，filecontent =; 
 ApacheHttpClientPost apacheHttpClientPost = new ApacheHttpClientPost（）; 
 try {
 DefaultHttpClient httpClient = new DefaultHttpClient（）; 
 HttpPost postRequest = new HttpPost（http：// localhost：9200 / versioneg / message / _percolate）; 
 filecontent = apacheHttpClientPost.readFileContent（path）; 
 System.out.println（filecontent）; 
 StringEntity input = new StringEntity（filecontent）; 
 input.setContentType（application / json）; 
 postRequest.setEntity（input）; 
 HttpResponse response = httpClient.execute（postRequest）; 
 if（response.getStatusLine（）。getStatusCode（）！= 201）{
 throw new RuntimeException（Failed：HTTP error code：+ response.getStatusLine（）。getStatusCode（））; 
} 
 BufferedReader br = new BufferedReader（new InputStreamReader（（response.getEntity（）。getContent（））））; 
字符串输出; 
 System.out.println（从服务器输出.... \\\
）; 
 while（（output = br.readLine（））！= null）{
 
 System.out.println（output）; 
} 
 httpClient.getConnectionManager（）。shutdown（）; 
} catch（MalformedURLException e）{
 e.printStackTrace（）; 
} catch（IOException e）{
 e.printStackTrace（）; 
} 
} 
 
 private String readFileContent（String pathname）throws IOException {
 
文件文件=新建文件（路径名）; 
 StringBuilder fileContents = new StringBuilder（（int）file.length（））; 
扫描仪扫描仪=新扫描仪（文件）; 
 String lineSeparator = System.getProperty（line.separator）; 
 
 try {
 while（scanner.hasNextLine（））{
 fileContents.append（scanner.nextLine（）+ lineSeparator）; 
} 
 return fileContents.toString（）; 
} finally {
 scanner.close（）; 
} 
} 
} 
  

控制台

  {
doc：{
os：Linux，
product 
name：abc，
version：10.1，
configversion：1，
arch：32，
license 商业，
db：{
@type：Oracle
} 
} 
} 
} 
 
线程main中的异常java.lang.RuntimeException：失败：HTTP错误代码：200 
在com.informatica.zanshin.esapi.utils.ApacheHttpClientPost.main（ApacheHttpClientPost.java:31）
  

这里是弹性搜索意义截图

解决方案

状态码200代表'OK'

检查w3c ref

您应该使用

  if（response .getStatusLine（）。getStatusCode（）！= 200）{
 //抛出异常或其他东西
} 
  

I am using Apache Http client for making use of ElasticSearch Rest Api, but I am always getting HTTP error code as 200. Please help

Java code

import java.io.BufferedReader;
import java.io.File;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.MalformedURLException;
import java.util.Scanner;

import org.apache.http.HttpResponse;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.DefaultHttpClient;

public class ApacheHttpClientPost {

    public static void main(String[] args) {
        String path="C:\\Tools\\ElasticSearchApi\\javadoc.txt", filecontent="";
        ApacheHttpClientPost apacheHttpClientPost = new ApacheHttpClientPost();
        try {
            DefaultHttpClient httpClient = new DefaultHttpClient();
            HttpPost postRequest = new HttpPost("http://localhost:9200/versioneg/message/_percolate");
            filecontent=apacheHttpClientPost.readFileContent(path);
            System.out.println(filecontent);
            StringEntity input = new StringEntity(filecontent);
            input.setContentType("application/json");
            postRequest.setEntity(input);
            HttpResponse response = httpClient.execute(postRequest);
            if (response.getStatusLine().getStatusCode() != 201) {
                throw new RuntimeException("Failed : HTTP error code : " + response.getStatusLine().getStatusCode());
            }
            BufferedReader br = new BufferedReader(new InputStreamReader((response.getEntity().getContent())));
            String output;
            System.out.println("Output from Server .... \n");
            while ((output = br.readLine()) != null) {

                System.out.println(output);
            }
            httpClient.getConnectionManager().shutdown();
        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    private String readFileContent(String pathname) throws IOException {

        File file = new File(pathname);
        StringBuilder fileContents = new StringBuilder((int)file.length());
        Scanner scanner = new Scanner(file);
        String lineSeparator = System.getProperty("line.separator");

        try {
            while(scanner.hasNextLine()) {        
                fileContents.append(scanner.nextLine() + lineSeparator);
            }
            return fileContents.toString();
        } finally {
            scanner.close();
        }
    }
}

Console

{
   "doc": {
      "os": "Linux",
      "product": {
         "name": "abc",
         "version": 10.1,
         "configversion": 1,
         "arch": 32,
         "license": "commercial",
         "db": {
            "@type": "Oracle"
         }
      }
   }
}

Exception in thread "main" java.lang.RuntimeException: Failed : HTTP error code : 200
    at com.informatica.zanshin.esapi.utils.ApacheHttpClientPost.main(ApacheHttpClientPost.java:31)

Here is elasticsearch sense screenshot

解决方案

Status Code 200 stands for 'OK'
check w3c ref

You should use

    if(response.getStatusLine().getStatusCode() != 200){
        // Throw exception or something else
    } 

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