I have a problem. I have the following PHP page:
<?php
header('Content-Type: application/json');
echo $_POST["agentid"];
?>
And my Java code is the following:
public String callWebpage(String strUrl, HashMap<String, String> data) throws InterruptedException, IOException {
var objectMapper = new ObjectMapper();
String requestBody = objectMapper
.writeValueAsString(data);
URL url = new URL(strUrl);
HttpURLConnection con = (HttpURLConnection) url.openConnection();
con.setRequestMethod("POST");
con.setDoOutput(true);
con.getOutputStream().write(requestBody.getBytes("UTF-8"));
String result = convertInputStreamToString(con.getInputStream());
return result;
}
To call the function, I use the following lines:
var values = new HashMap<String, String>() {{
put("agentid", String.valueOf(agentId));
}};
String jsonResponse = webAPI.callWebpage("https://www.test.org/test.php", values);
This does print the result of the page, but it gives me:
2021-02-28 00:40:16.735 Custom error: [8] Undefined index: agentid<br>2021-02-28 00:40:16.735 Error on line 5
The page is HTTPS and I do get a response, but why is my agentid variable not getting received by the page and how can I fix this?
Please, consider including the following code in your Java connection setup to indicate that you are POSTing JSON content before writing to the connection output stream:
con.setRequestProperty("Content-Type", "application/json");
In any way, I think the problem is in your PHP code: you are trying to process raw HTTP parameter information while you are receiving a JSON fragment as the request body instead.
In order to access the raw JSON information received in the HTTP request you need something like the following:
// Takes raw data from the request.
$json = file_get_contents('php://input');
// Converts it into a PHP object
$data = json_decode($json);
// Process information
echo $data->agentid;
Please, see this link and 'php://input` for more information.
Be aware that with the above-mentioned code you are returning a String to the Java client, although you indicated header('Content-Type: application/json'), maybe it could be the cause of any problem.
I think that you are using HTTP Request. Then, use exec instead of POST if that coder is persist the same as in the answer below.
Related
I am trying to perform a get request using Groovy using the below code:
String url = "url of endpoint"
def responseXml = new XmlSlurper().parse(url)
If the endpoint returns status as 200 then everything works good but there is one case where we have to validate the error response like below and status returned is 400:
<errors>
<error>One of the following parameters is required: xyz, abc.</error>
<error>One of the following parameters is required: xyz, mno.</error>
</errors>
In this case parse method throws :
java.io.IOException: Server returned HTTP response code: 400 for URL: "actual endpoint throwing error"
at sun.net.www.protocol.http.HttpURLConnection.getInputStream0(HttpURLConnection.java:1900)
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1498)
at com.sun.org.apache.xerces.internal.impl.XMLEntityManager.setupCurrentEntity(XMLEntityManager.java:646)
at com.sun.org.apache.xerces.internal.impl.XMLVersionDetector.determineDocVersion(XMLVersionDetector.java:150)
at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:831)
at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:796)
at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(XMLParser.java:142)
at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse(AbstractSAXParser.java:1216)
at com.sun.org.apache.xerces.internal.jaxp.SAXParserImpl$JAXPSAXParser.parse(SAXParserImpl.java:644)
at groovy.util.XmlSlurper.parse(XmlSlurper.java:205)
at groovy.util.XmlSlurper.parse(XmlSlurper.java:271)
Can anyone pls suggest how to handle if server give error message by throwing 400 status code?
In the question since we are getting 400 status code for GET request. So in built XmlSlurper().parse(URI) method does not work as it throw io.Exception.
Groovy also support HTTP methods for api request and response and the below worked for me:
def getReponseBody(endpoint) {
URL url = new URL(endpoint)
HttpURLConnection get = (HttpURLConnection)url.openConnection()
get.setRequestMethod("GET")
def getRC = get.getResponseCode()
BufferedReader br = new BufferedReader(new InputStreamReader(get.getErrorStream()))
StringBuffer xmlObject = new StringBuffer()
def eachLine
while((eachLine = br.readLine()) !=null){
xmlObject.append(eachLine)
}
get.disconnect()
return new XmlSlurper().parseText(xmlObject.toString())
}
Getting the response text from the HttpURLConnection class rather than implicitly through XmlSlurper allows you much more flexibility in handling unsuccessful responses. Try something like this:
def connection = new URL('https://your.url/goes.here').openConnection()
def content = { ->
try {
connection.content as String
} catch (e) {
connection.responseMessage
}
}()
if (content) {
def responseXml = new XmlSlurper().parseText(content)
doStuffWithResponseXml(responseXml)
}
Even better would be to use an actual full-featured HTTP client, like the Spring Framework's HttpClient or RestTemplate classes.
