java - 仅使用 Java 搜索文本

Question

我正在使用 ajax google API 来搜索针对 google 的字符串。它返回给我所有包含文本的所有标签的 HTML 文件。

如果我只想获取文本，我应该使用什么？

我的程序是用 Java 编写的。

问候

曼乔特

Answer 1

我做了一些谷歌搜索，发现了这个:

http://www.ajaxlines.com/ajax/stuff/article/using_google_is_ajax_search_api_with_java.php

这是那里的示例代码片段:

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.net.URL;
import java.net.URLConnection;
import java.net.URLEncoder;
import org.json.JSONArray;      // JSON library from http://www.json.org/java/
import org.json.JSONObject;

public class GoogleQuery {

 // Put your website here
 private final String HTTP_REFERER = "http://www.example.com/";

 public GoogleQuery() {
  makeQuery("questio verum");
  makeQuery("info:http://frankmccown.blogspot.com/");
  makeQuery("site:frankmccown.blogspot.com");
 }

 private void makeQuery(String query) {

  System.out.println(" Querying for " + query);

  try
  {
   // Convert spaces to +, etc. to make a valid URL
   query = URLEncoder.encode(query, "UTF-8");

   URL url = new URL("http://ajax.googleapis.com/ajax/services/search/web?start=0&rsz=large&v=1.0&q=" + query);
   URLConnection connection = url.openConnection();
   connection.addRequestProperty("Referer", HTTP_REFERER);

   // Get the JSON response
   String line;
   StringBuilder builder = new StringBuilder();
   BufferedReader reader = new BufferedReader(
     new InputStreamReader(connection.getInputStream()));
   while((line = reader.readLine()) != null) {
    builder.append(line);
   }

   String response = builder.toString();
   JSONObject json = new JSONObject(response);

   System.out.println("Total results = " +
     json.getJSONObject("responseData")
     .getJSONObject("cursor")
     .getString("estimatedResultCount"));

   JSONArray ja = json.getJSONObject("responseData")
   .getJSONArray("results");

   System.out.println(" Results:");
   for (int i = 0; i < ja.length(); i++) {
    System.out.print((i+1) + ". ");
    JSONObject j = ja.getJSONObject(i);
    System.out.println(j.getString("titleNoFormatting"));
    System.out.println(j.getString("url"));
   }
  }
  catch (Exception e) {
   System.err.println("Something went wrong...");
   e.printStackTrace();
  }
 }

 public static void main(String args[]) {
  new GoogleQuery();
 }
}

作为旁注，您应该注意不要违反 Google TOS:
“您明确同意不通过任何自动化方式(包括使用脚本或网络爬虫)访问(或试图访问)任何服务，并应确保您遵守任何 robots.txt 中规定的说明”
- http://www.google.com/accounts/TOS