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  • How do I start Nautilus as root?

    - by Pho swan
    I got problem with nautilus in Ubuntu 12.04 LTS when i run command "gksu nautilus /" password ask box appear and I type my super-user password , than password box is disappear and nautilus is not open. . when i try to open nautilus via normal user in command box . "nautilus" the folder is open up. when i try in terminal ... $sudo nautilus I got following error $ sudo nautilus / ** (nautilus:8523): WARNING **: Command line `dbus-launch --autolaunch=2c8ce9b7da2257c2609b749700000007 --binary-syntax --close-stderr' exited with non-zero exit status 1: Autolaunch error: X11 initialization failed.\n Could not parse arguments: Cannot open display: how can i fix this error..

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  • Are two database trips reasonable for a login system?

    - by Randolph Potter
    I am designing a login system for a project, and have an issue about it requiring two trips to the database when a user logs in. User types in username and password Database is polled and password hash is retrieved for comparative purposes (first trip) Code tests hash against entered password (and salt), and if verified, resets the session ID New session ID and username are sent back to the database to write a row to the login table, and generate a login ID for that session. EDIT: I am using a random salt. Does this design make sense? Am I missing something? Is my concern about two trips unfounded? Comments and suggestions are welcome.

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  • User Already Exists in the Current Database - SQL Server

    - by bullpit
    I was moving a lot of databases from one SQL Server to another, and my applications were giving me errors saying "Login failed for <user>". The user was already in the database with appropriate rights to allowed objects in the database. I tried mapping the user to the database and that's when I got this message: "User Already Exists in the Current Database"... I googled and found this very useful post about orphaned users when moving databases. These are the steps you should take to fix this issue: First, make sure that this is the problem. This will lists the orphaned users: EXEC sp_change_users_login 'Report' If you already have a login id and password for this user, fix it by doing: EXEC sp_change_users_login 'Auto_Fix', 'user' If you want to create a new login id and password for this user, fix it by doing: EXEC sp_change_users_login 'Auto_Fix', 'user', 'login', 'password'

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  • How does Requiring users to Periodically Change their Passwords Improve Security? [closed]

    - by Bob Kaufman
    I've had the same password for some sites for years with no regrets. Meanwhile, at work, I find myself being forced to change passwords every two to three months. My thinking is that if a password gets compromised, requiring that I change it several weeks out isn't going to protect me or the network very much. Moreover, I find that by being required to change passwords frequently, I degenerate into a predictable password pattern (e.g., BearsFan111, BearsFan222, ...) which results in easier to remember and easier to guess passwords. Is there a sound argument for requiring that passwords be changed periodically?

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  • auth_mysql and php [migrated]

    - by user1052448
    I have a directory with auth_mysql in a virtualhost file password protected using a mysql user/pass combo. The problem I have is one file inside that directory needs to be accessed without a user/pass. Is there a way I can pass the user/pass within the php file? Or excluse the one file? What would I put between the code below? <Location /password-protected> ...mysql password protection require valid-user </Location>

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  • Is it possible for an application (written in Mono C#) to run a console command?

    - by Razick
    I am wondering if a Mono C# application can somehow run a terminal command. For example, could the user give the program his or her password and then have the application run sudo apt-get install application-name (console requests password) password (console requests confirmation) y Preferably, this would be done without actually opening a terminal visible to the user, so that the application could provide the necessary feedback and manage the whole operation cleanly with as little user interaction as possible. Is there a way to do that? Let me know if clarification is needed. Thank you!

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  • How can I have my VPN connect automatically when the wireless connects?

    - by ams
    I have a working VPN connection using NetworkManager, OpenConnect, and the network-manager-openconnect-gnome package, but I have to start it manually every time I connect to a network, and I have to enter my password manually each time. How can I get it to connect automatically, and remember my password (securely)? I have checked the 'Connect Automatically' box on the Configure VPN page, but this seems to have no effect. I've also got the 'Start connecting automatically' box checked in the pop-up box, and that does avoid the need to press the connect button in that window, but seems to have no part in kicking off the whole process in the first place. There is no option to remember the password in the window, but maybe there's one somewhere else?

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  • Xfce session will not login from login prompt. Why?

    - by hydroparadise
    Well, not sure how I did it, but for some reason I am unable to get past the login screen on Xfce. What's odd is that I had set the machine up (from install) to login automatically. Then when I finish rebooting after installing sudo apt-get install openssh-server and sudo apt-get install vnc4server all the sudden I am getting prompted for my password. I proceed to type my password (successfully might I add) to login, the screen goes black for a second, then brings me right back to the same password prompt. I am able ssh in(now), but I get no love from workstation login. Maybe vnc4server is the culprit? What did I do? And can I fix it? EDIT: The guest account, however will let me login no problem. What gives? I have a sneaking feeling that this could be permission issue.

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  • Encrypted Ubuntu Partition HELP!

    - by user207984
    I had Zorin 7 Encrypted. I accidentally deleted important GRUB data trying to make more room in the boot folder i think? So the computer would no longer boot up. Anyways, I installed a second Linux OS on the same laptop, but I am unable to access all of my important information that is still on the encrypted partition. When i go to mount it to access everything, it asks for the encryption password... however, it tells me the password is incorrect, which I know 100% that it is the correct password. Someone please help me! Inform me how I can access this encrypted partition or restore the data so that I can boot the computer in Zorin 7 again.

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  • I can't log-in to WinSCP using username "root"

    - by Jessyle Ivy
    I can't log-in to WinSCP using username "root". I already change the password of "root" in Ubuntu, and I successfully log-in there. But in WinSCP, it goes like this Search for host... Connecting to host... Authenticating... Using username "root" Authenticating with pre-entered password. Access Denied. and I am need to re-type the password again. By the way I'm using VMware Player for Ubuntu. Thanks!

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  • 407 Proxy Authentication Required

    - by user38507
    When I try to install a software using Ubuntu Software center I get: Failed to download repository information Check Your Internet connection When I try to do a apt get-install something, I get: 407 Proxy Authentication Required I use a proxy server that requires a user-name and a password. I have set my systems proxy manually, by plugging in the required numbers in the Networks proxy and applied it system wide. I guess the problem now is plugging in my user-name and password. When I use INTERNET via Mozilla, it specifically asks me for my user-name and password.

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  • Nashorn ?? JDBC ? Oracle DB ?????·?? 3

    - by Homma
    ???? Nashorn ?? JavaScript ??????? JDBC ? Oracle DB ???????????????????? Oracle DB ????? SQL ??????????????? ???????????????????????????????? ????????? URL ? https://blogs.oracle.com/nashorn_ja/entry/nashorn_jdbc_3 ??? JDBC ??????????????? JDBC ????????????????? Nashorn ????? JavaScript ????????????? ???????? JDBC OCI ???????????????????????????????? ????? ?? Java ??????????????? Nashorn ? JavaScript ???????????????? // Invoke jjs with -scripting option. /* * This sample can be used to check the JDBC installation. * Just run it and provide the connect information. It will select * "Hello World" from the database. */ var OracleDataSource = Java.type("oracle.jdbc.pool.OracleDataSource"); function main() { // Prompt the user for connect information print("Please enter information to test connection to the database"); var user, password, database; user = readLine("user: "); slash_index = user.indexOf('/'); if (slash_index != -1) { password = user.substring(slash_index + 1) user = user.substring(0, slash_index); } else password = readLine("password: "); database = readLine("database(a TNSNAME entry): "); java.lang.System.out.print("Connecting to the database..."); java.lang.System.out.flush(); print("Connecting..."); // Open an OracleDataSource and get a connection var ods = new OracleDataSource(); ods.setURL("jdbc:oracle:oci:@" + database); ods.setUser(user); ods.setPassword(password); var conn = ods.getConnection(); print("connected."); // Create a statement var stmt = conn.createStatement(); // Do the SQL "Hello World" thing var rset = stmt.executeQuery("select 'Hello World' from dual"); while (rset.next()) print(rset.getString(1)); // close the result set, the statement and the connection rset.close(); stmt.close(); conn.close(); print("Your JDBC installation is correct."); } main(); oracle.jdbc.pool.OracleDataSource ? Java.type() ?????Nashorn ??????????????????????????? Java ? System.out.println() ? System.out.flush() ? java.lang. ???????????????? Java ?????????? readEntry() ????? Nashorn ? readLine() ???????????? Java ????????????????????????JavaScript ?????????????????? ?? Java ??????????????????????????? Java ???????????????? JavaScript ?????????????????? ???????? JDBC OCI ???????????????? LD_LIBRARY_PATH ????????????????? ???Nashorn ? readLine() ??????????jjs ????? -scripting ????????????????? $ export LD_LIBRARY_PATH=${ORACLE_HOME}/lib $ jjs -scripting -cp ${ORACLE_HOME}/jdbc/lib/ojdbc6.jar JdbcCheckup.js Please enter information to test connection to the database user: test password: test database(a TNSNAME entry): orcl Connecting to the database...Connecting... connected. Hello World Your JDBC installation is correct. JDBC OCI ????????????????? "select 'Hello World' from dual" ??? SQL ?????????????? ?????????????????database ???? :: ??????????? ??? ??? Oracle DB ????? SQL ???????????????? Java ? JDBC ??????????????????????????? Nashorn ??????????????????????????????????

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  • problem with network-manager-pptp

    - by Riuzaki90
    I've a problema with the VPA CAble connection of my university... on the website of the university there's a .sh file that set all the variables of the connection in ETC/PPP/PEERS and another .sh file that call the connection...I'm on ubuntu 11.10 and when I run the setup.sh I have this error: impossible to find network-manager-pptp these are the two file that I had talk about: #!/bin/bash echo "Creazione della connessione in corso attendere........." apt-get update apt-get install pptp-linux network-manager-pptp echo -n "Digitare la propria Username: " read USERNAME echo -n "Digitare la propria Password: " read PASSWORD pptpsetup --create UNICAL_Campus_Access --server 160.97.73.253 --username $USERNAME --password $PASSWORD echo 'pty "pptp 160.97.73.253 --nolaunchpppd"' >/etc/ppp/peers/UNICAL_Campus_Access echo 'require-mppe-128' >>/etc/ppp/peers/UNICAL_Campus_Access echo 'file /etc/ppp/options.pptp'>>/etc/ppp/peers/UNICAL_Campus_Access echo 'name '$USERNAME''>>/etc/ppp/peers/UNICAL_Campus_Access echo 'remotename PPTP'>>/etc/ppp/peers/UNICAL_Campus_Access echo 'ipparam UNICAL_Campus_Access'>>/etc/ppp/peers/UNICAL_Campus_Access echo $USERNAME' PPTP '$PASSWORD' *'>>/etc/ppp/chap-secrets rm /etc/ppp/options.pptp echo '###############################################################################'>/etc/ppp/options.pptp echo '# $Id: options.pptp,v 1.3 2006/03/26 23:11:05 quozl Exp $'>>/etc/ppp/options.pptp echo '#'>>/etc/ppp/options.pptp echo '# Sample PPTP PPP options file /etc/ppp/options.pptp'>>/etc/ppp/options.pptp echo '# Options used by PPP when a connection is made by a PPTP client.'>>/etc/ppp/options.pptp echo '# This file can be referred to by an /etc/ppp/peers file for the tunnel.'>>/etc/ppp/options.pptp echo '# Changes are effective on the next connection. See "man pppd".'>>/etc/ppp/options.pptp echo '#'>>/etc/ppp/options.pptp echo '# You are expected to change this file to suit your system. As'>>/etc/ppp/options.pptp echo '# packaged, it requires PPP 2.4.2 or later from http://ppp.samba.org/'>>/etc/ppp/options.pptp echo '# and the kernel MPPE module available from the CVS repository also on'>>/etc/ppp/options.pptp echo '# http://ppp.samba.org/, which is packaged for DKMS as kernel_ppp_mppe.'>>/etc/ppp/options.pptp echo '###############################################################################'>>/etc/ppp/options.pptp echo '# Lock the port'>>/etc/ppp/options.pptp echo 'lock'>>/etc/ppp/options.pptp echo '# Authentication'>>/etc/ppp/options.pptp echo '# We do not need the tunnel server to authenticate itself'>>/etc/ppp/options.pptp echo 'noauth'>>/etc/ppp/options.pptp echo '#We won"t do PAP, EAP, CHAP, or MSCHAP, but we will accept MSCHAP-V2'>>/etc/ppp/options.pptp echo '#(you may need to remove these refusals if the server is not using MPPE)'>>/etc/ppp/options.pptp echo 'refuse-pap'>>/etc/ppp/options.pptp echo 'refuse-eap'>>/etc/ppp/options.pptp echo 'refuse-chap'>>/etc/ppp/options.pptp echo 'refuse-mschap'>>/etc/ppp/options.pptp echo '# Compression Turn off compression protocols we know won"t be used'>>/etc/ppp/options.pptp echo 'nobsdcomp'>>/etc/ppp/options.pptp echo 'nodeflate'>>/etc/ppp/options.pptp echo '# Encryption'>>/etc/ppp/options.pptp echo '# (There have been multiple versions of PPP with encryption support,'>>/etc/ppp/options.pptp echo '# choose with of the following sections you will use. Note that MPPE'>>/etc/ppp/options.pptp echo '# requires the use of MSCHAP-V2 during authentication)'>>/etc/ppp/options.pptp echo '# http://ppp.samba.org/ the PPP project version of PPP by Paul Mackarras'>>/etc/ppp/options.pptp echo '# ppp-2.4.2 or later with MPPE only, kernel module ppp_mppe.o'>>/etc/ppp/options.pptp echo '#{{{'>>/etc/ppp/options.pptp echo '# Require MPPE 128-bit encryption'>>/etc/ppp/options.pptp echo '#require-mppe-128'>>/etc/ppp/options.pptp echo '#}}}'>>/etc/ppp/options.pptp echo '# http://polbox.com/h/hs001/ fork from PPP project by Jan Dubiec'>>/etc/ppp/options.pptp echo '#ppp-2.4.2 or later with MPPE and MPPC, kernel module ppp_mppe_mppc.o'>>/etc/ppp/options.pptp echo '#{{{'>>/etc/ppp/options.pptp echo '# Require MPPE 128-bit encryption'>>/etc/ppp/options.pptp echo '#mppe required,stateless'>>/etc/ppp/options.pptp echo '# }}}'>>/etc/ppp/options.pptp echo "setup di 'UNICAL Campus Access' terminato correttamente" echo "per connettersi eseguire lo script 'UNICAL_Campus_Access.sh' " and the second: #!/bin/bash echo "Connessione alla Rete del Centro Residenziale in corso attendere........." modprobe ppp_mppe pppd call UNICAL_Campus_Access sleep 30 tail -n 8 /var/log/messages echo "Connessione Stabilita" echo -n "Per terminare la connessione premere invio (in alternativa eseguire il commando 'killall pppd'):----> " read CONN killall pppd echo "Connessione terminata" I've correctly installed network-manager-pptp to the latest version...help?

