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  • FTP to SFTP in shell scripting

    - by Kimi
    This script is to connect to different servers and copy a file from a loaction defined. It is mandatory to use sftp and not ftp. #!/usr/bin/ksh -xvf Detail="jyotibo|snv4915|/tlmusr1/tlm/rt/jyotibo/JyotiBo/ jyotibo|snv4915|/tlmusr1/tlm/rt/jyotibo/JyotiBo/" password=Unix11! c_filename=import.log localpath1=`pwd` for i in $Detail do echo $i UserName=`echo $i | cut -d'|' -f1` echo $UserName remotehost=`echo $i | cut -d'|' -f2` echo $remotehost remote_path=`echo $i | cut -d'|' -f3` echo $remote_path { echo "open $remotehost user $UserName $password lcd $localpath1 cd $remote_path bi prompt mget $c_filename prompt " } |ftp -i -n -v 2>&1 done I want to do the similar thing using sftp instead of ftp.

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  • Verify windows log-in via smart card

    - by Ronen Rabinovitz
    Hi I need to verify in my WPF application if the user log in to his computer via password or via smart-card. Both login options are available in my company clients but my application need to open only in the smart-card login. All the clients are windows 7 OS. I look at some sites: http://technet.microsoft.com/en-us/library/ff404285(v=ws.10).aspx http://www.codeproject.com/Articles/240655/Using-a-Smart-Card-Certificate-with-NET-Security-i and I'm thinking I need to get the enhanced key usage (EKU) attribute field. If the EKU is empty = then the user was loged via password and not via smartcard. I only need this simple check, I do not care for creating/validations on certificates atc.

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  • WNetAddConnection2 from a Windows Service

    - by Flavio
    I'm trying to connect to a remote password protected shared folder from a Windows service, which runs as LocalSystem account. It seems that the LocalSystem account is unable to directly access password-protected network shares using WNetAddConnection2() or similar calls. Can anyone confirm this? I've read that impersonating an administrator user might be the way to go. I've tried using LogonUser() and ImpersonateLoggedOnUser() before WNetAddConnection2(), it appears that the mount of the network path succeeds, but then actual accesses (e.g. enumerating of files in remote folder) fail. Any ideas? Thanks.

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  • can't login to phpmyadmin

    - by user574383
    Hi, i am new at linux but i need phpmyadmin on my centos server. I did this: cd /var/www/html/ (document root of apache) wget http://sourceforge.net/projects/phpmyadmin/path/to/latest/version tar xvfz phpMyAdmin-3.3.9-all-languages.tar.gz mv phpMyAdmin-3.3.9-all-languages phpmyadmin rm phpMyAdmin-3.3.9-all-languages.tar.gz cd phpmyadmin/ cp config.sample.inc.php config.inc.php Ok so then i just got to a webbrowser and go to www.$ip/phpmyadmin and i am presented with a login screen asking for username and password. How can i get these credentials to log in? I'd like to log in as root i guess. But i don't know how to setup a root account and create a password for root using the cli and mysql. Please help? Thanks.

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  • Google Site Data fetching

    - by inTagger
    Hail! I want to fetch image from NOT PUBLIC Google Site's page. I'm using WebClient for this purposes. var uri = new Uri("http://sites.google.com/a/MYDOMAIN.COM/SITENAME/" + "_/rsrc/1234567890/MYIMAGE.jpg"); string fileName = "d:\\!temp\\MYIMAGE.jpg"; if (File.Exists(fileName)) File.Delete(fileName); using (var webClient = new WebClient()) { var networkCredential = new NetworkCredential("USERNAME", "PASSWORD"); var credentialCache = new CredentialCache { {new Uri("sites.google.com"), "Basic", networkCredential}, {new Uri("www.google.com"), "Basic", networkCredential} }; webClient.Credentials = credentialCache; webClient.DownloadFile(uri, fileName); } It doesn't download image, but html file with login form is downloaded. If i open this link in browser it shows me login form then i enter username and password and then i can see the image. How i must use my credentials to download file with WebClient or HttpWebRequest?

