Reputation: 310
I am trying to get a JSON response from Google Trends using HTTP. This is my code snippet:
public class TestClass {
public static void main(String[] args) throws Exception{
String address = "https://trends.google.com/trends/api/explore?hl=en-US&tz=240&req={\"comparisonItem\":[{\"keyword\":\"Miley\",\"geo\":\"US\",\"time\":\"2012-01-01 2014-01-01\"},{\"keyword\":\"Hannah Montana\",\"geo\":\"US\",\"time\":\"2012-01-01 2014-01-01\"}],\"category\":0,\"property\":\"\"}";
URL url = new URL(address);
HttpURLConnection con = (HttpURLConnection) url.openConnection();
con.setRequestMethod("GET");
int responseCode = con.getResponseCode();
System.out.println("URL is "+address);
System.out.println("Response code is " + responseCode); }
}
This is the output:
URL is https://trends.google.com/trends/api/explore?hl=en-US&tz=240&req={"comparisonItem":[{"keyword":"Miley","geo":"US","time":"2012-01-01 2014-01-01"},{"keyword":"Hannah Montana","geo":"US","time":"2012-01-01 2014-01-01"}],"category":0,"property":""}
Response code is 400
If I type the URL directly in my browser, Google gives me a JSON file with no problem. However, if I try to access that URL using Java, I am given a bad request. How can I solve this problem? Thanks in advance.
Upvotes: 1
Views: 1915
Reputation: 987
I solved your problem. I higly recomend http-request built on apache http api.
private static final HttpRequest<String> REQUEST =
HttpRequestBuilder.createGet("https://trends.google.com/trends/api/explore", String.class)
.addDefaultRequestParameter("hl", "en-US")
.addDefaultRequestParameter("tz", "240")
.responseDeserializer(ResponseDeserializer.ignorableDeserializer())
.build();
public void send() {
ResponseHandler<String> responseHandler = REQUEST.execute("req", "{\"comparisonItem\":[{\"keyword\":\"Miley\",\"geo\":\"US\",\"time\":\"2012-01-01 2014-01-01\"},{\"keyword\":\"Hannah Montana\",\"geo\":\"US\",\"time\":\"2012-01-01 2014-01-01\"}],\"category\":0,\"property\":\"\"}");
System.out.println(responseHandler.getStatusCode());
responseHandler.ifHasContent(System.out::println);
}
The code prints the response code 200 and response body which you get by browser.
Upvotes: 1
Reputation: 5855
You need to URL Encode the query string portion of your URL. Check out this question/answer for some ways to achieve this.
Upvotes: 3