Issue
I would like to run this specific curl command with a HTTP
POST
request in java
curl --location --request POST "http://106.51.58.118:5000/compare_faces?face_det=1" \
--header "user_id: myid" \
--header "user_key: thekey" \
--form "img_1=https://cdn.dnaindia.com/sites/default/files/styles/full/public/2018/03/08/658858-577200-katrina-kaif-052217.jpg" \
--form "img_2=https://cdn.somethinghaute.com/wp-content/uploads/2018/07/katrina-kaif.jpg"
I only know how to make simple POST
requests by passing a JSON
object, But i've never tried to POST
based on the above curl
command.
Here is a POST
example that I've made based on this curl
command:
curl -X POST TheUrl/sendEmail
-H 'Accept: application/json' -H 'Content-Type: application/json'
-d '{"emailFrom": "[email protected]", "emailTo":
["[email protected]"], "emailSubject": "Test email", "emailBody":
"708568", "generateQRcode": true}' -k
Here is how i did it using java
public void sendEmail(String url) {
try {
URL obj = new URL(url);
HttpURLConnection con = (HttpURLConnection) obj.openConnection();
//add reuqest header
con.setRequestMethod("POST");
con.setRequestProperty("Content-Type", "application/json; utf-8");
con.setRequestProperty("Accept", "application/json");
con.setDoOutput(true);
// Send post request
JSONObject test = new JSONObject();
test.put("emailFrom", emailFrom);
test.put("emailTo", emailTo);
test.put("emailSubject", emailSubject);
test.put("emailBody", emailBody);
test.put("generateQRcode", generateQRcode);
String jsonInputString = test.toString();
System.out.println(jsonInputString);
System.out.println("Email Response:" + returnResponse(con, jsonInputString));
} catch (Exception e) {
System.out.println(e);
}
System.out.println("Mail sent");
}
public String returnResponse(HttpURLConnection con, String jsonInputString) {
try (OutputStream os = con.getOutputStream()) {
byte[] input = jsonInputString.getBytes("utf-8");
os.write(input, 0, input.length);
} catch (Exception e) {
System.out.println(e);
}
try (BufferedReader br = new BufferedReader(new InputStreamReader(con.getInputStream(), "utf-8"))) {
StringBuilder response = new StringBuilder();
String responseLine = null;
while ((responseLine = br.readLine()) != null) {
response.append(responseLine.trim());
}
return response.toString();
} catch (Exception e) {
System.out.println("Couldnt read response from URL");
System.out.println(e);
return null;
}
}
I've found this useful link but i can't really understand how to use it in my example.
Is it any different from my example? and if yes how can i POST
the following data?
Note: Required Data
HEADERS:
user_id myid
user_key mykey
PARAMS:
face_det 1
boxes 120,150,200,250 (this is optional)
BODY:
img_1
multipart/base64 encoded image or remote url of image
img_2
multipart/base64 encoded image or remote url of image
Here is the complete documentation of the API
Solution
There are three things that your HttpURLConnection needs:
- The request method. You can set this with setRequestMethod.
- The headers. You can set them with setRequestProperty.
- The content type. The HTML specification requires that an HTTP request containing a form submission have
application/x-www-form-urlencoded
(ormultipart/form-data
) as its body’s content type. This is done by setting theContent-Type
header using thesetRequestProperty
method, just like the other headers.
It’s not clear what you’re trying to do here. As Boris Verkhovskiy points out, curl’s --form
option includes data as a part of a multipart request. In your command, the content of that request would be the characters of the URLs themselves. If you really want to submit URLs, not the images at those locations, you could use an application/x-www-form-urlencoded
request body to do it. The body itself needs to URL-encoded, as the content type indicates. The URLEncoder class exists for this purpose.
The steps look like this:
String img1 = "https://cdn.dnaindia.com/sites/default/files/styles/full/public/2018/03/08/658858-577200-katrina-kaif-052217.jpg";
String img2 = "https://cdn.somethinghaute.com/wp-content/uploads/2018/07/katrina-kaif.jpg";
con.setRequestMethod("POST");
con.setDoOutput(true);
con.setRequestProperty("user_id", myid);
con.setRequestProperty("user_key", thekey);
con.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
String body =
"img_1=" + URLEncoder.encode(img1, "UTF-8") + "&" +
"img_2=" + URLEncoder.encode(img2, "UTF-8");
try (OutputStream os = con.getOutputStream()) {
byte[] input = body.getBytes(StandardCharsets.UTF_8);
os.write(input);
}
However, if you want to submit the actual images, you will need to create a MIME request body. Java SE cannot do this, but the MimeMultipart class of JavaMail, which is part of the Java EE specification, can.
Multipart multipart = new MimeMultipart("form-data");
BodyPart part;
part = new MimeBodyPart();
part.setDataHandler(new DataHandler(new URL(img1)));
multipart.addBodyPart(part);
part = new MimeBodyPart();
part.setDataHandler(new DataHandler(new URL(img2)));
multipart.addBodyPart(part);
con.setRequestMethod("POST");
con.setDoOutput(true);
con.setRequestProperty("user_id", myid);
con.setRequestProperty("user_key", thekey);
con.setRequestProperty("Content-Type", multipart.getContentType());
try (OutputStream os = con.getOutputStream()) {
multipart.writeTo(os);
}
You should remove all catch
blocks from your code, and amend your method signatures to include throws IOException
(or throws IOException, MessagingException
). You don’t want users of your application to think the operation was successful if in fact it failed, right?
Answered By - VGR Answer Checked By - Mary Flores (WPSolving Volunteer)