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  • LexisNexis and Oracle Join Forces to Prevent Fraud and Identity Abuse

    - by Tanu Sood
    Author: Mark Karlstrand About the Writer:Mark Karlstrand is a Senior Product Manager at Oracle focused on innovative security for enterprise web and mobile applications. Over the last sixteen years Mark has served as director in a number of tech startups before joining Oracle in 2007. Working with a team of talented architects and engineers Mark developed Oracle Adaptive Access Manager, a best of breed access security solution.The world’s top enterprise software company and the world leader in data driven solutions have teamed up to provide a new integrated security solution to prevent fraud and misuse of identities. LexisNexis Risk Solutions, a Gold level member of Oracle PartnerNetwork (OPN), today announced it has achieved Oracle Validated Integration of its Instant Authenticate product with Oracle Identity Management.Oracle provides the most complete Identity and Access Management platform. The only identity management provider to offer advanced capabilities including device fingerprinting, location intelligence, real-time risk analysis, context-aware authentication and authorization makes the Oracle offering unique in the industry. LexisNexis Risk Solutions provides the industry leading Instant Authenticate dynamic knowledge based authentication (KBA) service which offers customers a secure and cost effective means to authenticate new user or prove authentication for password resets, lockouts and such scenarios. Oracle and LexisNexis now offer an integrated solution that combines the power of the most advanced identity management platform and superior data driven user authentication to stop identity fraud in its tracks and, in turn, offer significant operational cost savings. The solution offers the ability to challenge users with dynamic knowledge based authentication based on the risk of an access request or transaction thereby offering an additional level to other authentication methods such as static challenge questions or one-time password when needed. For example, with Oracle Identity Management self-service, the forgotten password reset workflow utilizes advanced capabilities including device fingerprinting, location intelligence, risk analysis and one-time password (OTP) via short message service (SMS) to secure this sensitive flow. Even when a user has lost or misplaced his/her mobile phone and, therefore, cannot receive the SMS, the new integrated solution eliminates the need to contact the help desk. The Oracle Identity Management platform dynamically switches to use the LexisNexis Instant Authenticate service for authentication if the user is not able to authenticate via OTP. The advanced Oracle and LexisNexis integrated solution, thus, both improves user experience and saves money by avoiding unnecessary help desk calls. Oracle Identity and Access Management secures applications, Juniper SSL VPN and other web resources with a thoroughly modern layered and context-aware platform. Users don't gain access just because they happen to have a valid username and password. An enterprise utilizing the Oracle solution has the ability to predicate access based on the specific context of the current situation. The device, location, temporal data, and any number of other attributes are evaluated in real-time to determine the specific risk at that moment. If the risk is elevated a user can be challenged for additional authentication, refused access or allowed access with limited privileges. The LexisNexis Instant Authenticate dynamic KBA service plugs into the Oracle platform to provide an additional layer of security by validating a user's identity in high risk access or transactions. The large and varied pool of data the LexisNexis solution utilizes to quiz a user makes this challenge mechanism even more robust. This strong combination of Oracle and LexisNexis user authentication capabilities greatly mitigates the risk of exposing sensitive applications and services on the Internet which helps an enterprise grow their business with confidence.Resources:Press release: LexisNexis® Achieves Oracle Validated Integration with Oracle Identity Management Oracle Access Management (HTML)Oracle Adaptive Access Manager (pdf)

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  • SQL SERVER – Importance of User Without Login

    - by pinaldave
    Some questions are very open ended and it is very hard to come up with exact requirements. Here is one question I was asked in recent User Group Meeting. Question: “In recent version of SQL Server we can create user without login. What is the use of it?” Great question indeed. Let me first attempt to answer this question but after reading my answer I need your help. I want you to help him as well with adding more value to it. Answer: Let us visualize a scenario. An application has lots of different operations and many of them are very sensitive operations. The common practice was to do give application specific role which has more permissions and access level. When a regular user login (not system admin), he/she might have very restrictive permissions. The application itself had a user name and password which means applications can directly login into the database and perform the operation. Developers were well aware of the username and password as it was embedded in the application. When developer leaves the organization or when the password was changed, the part of the application had to be changed where the same username and passwords were used. Additionally, developers were able to use the same username and password and login directly to the same application. In earlier version of SQL Server there were application roles. The same is later on replaced by “User without Login”. Now let us recreate the above scenario using this new “User without Login”. In this case, User will have to login using their own credentials into SQL Server. This means that the user who is logged in will have his/her own username and password. Once the login is done in SQL Server, the user will be able to use the application. Now the database should have another User without Login which has all the necessary permissions and rights to execute various operations. Now, Application will be able to execute the script by impersonating “user without login – with more permissions”. Here there is assumed that user login does not have enough permissions and another user (without login) there are more rights. If a user knows how the application is using the database and their various operations, he can switch the context to user without login making him enable for doing further modification. Make sure to explicitly DENY view definition permission on the database. This will make things further difficult for user as he will have to know exact details to get additional permissions. If a user is System Admin all the details which I just mentioned in above three paragraphs does not apply as admin always have access to everything. Additionally, the method describes above is just one of the architecture and if someone is attempting to damage the system, they will still be able to figure out a workaround. You will have to put further auditing and policy based management to prevent such incidents and accidents. I guess this is my answer. I read it multiple times but I still feel that I am missing something. There should be more to this concept than what I have just described. I have merely described one scenario but there will be many more scenarios where this situation will be useful. Now is your turn to help – please leave a comment with the additional suggestion where exactly “User without Login” will be useful as well did I miss anything when I described above scenario. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Query, SQL Security, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • why my code still cannot connect with database? [closed]

    - by Wen Teng
    package com.mems.travis; import java.util.ArrayList; import java.util.List; import org.apache.http.NameValuePair; import org.apache.http.message.BasicNameValuePair; import org.json.JSONObject; import android.app.Activity; import android.app.AlertDialog; import android.content.DialogInterface; import android.content.Intent; import android.os.AsyncTask; import android.os.Bundle; import android.util.Log; import android.view.View; import android.widget.Button; import android.widget.EditText; import android.widget.RadioButton; public class UserRegister extends Activity { JSONParser jsonParser = new JSONParser(); EditText inputName; EditText inputUsername; EditText inputEmail; EditText inputPassword; RadioButton button1; RadioButton button2; Button button3; int success = 0; // url to create new product private static String url_register_user = "http://192.168.1.100/MEMS/add_user.php"; // JSON Node names private static final String TAG_SUCCESS = "success"; @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.activity_user_register); // Edit Text inputName = (EditText) findViewById(R.id.nameTextBox); inputUsername = (EditText) findViewById(R.id.usernameTextBox); inputEmail = (EditText) findViewById(R.id.emailTextBox); inputPassword = (EditText) findViewById(R.id.pwTextBox); // Create button //RadioButton button1 = (RadioButton) findViewById(R.id.studButton); // RadioButton button2 = (RadioButton) findViewById(R.id.shopownerButton); Button button3 = (Button) findViewById(R.id.regSubmitButton); // button click event button3.setOnClickListener(new View.OnClickListener() { public void onClick(View view) { String name = inputName.getText().toString(); String username = inputUsername.getText().toString(); String email = inputEmail.getText().toString(); String password = inputPassword.getText().toString(); if (name.contentEquals("")||username.contentEquals("")||email.contentEquals("")||password.contentEquals("")) { AlertDialog.Builder builder = new AlertDialog.Builder(UserRegister.this); // 2. Chain together various setter methods to set the dialog characteristics builder.setMessage(R.string.nullAlert) .setTitle(R.string.alertTitle); builder.setPositiveButton(R.string.ok, new DialogInterface.OnClickListener() { public void onClick(DialogInterface dialog, int id) { // User clicked OK button } }); // 3. Get the AlertDialog from create() AlertDialog dialog = builder.show(); } else { new RegisterNewUser().execute(); } } }); } class RegisterNewUser extends AsyncTask<String, String, String>{ protected String doInBackground(String... args) { String name = inputName.getText().toString(); String username = inputUsername.getText().toString(); String email = inputEmail.getText().toString(); String password = inputPassword.getText().toString(); // Building Parameters List<NameValuePair> params = new ArrayList<NameValuePair>(); params.add(new BasicNameValuePair("name", name)); params.add(new BasicNameValuePair("username", username)); params.add(new BasicNameValuePair("email", email)); params.add(new BasicNameValuePair("password", password)); // getting JSON Object // Note that create product url accepts POST method JSONObject json = jsonParser.makeHttpRequest(url_register_user, "GET", params); // check log cat for response Log.d("Send Notification", json.toString()); try { int success = json.getInt(TAG_SUCCESS); if (success == 1) { // successfully created product Intent i = new Intent(getApplicationContext(), StudentLogin.class); startActivity(i); finish(); } else { // failed to register } } catch (Exception e) { e.printStackTrace(); } return null; } } }

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  • Why do I get "Invalid Column Name" errors in EF4?