You should check the return code and than obtain the error stream from http request instance in case of an error. The problem itself has nothing to do with JsonSlurper, as no instance of "input stream" is returned from http request instance if service returns not successfull return codes (400, 401, 500 etc.) POST example can be seen below:
http= new URL("yourUrl").openConnection() as HttpURLConnection
http.setRequestMethod('POST')
http.setDoOutput(true)
http.setRequestProperty("Content-Type", 'application/json')
http.setRequestProperty("Accept", 'application/json')
http.setRequestProperty("Authorization", "Bearer $yourTokenVariable")
http.outputStream.write(data.getBytes("UTF-8"))
http.connect()
if(http.getResponseCode() != 200 && http.getResponseCode() != 201){
throw new InvalidInputException("There was an error: " + http.getErrorStream().getText("UTF-8"))
} else {
//You can take input stream here
}
I request POST by this code
URL url = new URL("adress/discordnotifi");
HttpURLConnection httpURLConnection = (HttpURLConnection) url.openConnection();
httpURLConnection.setReadTimeout(5000);
httpURLConnection.setConnectTimeout(5000);
httpURLConnection.setRequestMethod("POST");
httpURLConnection.connect();
JSONObject object = new JSONObject();
myPWord = ((EditText) (findViewById(R.id.edit_Id))).getText().toString();
object.put("token", tokens);
object.put("discordid", mydiscord);
OutputStream outputStream = httpURLConnection.getOutputStream();
outputStream.write(object.toString().getBytes("UTF-8"));
Log.d("debug",object.toString());
outputStream.flush();
outputStream.close();
and my tokens and mydiscord is the information that I want to send as JSON format like
{"token":"tokens","discordid":"mydiscord"}
and from python flask
#app.route('/discordnotifi', methods=['POST'])
def post():
content = request.json
print(content)
Discordid = int(content["discordid"])
token = content['token']
"""Discordid = request.form.get("discordid")
token = request.form.get("token")"""
print(Discordid, token)
return ("Thx")
at print(content) I get None I really don't know whts wrong here. I was planning to send Json with information but I getting None from Json.
The very least you're missing is enabling output on the connection. Add:
httpURLConnection.setDoOutput(true);
The server may also require that you set some headers, for example content-type to tell it that you are sending JSON.
httpURLConnection.setRequestProperty("Content-Type", "application/json; utf-8");
If you can use Java 11, consider using the new HttpClient class instead of HttpUrlConnection. It simplifies creating correct requests.
NOTICE UPDATE!!
The problem got solved and i added my own answer in the thread
In short, I have attempted to add the parameter "scan_id" value but since it is a POST i can't add the value directly in the url path.
using the code i already have, how would i go about modifying or adding so that the url is correct, that is, so that it accepts my POST?.
somehow i have been unable to find any examples that have helped me in figuring out how i would go about doing this..
I know how to do a POST with a payload, a GET with params. but a post with Params is very confusing to me.
Appreciate any help. (i'd like to continue using HttpUrlConnection unless an other example is provided that also tells me how to send the request and not only configuring the path.
I've tried adding it to the payload.
I've tried UriBuilder but found it confusing and in contrast with the rest of my code, so wanted to ask for help with HttpUrlConnection.
URL url = new URL("http://localhost/scans/{scan_id}/launch");
HttpURLConnection con = (HttpURLConnection) url.openConnection();
con.setRequestMethod("POST");
con.setRequestProperty("tmp_value_dont_mind_this", "432432");
con.setRequestProperty("X-Cookie", "token=" + "43432");
con.setRequestProperty("X-ApiKeys", "accessKey="+"43234;" + " secretKey="+"43234;");
con.setDoInput(true);
con.setDoOutput(true); //NOT NEEDED FOR GETS
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
//First example of writing (works when writing a payload)
OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream(), "UTF-8");
writer.write(payload);
writer.close();
//second attemp at writing, doens't work (wanted to replace {scan_id} in the url)
DataOutputStream writer = new DataOutputStream(con.getOutputStream());
writer.writeChars("scan_id=42324"); //tried writing directly
//writer.write(payload);
writer.close();
Exception:
Exception in thread "main" java.io.IOException: Server returned HTTP response code: 400 for URL: http://localhost/scans/launch
I'd like one of the three response codes because then i know the Url is correct:
200 Returned if the scan was successfully launched.
403 Returned if the scan is disabled.
404 Returned if the scan does not exist.
I've tried several urls
localhost/scans/launch,
localhost/scans//launch,
localhost/scans/?/launch,
localhost/scans/{scan_id}/launch,
So with the help of a friend and everyone here i solved my problem.
The below code is all the code in an entire class explained bit by bit. at the bottom you have the full class with all its syntax etc, that takes parameters and returns a string.
in a HTTP request there are certain sections.
Such sections include in my case, Request headers, parameters in the Url and a Payload.
depending on the API certain variables required by the API need to go into their respective category.
My ORIGINAL URL looked like this: "http://host:port/scans/{scan_id}/export?{history_id}"
I CHANGED to: "https://host:port/scans/" + scan_Id + "/export?history_id=" + ID;
and the API i am calling required an argument in the payload called "format" with a value.
String payload = "{\"format\" : \"csv\"}";
So with my new URL i opened a connection and set the request headers i needed to set.
HttpsURLConnection con = (HttpsURLConnection) url.openConnection();
The setDoOutput should be commented out when making a GET request.
con.setDoInput(true);
con.setDoOutput(true);
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
con.setRequestProperty("X-Cookie", "token=" + token);
con.setRequestProperty("X-ApiKeys", "accessKey="+"23243;" +"secretKey="+"45543;");
Here i write to the payload.
//WRITING THE PAYLOAD to the http call
OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream(), "UTF-8");
writer.write(payload);
writer.close();
After i've written the payload i read whatever response i get back (this depends on the call, when i do a file download (GET Request) i don't have a response to read as i've already read the response through another piece of code).
I hope this helps anyone who might encounter this thread.
public String requestScan(int scan_Id, String token, String ID) throws MalformedInputException, ProtocolException, IOException {
try {
String endpoint = "https://host:port/scans/" + scan_Id + "/export?history_id=" ID;
URL url = new URL(endpoint);
String payload= "{\"format\" : \"csv\"}";
HttpsURLConnection con = (HttpsURLConnection) url.openConnection();
con.setDoInput(true);
con.setDoOutput(true);
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
con.setRequestProperty("X-Cookie", "token=" + token);
con.setRequestProperty("X-ApiKeys", "accessKey="+"324324;" +
"secretKey="+"43242;");
//WRITING THE PAYLOAD to the http call
OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream(), "UTF-8");
writer.write(payload);
writer.close();
//READING RESPONSE
BufferedReader br = new BufferedReader(new InputStreamReader(con.getInputStream()));
StringBuffer jsonString = new StringBuffer();
String line;
while ((line = br.readLine()) != null) {
jsonString.append(line);
}
br.close();
con.disconnect();
return jsonString.toString();
} catch (Exception e) {
throw new RuntimeException(e.getMessage());
}
}
As discussed here the solution would be to change the content type to application/x-www-form-urlencoded, but since you are already using application/json; charset=UTF-8 (which I am assuming is a requirement of your project) you have no choise to redesign the whole thing. I suggest you one of the following:
Add another GET service;
Add another POST service with content type application/x-www-form-urlencoded;
Replace this service with one of the above.
Do not specify the content type at all so the client will accept anything. (Don't know if possible in java)
If there are another solutions I'm not aware of, I don't know how much they would be compliant to HTTP protocol.
(More info)
Hope I helped!
Why you are not using like this. Since you need to do a POST with HttpURLConnection, you need to write the parameters to the connection after you have opened the connection.
String urlParameters = "scan_id=42324";
byte[] postData = urlParameters.getBytes(StandardCharsets.UTF_8);
DataOutputStream dataOutputStream = new DataOutputStream(conn.getOutputStream());
dataOutputStream.write(postData);
Or if you have launch in the end, just change the above code to the following,
String urlParameters = "42324/launch";
byte[] postData = urlParameters.getBytes(StandardCharsets.UTF_8);
DataOutputStream dataOutputStream = new DataOutputStream(conn.getOutputStream());
dataOutputStream.write(postData);
URL url = new URL("http://localhost/scans/{scan_id}/launch");
That line looks odd to me; it seems you are trying to use a URL where you are intending the behavior of a URI Template.