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  • Ios Parse Login

    - by user3806600
    I am programming an IOS app that will use parse to login a user. Using storyboard I have the login button connected to another view controller with the push segue. Whether the username and password are correct the button selection always goes to the new view controller. I might be doing this all wrong. Any help is appreciated. Here is my code: - (IBAction)signIn:(id)sender { [PFUser logInWithUsernameInBackground:self.emailField.text password:self.passwordField.text block:^(PFUser *user, NSError *error) { if (!error) { [self performSegueWithIdentifier:@"signIn" sender:nil]; } else { // The login failed. Check error to see why. } }]; }

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  • http_post_data basic authentication?

    - by kristian nissen
    I have a remote service that I need to access, according to the documentation it's restricted using basic authentication and all requests have to be posted (HTTP POST). The documentation contains this code example - VB script: Private Function SendRequest(ByVal Url, ByVal Username, ByVal Password, ByVal Request) Dim XmlHttp Set XmlHttp = CreateObject("MSXML2.XmlHttp") XmlHttp.Open "POST", Url, False, Username, Password XmlHttp.SetRequestHeader "Content-Type", "text/xml" XmlHttp.Send Request Set SendRequest = XmlHttp End Function how can I accomplish this in PHP? When I post data to the remote server it replies: 401 Unauthorized Access which is fine because I'm not posting my user/pass just the data. Bu when I add my user/pass as it's describe here: http://dk.php.net/manual/en/http.request.options.php like this: $res = http_post_data('https://example.com', $data, array( 'Content-Type: "text/xml"', 'httpauth' => base64_encode('user:pass'), 'httpauthtype' => HTTP_AUTH_BASIC ) ); the protocol is https - I get a runtime error in return (it's a .Net service). I have tried it without the base64_encode but with the same result.

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  • Creating keystore for jarsigner programmatically

    - by skayred
    I'm trying to generate keystore with certificate to use it with JarSigner. Here is my code: System.out.println("Keystore generation..."); Security.addProvider(new BouncyCastleProvider()); String domainName = "example.org"; KeyPairGenerator keyGen = KeyPairGenerator.getInstance("RSA"); SecureRandom random = SecureRandom.getInstance("SHA1PRNG", "SUN"); keyGen.initialize(1024, random); KeyPair pair = keyGen.generateKeyPair(); X509V3CertificateGenerator v3CertGen = new X509V3CertificateGenerator(); int serial = new SecureRandom().nextInt(); v3CertGen.setSerialNumber(BigInteger.valueOf(serial < 0 ? -1 * serial : serial)); v3CertGen.setIssuerDN(new X509Principal("CN=" + domainName + ", OU=None, O=None L=None, C=None")); v3CertGen.setNotBefore(new Date(System.currentTimeMillis() - 1000L * 60 * 60 * 24 * 30)); v3CertGen.setNotAfter(new Date(System.currentTimeMillis() + (1000L * 60 * 60 * 24 * 365*10))); v3CertGen.setSubjectDN(new X509Principal("CN=" + domainName + ", OU=None, O=None L=None, C=None")); v3CertGen.setPublicKey(pair.getPublic()); v3CertGen.setSignatureAlgorithm("MD5WithRSAEncryption"); X509Certificate PKCertificate = v3CertGen.generateX509Certificate(pair.getPrivate()); FileOutputStream fos = new FileOutputStream("/Users/dmitrysavchenko/testCert.cert"); fos.write(PKCertificate.getEncoded()); fos.close(); KeyStore ks = KeyStore.getInstance(KeyStore.getDefaultType()); char[] password = "123".toCharArray(); ks.load(null, password); ks.setCertificateEntry("hive", PKCertificate); fos = new FileOutputStream("/Users/dmitrysavchenko/hive-keystore.pkcs12"); ks.store(fos, password); fos.close(); It works, but when I'm trying to sign my JAR with this keystore, I get the following error: jarsigner: Certificate chain not found for: hive. hive must reference a valid KeyStore key entry containing a private key and corresponding public key certificate chain. I've discovered that there must be a private key, but I don't know how to add it to certificate. Can you help me?

