I'm doing an app that logs users into websites that have an authentication popup using webView.
The problem comes when logging in and then going back to log in with different credentials, causing to skip HttpAuthRequest and therefore being logged in with the first credentials
What should I do?
The below lines of code is enough for clearing the cookies of Web View
CookieSyncManager.createInstance(context);
CookieManager cookieManager = CookieManager.getInstance();
cookieManager.removeAllCookie();
CookieSyncManager.getInstance().sync();
Happy Coding! Thanks.
I did the trick logging in like this:
// Prefix http:// or https://
String prefix = url.substring(0, url.indexOf("/") + 2);
// Prefix + username +: + password + # + url without prefix
String finalUrl = prefix + username + ":" + password + "#" + url.substring(url.indexOf("/") + 2);
webView.loadUrl( finalUrl );
Instead of using onReceivedHttpAuthRequest
Related
I'm trying to filter in my presignup lambda for a few params (in the same call) with Cognito ListUserRequest but it doesn't work.
Somebody knows a workaround without making several calls?
Thanks a lot!
ListUsersRequest listUsersRequest = new ListUsersRequest();
listUsersRequest.withUserPoolId(USER_POOL_ID);
listUsersRequest.withFilter("username = \"" + USER_NAME + "\" and email = \"" + EMAIL + "\""");
I'm trying to use the Authentication::login() API call in the DocuSign Java SDK and am receiving an error. Here's some code:
#Component
public class TestClass {
private ApiClient apiClient;
public void authenticate() {
this.apiClient = new ApiClient("account-d.docusign.com", "docusignAccessCode",
"mySecretIntegratorKey", "myClientSecret");
final AuthenticationApi authenticationApi = new AuthenticationApi(this.apiClient);
try {
// ERROR ON THE LINE BELOW
final LoginInformation loginInformation = authenticationApi.login();
} catch (final ApiException e) {
// do something appropriate
}
}
}
The mySecretIntegratorKey and myClientSecret values are not the real values I'm sending in obviously, but the other ones are.
Here is the error I am receiving when making the login() call:
Caused by: org.apache.oltu.oauth2.common.exception.OAuthSystemException: Missing grant_type/code
at com.docusign.esign.client.auth.OAuth$OAuthJerseyClient.execute(OAuth.java:184)
at org.apache.oltu.oauth2.client.OAuthClient.accessToken(OAuthClient.java:65)
at org.apache.oltu.oauth2.client.OAuthClient.accessToken(OAuthClient.java:55)
at org.apache.oltu.oauth2.client.OAuthClient.accessToken(OAuthClient.java:71)
at com.docusign.esign.client.auth.OAuth.updateAccessToken(OAuth.java:92)
... 123 common frames omitted
I realize that this is using the older legacy authentication, however I have a limitation that won't allow me to upgrade to the newer method of authentication until the first of the year. So for now I need to use this legacy method using SDK Version 2.2.1.
Any ideas what I'm doing wrong here? I'm sure it is something simple...
Thank you for your time.
You want to use Legacy authentication?
In that case you need to make a number of updates to your code.
Only call new ApiClient(base_url)
Set the X-DocuSign-Authentication header--
From an old Readme:
String authHeader = "{\"Username\":\"" + username +
"\",\"Password\":\"" + password +
"\",\"IntegratorKey\":\"" + integratorKey + "\"}";
apiClient.addDefaultHeader("X-DocuSign-Authentication", authHeader);
Then use the authenticationApi.login to look up the user's Account ID(s) and matching base urls.
The authenticationApi.login doe not actually log you in. (!)
Rather, that method just gives you information about the current user.
There is no login with the API since it does not use sessions. Instead, credentials are passed with every API call. The credentials can be an Access Token (preferred), or via Legacy Authentication, a name / password / integration key triplet.
When using Legacy Authentication, the client secret is not used.
More information: see the Readme section for using username/password in this old version of the repo.
Just in case someone was looking for complete legacy code that works! The below C# code snippet works. This is production ready code. I've tested it and it works. You will have to create an EnvelopeDefinition separately as this code is not included. However, the piece below will authenticate the user and will successfully send an envelope and get back the Envelope ID:
string username = "john.bunce#mail.com";
string password = "your_password";
string integratorKey = "your_integration_key";
ApiClient apiClient = new ApiClient("https://www.docusign.net/restapi");
string authHeader = "{\"Username\":\"" + username + "\", \"Password\":\"" + password + "\", \"IntegratorKey\":\"" + integratorKey + "\"}";
apiClient.Configuration.AddDefaultHeader("X-DocuSign-Authentication", authHeader);
AuthenticationApi authApi = new AuthenticationApi(apiClient.Configuration);
LoginInformation loginInfo = authApi.Login();
string accountId = loginInfo.LoginAccounts[0].AccountId;
string baseURL = loginInfo.LoginAccounts[0].BaseUrl;
string[] baseUrlArray= Regex.Split(baseURL, "/v2");
ApiClient apiClient2 = new ApiClient(baseUrlArray[0]);
string authHeader2 = "{\"Username\":\"" + username + "\", \"Password\":\"" + password + "\", \"IntegratorKey\":\"" + integratorKey + "\"}";
apiClient2.Configuration.AddDefaultHeader("X-DocuSign-Authentication", authHeader2);
EnvelopesApi envelopesApi = new EnvelopesApi(apiClient2.Configuration);
EnvelopeSummary results = envelopesApi.CreateEnvelope(accountId, envelopeDefinition);
string envelopeID = results.EnvelopeId;
I'm looking for a was to get the groupId. To be more specific, I want to get the site ID of a community or organisation when a user signs in so I can redirect the user to the right "site".
I tried looking into PortalUtil in Liferay's documentation but it doesn't offer an easy function to get that ID.
I also tried ThemeDisplay but that only works for portlets.
Here is an excerpt from a LoginPostAction in hook which serves your needs.
User user = PortalUtil.getUser(request);
List<Organization> orgList = OrganizationLocalServiceUtil.getUserOrganizations(user.getUserId());
for (Organization org : orgList) {
String orgFriendlyURL = org.getGroup().getFriendlyURL();
.
.//some custom code
.
String myPath = "/" + language + "/group" + orgFriendlyURL + "/home";
LastPath lastPath = new LastPath(StringPool.BLANK, myPath);
HttpSession session = request.getSession();
session.setAttribute(WebKeys.LAST_PATH, lastPath);
_log.debug("Last Path for current User[" + user.getScreenName() + "] is : " + lastPath);
break;
}
I am trying to get just the domain name (http://www.example.com) out of log files that looks like this:
http://maps.google.com/maps?hl=en&tab=wl
http://l.macys.com/simi-valley-ca?cm_mmc=macys_
https://www.google.co.in/
https://www.google.ca/
I want just
http://maps.google.com/
http://l.macys.com/
https://www.google.co.in/
https://www.google.ca/
Any ideas?
How about
URL url = new URL("http://maps.google.com/maps?hl=en&tab=wl");
System.out.println(url.getProtocol()+"://"+url.getHost());
Output
http://maps.google.com
If you don't want to handle it yourself, then a full proof way is following:
URL url = new URL("http://l.macys.com/simi-valley-ca?cm_mmc=macys_");
System.out.println(url.getProtocol() + "://" + url.getHost() + ((url.getPort()==-1)?"" : ":" + url.getPort()) + "/" );
You can skip url.getPort if you are sure that there will never be a port type url!!
Cheers
From the following URL I need to get (http://localhost:9090/dts) alone.
That is I need to remove (documents/savedoc) (OR)
need to get only - (http://localhost:9090/dts)
http://localhost:9090/dts/documents/savedoc
Is there any method available in request to get the above?
I tried the following and got the result. But still trying.
System.out.println("URL****************"+request.getRequestURL().toString());
System.out.println("URI****************"+request.getRequestURI().toString());
System.out.println("ContextPath****************"+request.getContextPath().toString());
URL****************http://localhost:9090/dts/documents/savedoc
URI****************/dts/documents/savedoc
ContextPath****************/dts
Can anyone please help me in fixing this?
You say you want to get exactly:
http://localhost:9090/dts
In your case, the above string consist of:
scheme: http
server host name: localhost
server port: 9090
context path: dts
(More info about the elements of a request path can be found in the official Oracle Java EE Tutorial: Getting Information from Requests)
##First variant:###
String scheme = request.getScheme();
String serverName = request.getServerName();
int serverPort = request.getServerPort();
String contextPath = request.getContextPath(); // includes leading forward slash
String resultPath = scheme + "://" + serverName + ":" + serverPort + contextPath;
System.out.println("Result path: " + resultPath);
##Second variant:##
String scheme = request.getScheme();
String host = request.getHeader("Host"); // includes server name and server port
String contextPath = request.getContextPath(); // includes leading forward slash
String resultPath = scheme + "://" + host + contextPath;
System.out.println("Result path: " + resultPath);
Both variants will give you what you wanted: http://localhost:9090/dts
Of course there are others variants, like others already wrote ...
It's just in your original question you asked about how to get http://localhost:9090/dts, i.e. you want your path to include scheme.
In case you still doesn't need a scheme, the quick way is:
String resultPath = request.getHeader("Host") + request.getContextPath();
And you'll get (in your case): localhost:9090/dts
AFAIK for this there is no API provided method, need to customization.
String serverName = request.getServerName();
int portNumber = request.getServerPort();
String contextPath = request.getContextPath();
// try this
System.out.println(serverName + ":" +portNumber + contextPath );
Just remove URI from URL and then append context path to it. No need to fiddle with loose schemes and ports which is only more tedious when you're dealing with default port 80 which don't need to appear in URL at all.
StringBuffer url = request.getRequestURL();
String uri = request.getRequestURI();
String ctx = request.getContextPath();
String base = url.substring(0, url.length() - uri.length() + ctx.length());
// ...
See also:
Browser can't access/find relative resources like CSS, images and links when calling a Servlet which forwards to a JSP (for the JSP/JSTL variant of composing the base URL)
In my understanding, you need the domain part and Context path only. Based on this understanding, You can use this method to get the required string.
String domain = request.getRequestURL().toString();
String cpath = request.getContextPath().toString();
String tString = domain.subString(0, domain.indexOf(cpath));
tString = tString + cpath;
For those who want to get, in their endpoint, the URL of the front page which targeted the endpoint. You can use this:
request.getHeader("referer")
Usually I have a method like this:
public String getAbsoluteContextPath() throws MalformedURLException {
ExternalContext context = FacesContext.getCurrentInstance().getExternalContext();
HttpServletRequest request = (HttpServletRequest) context.getRequest();
URL url = new URL(request.getRequestURL().toString());
return url.getProtocol() + "://" + url.getAuthority() + context.getRequestContextPath();
}
This method will return what you want, with the port number only if it exists in the current request. In your case it will return: http://localhost:9090/dts