The exact syntax will depend on which template implementation you choose; an implementation using the Spring libraries might look like:
import org.springframework.web.util.UriTemplate;
import java.net.url;
// Warning - UNTESTED code ahead
UriTemplate template = new UriTemplate("http://localhost/scans/{scan_id}/launch");
Map<String,String> uriVariables = Collections.singletonMap("scan_id", "42324");
URI uri = template.expand(uriVariables);
URL url = uri.toURL();
I'm trying to do a "POST" method in Java. I create my output with the OrientDB method like this:
"http://xxxxxxxxxxx:2480/command/mydb/sql/CREATE VERTEX V SET name = ' datoAletarorio'"
I need to use the write and flush methods to send the command.
My DB is empty with this method.
Where is my error? Here is my code:
//...
PrintWriter out = null;
//...
conexion = (HttpURLConnection) url.openConnection();
conexion.setDoOutput(true);
conexion.setRequestMethod("POST");
out = new PrintWriter(conexion.getOutputStream());
conexion.connect();
//...
String cumuloDatos1 = "http://xxxxxxxxxxx:2480/command/mydb/sql/CREATE VERTEX V SET name = ' datoAletarorio'"
out.write(cumuloDatos1);
out.flush();
//..
conexion.disconnect();
Thank in advance.
The docs says:
The command-text can appear in either the URL or the content of the
POST transmission. Where the command-text is included in the URL, it
must be encoded as per normal URL encoding.
So you probably have to encode the URL before sending the request:
String cumuloDatos1 =
"http://xxxxxxxxxxx:2480/command/mydb/sql/" +
"CREATE%20VERTEX%20V%20SET%20name%20%3D%20%27%20datoAletarorio%27"
Anyway, you should see messages in the logs for a 400 or similiar in the server, if the request isn't valid.
I am trying to send a post request to a url using HttpURLConnection (for using cUrl in java).
The content of the request is xml and at the end point, the application processes the xml and stores a record to the database and then sends back a response in form of xml string. The app is hosted on apache-tomcat locally.
When I execute this code from the terminal, a row gets added to the db as expected. But an exception is thrown as follows while getting the InputStream from the connection
java.io.FileNotFoundException: http://localhost:8080/myapp/service/generate
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1401)
at org.kodeplay.helloworld.HttpCurl.main(HttpCurl.java:30)
Here is the code
public class HttpCurl {
public static void main(String [] args) {
HttpURLConnection con;
try {
con = (HttpURLConnection) new URL("http://localhost:8080/myapp/service/generate").openConnection();
con.setRequestMethod("POST");
con.setDoOutput(true);
con.setDoInput(true);
File xmlFile = new File("test.xml");
String xml = ReadWriteTextFile.getContents(xmlFile);
con.getOutputStream().write(xml.getBytes("UTF-8"));
InputStream response = con.getInputStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(response));
for (String line ; (line = reader.readLine()) != null;) {
System.out.println(line);
}
reader.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
}
Its confusing because the exception is traced to the line InputStream response = con.getInputStream(); and there doesn't seem to be any file involved for a FileNotFoundException.
When I try to open a connection to an xml file directly, it doesn't throw this exception.
The service app uses spring framework and Jaxb2Marshaller to create the response xml.
The class ReadWriteTextFile is taken from here
Thanks.
Edit:
Well it saves the data in the DB and sends back a 404 response status code at the same time.
I also tried doing a curl using php and print out the CURLINFO_HTTP_CODE which turns out to be 200.
Any ideas on how do I go about debugging this ? Both service and client are on the local server.
Resolved:
I could solve the problem after referring to an answer on SO itself.
It seems HttpURLConnection always returns 404 response when connecting to a url with a non standard port.
Adding these lines solved it
con.setRequestProperty("User-Agent","Mozilla/5.0 ( compatible ) ");
con.setRequestProperty("Accept","*/*");
I don't know about your Spring/JAXB combination, but the average REST webservice won't return a response body on POST/PUT, just a response status. You'd like to determine it instead of the body.
Replace
InputStream response = con.getInputStream();
by
int status = con.getResponseCode();
All available status codes and their meaning are available in the HTTP spec, as linked before. The webservice itself should also come along with some documentation which overviews all status codes supported by the webservice and their special meaning, if any.
If the status starts with 4nn or 5nn, you'd like to use getErrorStream() instead to read the response body which may contain the error details.
InputStream error = con.getErrorStream();
FileNotFound is just an unfortunate exception used to indicate that the web server returned a 404.
To anyone with this problem in the future, the reason is because the status code was a 404 (or in my case was a 500). It appears the InpuStream function will throw an error when the status code is not 200.
In my case I control my own server and was returning a 500 status code to indicate an error occurred. Despite me also sending a body with a string message detailing the error, the inputstream threw an error regardless of the body being completely readable.
If you control your server I suppose this can be handled by sending yourself a 200 status code and then handling whatever the string error response was.
For anybody else stumbling over this, the same happened to me while trying to send a SOAP request header to a SOAP service. The issue was a wrong order in the code, I requested the input stream first before sending the XML body. In the code snipped below, the line InputStream in = conn.getInputStream(); came immediately after ByteArrayOutputStream out = new ByteArrayOutputStream(); which is the incorrect order of things.
ByteArrayOutputStream out = new ByteArrayOutputStream();
// send SOAP request as part of HTTP body
byte[] data = request.getHttpBody().getBytes("UTF-8");
conn.getOutputStream().write(data);
if (conn.getResponseCode() != HttpURLConnection.HTTP_OK) {
Log.d(TAG, "http response code is " + conn.getResponseCode());
return null;
}
InputStream in = conn.getInputStream();
FileNotFound in this case was an unfortunate way to encode HTTP response code 400.
FileNotFound in this case means you got a 404 from your server - could it be that the server does not like "POST" requests?
FileNotFound in this case means you got a 404 from your server
You Have to Set the Request Content-Type Header Parameter
Set “content-type” request header to “application/json” to send the request content in JSON form.
This parameter has to be set to send the request body in JSON format.
Failing to do so, the server returns HTTP status code “400-bad request”.
con.setRequestProperty("Content-Type", "application/json; utf-8");
Full Script ->
public class SendDeviceDetails extends AsyncTask<String, Void, String> {
#Override
protected String doInBackground(String... params) {
String data = "";
String url = "";
HttpURLConnection con = null;
try {
// From the above URL object,
// we can invoke the openConnection method to get the HttpURLConnection object.
// We can't instantiate HttpURLConnection directly, as it's an abstract class:
con = (HttpURLConnection)new URL(url).openConnection();
//To send a POST request, we'll have to set the request method property to POST:
con.setRequestMethod("POST");
// Set the Request Content-Type Header Parameter
// Set “content-type” request header to “application/json” to send the request content in JSON form.
// This parameter has to be set to send the request body in JSON format.
//Failing to do so, the server returns HTTP status code “400-bad request”.
con.setRequestProperty("Content-Type", "application/json; utf-8");
//Set Response Format Type
//Set the “Accept” request header to “application/json” to read the response in the desired format:
con.setRequestProperty("Accept", "application/json");
//To send request content, let's enable the URLConnection object's doOutput property to true.
//Otherwise, we'll not be able to write content to the connection output stream:
con.setDoOutput(true);
//JSON String need to be constructed for the specific resource.
//We may construct complex JSON using any third-party JSON libraries such as jackson or org.json
String jsonInputString = params[0];
try(OutputStream os = con.getOutputStream()){
byte[] input = jsonInputString.getBytes("utf-8");
os.write(input, 0, input.length);
}
int code = con.getResponseCode();
System.out.println(code);
//Get the input stream to read the response content.
// Remember to use try-with-resources to close the response stream automatically.
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());
}
System.out.println(response.toString());
}
} catch (Exception e) {
e.printStackTrace();
} finally {
if (con != null) {
con.disconnect();
}
}
return data;
}
#Override
protected void onPostExecute(String result) {
super.onPostExecute(result);
Log.e("TAG", result); // this is expecting a response code to be sent from your server upon receiving the POST data
}
and call it
new SendDeviceDetails().execute("");
you can find more details in this tutorial
https://www.baeldung.com/httpurlconnection-post
The solution:
just change localhost for the IP of your PC
if you want to know this: Windows+r > cmd > ipconfig
example: http://192.168.0.107/directory/service/program.php?action=sendSomething
just replace 192.168.0.107 for your own IP (don't try 127.0.0.1 because it's same as localhost)
Please change
con = (HttpURLConnection) new URL("http://localhost:8080/myapp/service/generate").openConnection();
To
con = (HttpURLConnection) new URL("http://YOUR_IP:8080/myapp/service/generate").openConnection();