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  • Authenticating wcf service

    - by Muhammad Jamal Shaikh
    hi , i want to implement a web service which is both in java and .net.but i will first create a wcf service and later convert the code to java . for securing the webservice , i have this in my mind. i shall be implementing asp.net form authentication type of stuff . i intent to sent a token to the client on providing valid userID and password to the login method. in return i would send a token. on all other service operations i shall check the token . Now i have a wcf client and a wcf service . what should be the shortest way forward ?should i use soap auth header to send the user ID and password and how should the token be transported ( in which field ? http or soap's ?) . P.S: of course i shall enable ssl later.

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  • Inserting the record into Data Base through JPA

    - by vinay123
    In my code I am using JSF - Front end , EJB-Middile Tier and JPA connect to DB.Calling the EJB using the Webservices.Using MySQL as DAtabase. I have created the Voter table in which I need to insert the record. I ma passing the values from the JSF to EJB, it is working.I have created JPA controller class (which automatcally generates the persistence code based on the data base classes) Ex: getting the entity manager etc., em = getEntityManager(); em.getTransaction().begin(); em.persist(voter); em.getTransaction().commit(); I have created the named query also: @NamedQuery(name = "Voter.insertRecord", query = "INSERT INTO Voter v values v.voterID = :voterID,v.password = :password,v.partSSN = :partSSN,v.address = :address, v.zipCode = :zipCode,v.ssn = :ssn, v.vFirstName = :vFirstName,v.vLastName = :vLastName,v.dob = :dob"),But still not able to insert the record? Can anyone help me in inserting the record into the Data base through JPA.(Persistence object)?

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  • Can access maven repository from behind proxy, need help.

    - by Digambar Daund
    I am trying to access maven repository from behind proxy. I configured settings.xml correctly (i guess so...) true http username password 12.34.56.78 8080 But still I am error message like... if i dont configure userid/password gets correct error message which is HTTP response code 407 - saying authentication required. But If I configure correct/incorrect proxy authentication it always prints below error message.... Downloading: http://repo1.maven.org/maven2/org/apache/maven/plugins/maven-clean-plugin/2.2/maven-clean-plugin-2.2.pom [WARNING] Unable to get resource 'org.apache.maven.plugins:maven-clean-plugin:pom:2.2' from repository central (http://repo1.maven.org/maven2): Error trans ferring file: Server redirected too many times (20) Downloading: http://repo1.maven.org/maven2/org/apache/maven/plugins/maven-clean-plugin/2.2/maven-clean-plugin-2.2.pom [WARNING] Unable to get resource 'org.apache.maven.plugins:maven-clean-plugin:pom:2.2' from repository central (http://repo1.maven.org/maven2): Error trans ferring file: Server redirected too many times (20) [INFO] ------------------------------------------------------------------------ [ERROR] BUILD ERROR

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  • Why can't Perl's WWW::Machanize find the form by field names?

    - by FRESHTER
    #!/usr/bin/perl use WWW::Mechanize; use Compress::Zlib; my $mech = WWW::Mechanize->new(); my $username = ""; #fill in username here my $keyword = ""; #fill in password here my $mobile = $ARGV[0]; my $text = $ARGV[1]; $deb = 1; print length($text)."\n" if($deb); $text = $text."\n\n\n\n\n" if(length($text) < 135); $mech->get("http://wwwl.way2sms.com/content/index.html"); unless($mech->success()) { exit; } $dest = $mech->response->content; print "Fetching...\n" if($deb); if($mech->response->header("Content-Encoding") eq "gzip") { $dest = Compress::Zlib::memGunzip($dest); $mech->update_html($dest); } $dest =~ s/<form name="loginForm"/<form action='..\/auth.cl' name="loginForm"/g; $mech->update_html($dest); $mech->form_with_fields(("username","password")); $mech->field("username",$username); $mech->field("password",$keyword); print "Loggin...\n" if($deb); $mech->submit_form(); $dest= $mech->response->content; if($mech->response->header("Content-Encoding") eq "gzip") { $dest = Compress::Zlib::memGunzip($dest); $mech->update_html($dest); } $mech->get("http://wwwl.way2sms.com//jsp/InstantSMS.jsp?val=0"); $dest= $mech->response->content; if($mech->response->header("Content-Encoding") eq "gzip") { $dest = Compress::Zlib::memGunzip($dest); $mech->update_html($dest); } print "Sending ... \n" if($deb); $mech->form_with_fields(("MobNo","textArea")); $mech->field("MobNo",$mobile); $mech->field("textArea",$text); $mech->submit_form(); if($mech->success()) { print "Done \n" if($deb); } else { print "Failed \n" if($deb); exit; } $dest = $mech->response->content; if($mech->response->header("Content-Encoding") eq "gzip") { $dest = Compress::Zlib::memGunzip($dest); #print $dest if($deb); } if($dest =~ m/successfully/sig) { print "Message sent successfully" if($deb); } exit; When run this code halts with an error saying: There is no form with the requested fields at ./sms.pl line 65 Can't call method "value" on an undefined value at /usr/share/perl5/vendor_perl/WWW/Mechanize.pm line 1348.

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  • How to retrieve the GUID for Yahoo's Contacts API

    - by Jack Marchetti
    I'm attemping to use the Yahoo Contacts API to add an "invite your friends" feature on a site I'm building. I've found the correct web service to call (http://social.yahooapis.com/v1/user/{guid}/contacts) but it is asking for the user's GUID, not their username/password. I've searched, and am unable to find a "lookup" feature through the Yahoo API which lets me get the user's guid from their username/password. Does anyone have any experience with the Contacts API. I've reaad over the documentation, and looked at YQL as well, but I still haven't found how to get the user's guid. Thanks guys.

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  • how to add a entry to tables with relationships?

    - by siulamvictor
    I have 2 models, Users & Accounts. They are in one-to-many relationship, i.e. each accounts have many users. Accounts company_id company_name company_website Users user_id user_name password company_id email How can I add these entries to database using ActiveRecord? Supposed I don't is the company existed in the database when I add a new entry. Name Email Password Company ----------------------------------------------------------------------------- Albert [email protected] 123456 ABC Company Betty [email protected] 234567 ABC Company Carmen [email protected] 765432 XXX Company David [email protected] 654321 ABC Company

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  • How can I create a user registration form using datastore - google appengine ??

    - by srivigneshwar
    I am very new to jsp and google appengine , but still I can do something if I get some basic idea , I wanna create an user registration form with fields like name, user id , password, confirm password, etc,. using google appengine datastore feature. and I wanna retrieve user information . Please help me .. Thanks in advance. Thanks for the replies, I need to get some information from user and I want to store it in database, and I will use that information for my application. How can I do that?

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  • automatic enter login credentials while testing my app (from visual studio 2008)

    - by Michel
    Hi all, this is something that i don't want to program, but i was looking for a handy way of logging on to my web app. i'm building (and testing and running) my webapp over and over again, and here i've been provided a strong password. needless to say it's not so nice to enter my full user name + strong password 30 times a day. is there a nifty tool which lives in the background and when i open page localhost/mytestpage.aspx, it will say: "hey, let me type in michel and sdfs%^%gfhg in these two textboxes"?

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  • How to get the result for return statement from JSON parsing?

    - by blankon91
    I've follow the code for parsing the value with JSON from here, but I get the problem in my return statement. I want to put the parsing result into my return statement. How to do that? Here is my code: public String MASUK(String user, String password) { SoapObject request = new SoapObject(WSDL_TARGET_NAMESPACE,OPERATION_NAME); PropertyInfo pi = new PropertyInfo(); pi.setName("ccduser"); pi.setValue(user); pi.setType(String.class); request.addProperty(pi); PropertyInfo pi2 = new PropertyInfo(); pi2.setName("password"); pi2.setValue(password); pi2.setType(String.class); request.addProperty(pi2); SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11); envelope.dotNet = true; envelope.setOutputSoapObject(request); HttpTransportSE httpTransport = new HttpTransportSE(SOAP_ADDRESS); try { httpTransport.call(SOAP_ACTION, envelope); SoapObject resultSOAP = (SoapObject) envelope.bodyIn; /* gets our result in JSON String */ String ResultObject = resultSOAP.getProperty(0).toString(); resultSOAP = (SoapObject) envelope.bodyIn; ResultObject = resultSOAP.getProperty(0).toString(); if (ResultObject.startsWith("{")) { // if JSON string is an object JSONObj = new JSONObject(ResultObject); Iterator<String> itr = JSONObj.keys(); while (itr.hasNext()) { String Key = (String) itr.next(); String Value = JSONObj.getString(Key); BundleResult.putString(Key, Value); // System.out.println(bundleResult.getString(Key)); } } else if (ResultObject.startsWith("[")) { // if JSON string is an array JSONArr = new JSONArray(ResultObject); System.out.println("length" + JSONArr.length()); for (int i = 0; i < JSONArr.length(); i++) { JSONObj = (JSONObject) JSONArr.get(i); BundleResult.putString(String.valueOf(i), JSONObj.toString()); // System.out.println(bundleResult.getString(i)); } } } catch (Exception exception) { } return null; }

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  • Why always fires OnFailure when return View() to Ajax Form ?

    - by Wahid Bitar
    I'm trying to make a log-in log-off with Ajax supported. I made some logic in my controller to sign the user in and then return simple partial containing welcome message and log-Off ActionLink my Action method looks like this : public ActionResult LogOn(LogOnModel model, string returnUrl) { if (ModelState.IsValid) { if (MembershipService.ValidateUser(model.UserName, model.Password)) { FormsService.SignIn(model.UserName, model.RememberMe); if (Request.IsAjaxRequest()) { //HERE IS THE PROBLEM :( return View("LogedInForm"); } else { if (!String.IsNullOrEmpty(returnUrl)) return Redirect(returnUrl); else return RedirectToAction("Index", "Home"); } } else { ModelState.AddModelError("", "The user name or password provided is incorrect."); if (Request.IsAjaxRequest()) { return Content("There were an error !"); } } } return View(model); } and I'm trying to return this simple partial : Welcome <b><%= Html.Encode(Model.UserName)%></b>! <%= Html.ActionLink("Log Off", "LogOff", "Account") %> and of-course the two partial are strongly-typed to LogOnModel. But if i returned View("PartialName") i always get OnFailure with status code 500. While if i returned Content("My Message") everything is going right. so please tell me why i always get this "StatusCode = 500" ??. where is the big mistake ??. By the way in my Site MasterPage i rendered partial to show long-on simple form this partial looks like this : <script type="text/javascript"> function ShowErrorMessage(ajaxContext) { var response = ajaxContext.get_response(); var statusCode = response.get_statusCode(); alert("Sorry, the request failed with status code " + statusCode); } function ShowSuccessMessage() { alert("Hey everything is OK!"); } </script> <div id="logedInDiv"> </div> <% using (Ajax.BeginForm("LogOn", "Account", new AjaxOptions { UpdateTargetId = "logedInDiv", InsertionMode = InsertionMode.Replace, OnSuccess = "ShowSuccessMessage", OnFailure = "ShowErrorMessage" })) { %> <%= Html.TextBoxFor(m => m.UserName)%> <%= Html.PasswordFor(m => m.Password)%> <%= Html.CheckBoxFor(m => m.RememberMe)%> <input type="submit" value="Log On" /> < <% } %>

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  • PHP multiuser login class or script

    - by FFish
    I am looking for a simple but secure login script with mySQL PHP: sessions, MD5 that I can use with my exsisting database. Cookies to store password + password recovery by email. Change login/pass. I do not need registering, I register the user myself with temp login/pass. table agents agent1 agent2 table albums album1, owner: agent1 album2, owner: agent1 album3, owner: agent2 ... login.php agent1 logs in and has access to his albums: - album1 - album2 agent1 can edit his albums: edit.php?ref=album1 but NOT edit.php?ref=album3 by changing the ?ref variable

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  • Killing Mysql prcoesses staying in sleep command.

    - by Shino88
    Hey I am connecting a MYSQL database through hibernate and i seem to have processes that are not being killed after they are finished in the session. I have called flush and close on each session but when i check the server the last processes are still there with a sleep command. This is a new problem which i am having and was not the case yesterday. Is there any way i can ensure the killng of theses processes when i am done with a session. Below is an example of one of my classes. public JSONObject check() { //creates a new session needed to add elements to a database Session session = null; //holds the result of the check in the database JSONObject check = new JSONObject(); try{ //creates a new session needed to add elements to a database SessionFactory sessionFactory = new Configuration().configure().buildSessionFactory(); session = sessionFactory.openSession(); if (justusername){ //query created to select a username from user table String hquery = "Select username from User user Where username = ? "; //query created Query query = session.createQuery(hquery); //sets the username of the query the values JSONObject contents query.setString(0, username); // executes query and adds username string variable String user = (String) query.uniqueResult(); //checks to see if result is found (null if not found) if (user == null) { //adds false to Jobject if not found check.put("indatabase", "false"); } else { check.put("indatabase", "true"); } //adds check to Jobject to say just to check username check.put("justusername", true); } else { //query created to select a username and password from user table String hquery = "Select username from User user Where username = :user and password = :pass "; Query query = session.createQuery(hquery); query.setString("user", username); query.setString("pass", password); String user = (String) query.uniqueResult(); if(user ==null) { check.put("indatabase", false); } else { check.put("indatabase", true); } check.put("justusername", false); } }catch(Exception e){ System.out.println(e.getMessage()); //logg.log(Level.WARNING, " Exception", e.getMessage()); }finally{ // Actual contact insertion will happen at this step session.flush(); session.close(); } //returns Jobject return check; }

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  • How to read an XML file with Java?

    - by Yatendra Goel
    I don't need to read complex XML files. I just want to read the following configuration file with a simplest XML reader <config> <db-host>localhost</db-host> <db-port>3306</db-port> <db-username>root</db-username> <db-password>root</db-password> <db-name>cash</db-name> </config> How to read the above XML file with a XML reader through Java?

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  • how to show all method and data when the object not has "__iter__" function in python..

    - by zjm1126
    i find a way : (1):the dir(object) is : a="['__class__', '__contains__', '__delattr__', '__delitem__', '__dict__', '__doc__', '__getattribute__', '__getitem__', '__hash__', '__init__', '__iter__', '__metaclass__', '__module__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__setitem__', '__str__', '__weakref__', '_errors', '_fields', '_prefix', '_unbound_fields', 'confirm', 'data', 'email', 'errors', 'password', 'populate_obj', 'process', 'username', 'validate']" (2): b=eval(a) (3)and it became a list of all method : ['__class__', '__contains__', '__delattr__', '__delitem__', '__dict__', '__doc__', '__getattribute__', '__getitem__', '__hash__', '__init__', '__iter__', '__metaclass__', '__module__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__setitem__', '__str__', '__weakref__', '_errors', '_fields', '_prefix', '_unbound_fields', 'confirm', 'data', 'email', 'errors', 'password', 'populate_obj', 'process', 'username', 'validate'] (3)then show the object's method,and all code is : s='' a=eval(str(dir(object))) for i in a: s+=str(i)+':'+str(object[i]) print s but it show error : KeyError: '__class__' so how to make my code running . thanks

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  • Pros / Cons displaying list of users at login page

    - by Radu094
    We seem to have a lot of clients asking us to change the login screen in this manner: Display a list of all available users (thumbnail picture + name) User selects a username from the list A password prompt appears near the username User enters password then presses enter This sounds remarcably similar to the Windows XP login, which is probably where they got the ideea in the first place. There are only about 4 - 5 different users that can login at any given station, so implementing that list on one screen is feasable. So I was wondering if there are any usability experts with some word on this method of login. As far as I can tell, MS droped this behaviour in Vista/Win7, didn't they?

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