简体   繁体   中英

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.

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM