You need to add "&rsz=NUMBER" to your url, where NUMBER is a number from 1 - 8, inclusive. Beyond that, you can also use the start parameter to pull additional "pages" of results with subsequent requests.
That said, it is important to note that the AJAX Search API (all varieties) has been deprecated and scheduled to be discontinued no later than May 2014. The Web Search API, in particular, will be discontinued no later than November 2013. You may want to utilize the Custom Search API or consider other options, depending on your application needs. At the very least, be aware of the pending doom of the AJAX APIs. Jeremy R. Geerdes Generally Cool Guy Des Moines, IA For more information or a project quote: [email protected] If you're in the Des Moines, IA, area, check out Debra Heights Wesleyan Church! On Jun 29, 2011, at 8:08 AM, assma.m wrote: > Hi, > I'm working in Java (Eclipse Platform). > I'm a new user of Google Ajax APIs. > I need to get results from google when giving a query. > But I was surprised that I got only four results. I don't know how to > get more results . > This is the code (I developed it using : > http://code.google.com/intl/fr/apis/websearch/docs/#The_Basics : the > JAVA Access Section) : > > import java.io.BufferedReader; > import java.io.InputStreamReader; > import java.net.URL; > import java.net.URLConnection; > import java.net.URLEncoder; > > > import net.sf.json.*; > > > public class GoogleQuery { > > private final String HTTP_REFERER = "http:// > www.mygooglesearchssite.com/AsmaSite/"; > > public GoogleQuery() { > makeQuery("life"); > } > > private void makeQuery(String query) { > > System.out.println("\nQuerying for " + query); > > try > { > // Convert spaces to +, etc. to make a valid URL > query = URLEncoder.encode(query, "UTF-8"); > > URL url = new URL( > "https://ajax.googleapis.com/ajax/services/search/web?v=1.0&"; > + "q="+query > +"&key=MYKEY&userip=MYUSERIP"); > > 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) { > System.out.println(line); > builder.append(line); > } > > String response = builder.toString(); > JSONObject json = new JSONObject(builder.toString()); > JSONObject jsonObject = new JSONObject(); > > System.out.println("Total results = " + > json.getJSONObject("responseData") > .getJSONObject("cursor") > .getString("estimatedResultCount")); > > JSONArray ja = > json.getJSONObject("responseData").getJSONArray("results"); > > System.out.println("\nResults:"); > 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(); > } > } > > I need your help. > Regards, > Asma > > -- > You received this message because you are subscribed to the Google Groups > "Google AJAX APIs" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/google-ajax-search-api?hl=en. > -- You received this message because you are subscribed to the Google Groups "Google AJAX APIs" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/google-ajax-search-api?hl=en.
