General Answer The general answer to your question is that it depends. And you get to decide by specifying what your “Content-Type” is in the HTTP headers. A value of “application/x-www-form-urlencoded” means that your POST body will need to be URL encoded just like a GET parameter string. A value of “multipart/form-data” means that you’ll … Read more
From: https://requests.readthedocs.io/en/latest/user/quickstart/#post-a-multipart-encoded-file Requests makes it very simple to upload Multipart-encoded files: with open(‘report.xls’, ‘rb’) as f: r = requests.post(‘http://httpbin.org/post’, files={‘report.xls’: f}) That’s it. I’m not joking – this is one line of code. The file was sent. Let’s check: >>> r.text { “origin”: “179.13.100.4”, “files”: { “report.xls”: “<censored…binary…data>” }, “form”: {}, “url”: “http://httpbin.org/post”, “args”: {}, … Read more
Update in 2018: I just got some points for this old answer, and just wanted to add that the best solution would be to make the operation idempotent so that duplicate submissions are harmless. Eg, if the form creates an order, put a unique ID in the form. The first time the server sees an … Read more
using (WebClient client = new WebClient()) { byte[] response = client.UploadValues(“http://dork.com/service”, new NameValueCollection() { { “home”, “Cosby” }, { “favorite+flavor”, “flies” } }); string result = System.Text.Encoding.UTF8.GetString(response); } You will need these includes: using System; using System.Collections.Specialized; using System.Net; If you’re insistent on using a static method/class: public static class Http { public static byte[] … Read more
<form target=”_blank” [….] will submit the form in a new tab… I am not sure if is this what you are looking for, please explain better…
System.Net.HttpWebRequest adds the header ‘HTTP header “Expect: 100-Continue”‘ to every request unless you explicitly ask it not to by setting this static property to false: System.Net.ServicePointManager.Expect100Continue = false; Some servers choke on that header and send back the 417 error you’re seeing. Give that a shot.
The key is that you want to: set the httpMethod to POST; optionally, set the Content-Type header, to specify how the request body was encoded, in case server might accept different types of requests; optionally, set the Accept header, to request how the response body should be encoded, in case the server might generate different … Read more
You have to put together the x-www-form-urlencoded payload yourself, like this: var details = { ‘userName’: ‘test@gmail.com’, ‘password’: ‘Password!’, ‘grant_type’: ‘password’ }; var formBody = []; for (var property in details) { var encodedKey = encodeURIComponent(property); var encodedValue = encodeURIComponent(details[property]); formBody.push(encodedKey + “=” + encodedValue); } formBody = formBody.join(“&”); fetch(‘https://example.com/login’, { method: ‘POST’, headers: { … Read more
disabled input will not submit data. Use the readonly attribute: <input type=”text” readonly /> Source here
You can get output stream for the connection and write the parameter query string to it. URL url = new URL(“http://yoururl.com”); HttpsURLConnection conn = (HttpsURLConnection) url.openConnection(); conn.setReadTimeout(10000); conn.setConnectTimeout(15000); conn.setRequestMethod(“POST”); conn.setDoInput(true); conn.setDoOutput(true); List<NameValuePair> params = new ArrayList<NameValuePair>(); params.add(new BasicNameValuePair(“firstParam”, paramValue1)); params.add(new BasicNameValuePair(“secondParam”, paramValue2)); params.add(new BasicNameValuePair(“thirdParam”, paramValue3)); OutputStream os = conn.getOutputStream(); BufferedWriter writer = new BufferedWriter( new … Read more