Convert Curl to Java - java

I am having trouble converting this to Java. So the curl line is
-d '{"skills":{"__op":"AddUnique","objects":["flying","kungfu"]}}' \
Normally what I do to a curl command such as this:
-d '{"score":73453}' \
would be:
httpRequest.setContent("{\"" +
"score\": 73452 "+
"}");

Not entirely sure what your issue actually is but a reasonably straight forward way to match that command in Java is this:
String url = "whatever.your.url.is";
HttpURLConnection connection = (HttpURLConnection) new URL(url).openConnection();
connection .setRequestProperty("Content-Type", "application/json");
connection .setRequestMethod("POST");
JSONObject json =new JSONObject();
JSONObject skills =new JSONObject();
skills.put("__op", "AddUnique");
skills.put("objects", new JSONArray(Arrays.asList("flying", "kungfu"));
json.put("skills": skills);
OutputStreamWriter wr= new OutputStreamWriter(connection.getOutputStream());
wr.write(json.toString());

Related

How to transform a curl command to HTTP POST request using java

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": "smth#domain.com", "emailTo":
["smth#gmail.com"], "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
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 (or multipart/form-data) as its body’s content type. This is done by setting the Content-Type header using the setRequestProperty 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?

cURL DELETE on java

I use command "curl -X DELETE --header 'Accept: application/json' 'http://10.10.1.29:8181/onos/v1/flows/application/olsrflow' -u karaf:karaf"
It Can work.
but use JAVA Code does't work.
have any problem in my JAVA code?
URL dc0ContrailUrl2 = new URL("http://10.10.1.29:8181/onos/v1/flows/application/olsrflow");
HttpURLConnection dcConn2 = (HttpURLConnection) dc0ContrailUrl2.openConnection();
dcConn2.setDoOutput(true);
String login = "karaf:karaf";
String content = URLEncoder.encode (login) ;
String basicAuth = "Basic " + new String(new Base64().encode(login.getBytes()));
dcConn2.setRequestProperty("Authorization",basicAuth);
dcConn2.setRequestProperty("Content-Type", "application/json");
dcConn2.setRequestMethod("DELETE");
BufferedReader in2 = new BufferedReader(new InputStreamReader(dcConn2.getInputStream()));
String inputLine2;
while ((inputLine2 = in2.readLine()) != null){ //while response is not null, assign response to inputLine and print inputLine
System.out.println(inputLine2);
}
in2.close();
error HTTP response code: 415
You did not add Accept type the same as curl. Add the following line and remove the Content-Type:
dcConn2.setRequestProperty("Accept", "application/json");
Typically it means that server does not support MediaType has been passed. You have to figure out what does it expect and setup it in your request accordingly.

HTTP Post message for firebase in Java

I would like to send a notification from my Java program to Firebase. But I don't know how to do that.
In order to send a notification using curl:
# api_key=YOUR_SERVER_KEY
# curl --header "Authorization: key=$api_key" \
--header Content-Type:"application/json" \
https://fcm.googleapis.com/fcm/send \
-d "{\"registration_ids\":[\"ABC\"]}"
I want the equivalent of that in Java. I tried the following but it is not correct
String rawData = "{\"registration_ids\":[\"ABC\"]}";
String encodedData = URLEncoder.encode( rawData, "UTF-8" );
URL u = new URL("https://fcm.googleapis.com/fcm/send");
HttpURLConnection conn = (HttpURLConnection) u.openConnection();
conn.setDoOutput(true);
conn.setRequestMethod("POST");
conn.setRequestProperty("Authorization", "key="+api_key);
conn.setRequestProperty("Content-Type", "application/json");
OutputStream os = conn.getOutputStream();
os.write(encodedData.getBytes());

Convert Curl to Java equivalent

I'm working with New Relic REST API for the first time, I have a curl command:
curl -X GET 'https://api.newrelic.com/v2/applications/appid/metrics/data.json' \
-H 'X-Api-Key:myApiKey' -i \
-d 'names[]=EndUser/WebTransaction/WebTransaction/JSP/index.jsp'
I want to send this command in a java servlet and get a JSON object from the response ready for parsing, What is the best solution?
HttpURLConnection?
Apache httpclient?
I've tried a few different solutions, but nothing has worked so far and most examples I could find are using the depreciated DefaultHttpClient
Here is an example of one of my attempts:
String url = "https://api.newrelic.com/v2/applications.json";
HttpURLConnection conn = (HttpURLConnection) new URL(url).openConnection();
conn.setRequestProperty("Content-Type", "application/json");
conn.setRequestProperty("X-Api-Key", "myApiKey");
conn.setRequestMethod("GET");
JSONObject names =new JSONObject();
try {
names.put("names[]=", "EndUser/WebTransaction/WebTransaction/JSP/index.jsp");
} catch (JSONException e) {
e.printStackTrace();
}
OutputStreamWriter wr= new OutputStreamWriter(conn.getOutputStream());
wr.write(names.toString());
Edit
I've modified the code a bit, it's working now thanks.
String names = "names[]=EndUser/WebTransaction/WebTransaction/JSP/index.jsp";
String url = "https://api.newrelic.com/v2/applications/myAppId/metrics/data.json";
String line;
try (PrintWriter writer = response.getWriter()) {
HttpURLConnection conn = (HttpURLConnection) new URL(url).openConnection();
conn.setRequestProperty("Accept", "application/json");
conn.setRequestProperty("X-Api-Key", "myApiKey");
conn.setRequestMethod("GET");
conn.setDoOutput(true);
conn.setDoInput(true);
OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
wr.write(names);
wr.flush();
BufferedReader reader = new BufferedReader(new
InputStreamReader(conn.getInputStream()));
while ((line = reader.readLine()) != null) {
System.out.println(line);
writer.println(HTML_START + "<h2> NewRelic JSON Response:</h2><h3>" + line + "</h3>" + HTML_END);
}
wr.close();
reader.close();
}catch(MalformedURLException e){
e.printStackTrace();
}
curl -d sends whatever you specify without formatting it in any way. Just send the string names[]=EndUser/... in the OutputStream, without wrapping it in a JSONObject. Don't forget to call wr.flush() after writing the string. And of course, after that, you need to get the InputStream and start reading from it (I only mention this because it's not in your snippet).

convert curl request into URLConnection

I have this cURL request:
curl -H 'Accept: application/vnd.twitchtv.v3+json' -H 'Authorization: OAuth <access_token>' \
-X PUT https://api.twitch.tv/kraken/users/<bot_name>/follows/channels/<channel_name>
I need to turn it into a Java URLConnection request. This is what I have so far:
String url = "https://api.twitch.tv/kraken/?oauth_token=" + bot.botOAuth.substring("oauth:".length());
URL obj = new URL(url);
HttpURLConnection conn = (HttpURLConnection) obj.openConnection();
conn.setRequestProperty("Content-Type", "application/json");
conn.setDoOutput(true);
conn.setRequestMethod("PUT");
OutputStreamWriter out = new OutputStreamWriter(conn.getOutputStream());
out.write("https://api.twitch.tv/kraken/users/" + bot.botName + "/follows/channels/" + gamrCorpsTextField.getText());
out.close();
new InputStreamReader(conn.getInputStream());
Any help will be appreciated!
The URL you are preparing to open in this code:
String url = "https://api.twitch.tv/kraken/?oauth_token=" + bot.botOAuth.substring("oauth:".length());
does not match your curl request URL:
https://api.twitch.tv/kraken/users/<bot_name>/follows/channels/<channel_name>
You appear to want something more like this:
URL requestUrl = new URL("https://api.twitch.tv/kraken/users/" + bot.botName
+ "/follows/channels/" + gamrCorpsTextField.getText());
HttpURLConnection connection = (HttpUrlConnection) requestUrl.openConnection();
connection.setRequestMethod("PUT");
connection.setRequestProperty("Accept", "application/vnd.twitchtv.v3+json");
connection.setRequestProperty("Authorization", "OAuth <access_token>");
connection.setDoInput(true);
connection.setDoOutput(false);
That sets up a "URLConnection request" equivalent to the one the curl command will issue, as requested. From there you get the response code, read response headers and body, and so forth via the connection object.

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