    - by camainc
    I am trying to learn Entity Framework 4.0. Disclaimer 1: I am brand new to Entity Framework. I have successfully used LinqToSQL. Disclaimer 2: I am really a VB.Net programmer, so the problem could be in the C# code. Given this code snippet: public int Login(string UserName, string Password) { return _dbContext.Memberships .Where(membership => membership.UserName.ToLower() == UserName.ToLower() && membership.Password == Password) .SingleOrDefault().PrimaryKey; } Why do you suppose I get "Invalid column name" errors? {"Invalid column name 'UserName'.\r\nInvalid column name 'Password'.\r\nInvalid column name 'UserName'.\r\nInvalid column name 'Password'."} Those column names are spelled and cased correctly. I also checked the generated code for the entity in question, and those columns are properties in the entity. The intellisense and code completion also puts the column names into the expression just as they are here. I am stumped by this. Any help would be much appreciated. https://docs.google.com/leaf?id=0B-xLbzoqGvXvNjBmZmNjNDAtY2RhNC00NDA2LWIxNzMtYjhjNTYxMDIyZmZl&hl=en

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  • VBA WinHTTPRequest and submitting forms

    - by Hazerider
    Hi. I spent all day yesterday trying to figure out how to submit a form using WinHTTPRequest. I can do it pretty easily with an InternetExplorer object, but the problem is that I need to save a PDF file that gets returned, and I am not sure how to do this with the IE object. Here is the relevant HTML code snippet: <div class="loginHome-left"> <fieldset> <h3>Log in Using</h3> <form> <label for="standardLogin" accesskey="s"> <input name="useLogin" id="standardLogin" value="standard" type="radio" checked="true">Standard Login</label> &nbsp; <label for="rsaSecurID" accesskey="r"> <input name="useLogin" value="rsaSecur" type="radio" id="rsaSecurID" onclick="redirectLogin('ct_logon_securid');return false;">RSA SecurID</label> &nbsp; <label for="employeeNTXP" accesskey="e"> <input name="useLogin" id="employeeNTXP" value="employee" type="radio" onclick="redirectLogin('ct_logon_external_nt');return false; "> Employee Windows Login<br></label> </form> <br> <div class="error">Error: ...</div><br> <form onSubmit="if(validate(this)) {formSubmit();} return false;" name="passwdForm" method="post" action="/UAB/ct_logon"> <input value="custom" name="pageId" type="hidden"> <input value="custom" name="auth_mode" type="hidden"> <input value="/UAB/ct_logon" name="ct_orig_uri" type="hidden"> <INPUT VALUE="" NAME="orig_url" TYPE="hidden"> <input value="" name="lpSp" type="hidden"> <label for="user"> <strong>Username</strong> </label> <input autocomplete="off" name="user" type="text" value="" class="txtFld" onkeypress="return handleEnter(this, event);"> <br> <label for="EnterPassword"> <strong>Password</strong>&nbsp;&nbsp;(<a tabindex="-1" href="/UAB/BCResetWithSecrets">Forgot Your Password?</a>) </label> <input autocomplete="off" name="password" type="password" class="txtFld" onkeypress="return handleEnter(this, event);"> <INPUT id="rememberLogin" name="lpCookie" type="checkbox"> <label for="rememberLogin">Remember My Login Information</label><br> </form> <div class="right"> <br> <input type="image" src="/BC_S/images/bclogin/btn_login.gif" name="" value="Submit" onClick="if(validate(document.forms['passwdForm'])){formSubmit();}return false;"> </div> <div class="clearfix"></div> </fieldset> </div> In order to log in through InternetExplorer, I do the following: Sub TestLogin() Dim ie As InternetExplorer, doc As HTMLDocument, form As HTMLFormElement, inp As Variant Set ie = New InternetExplorer ie.Visible = True ie.navigate "https://URL of the login page" Do Until ie.readyState = READYSTATE_COMPLETE Loop Set doc = ie.document For Each form In doc.forms If InStr(form.innerText, "Password") <> 0 Then form.elements("user").Value = "my_name" form.elements("password").Value = "my_password" Exit For Else End If Next 'This is the unnamed input with an image that is used to submit the form' doc.all(78).Click ie.navigate "https://url of the PDF" Do Until ie.readyState = READYSTATE_COMPLETE Loop Dim filename As String, filenum As Integer filename = "somefile.pdf" filenum = FreeFile Open filename For Binary Access Write As #filenum Write #filenum, doc.DocumentElement.innerText Close #filenum ie.Quit Debug.Print Set ie = Nothing End Sub What I really would like to do is something along the lines of the following: Sub TestLogin3() Dim whr As New WinHttpRequest, postData As String whr.Open "POST", "https://live.barcap.com/UAB/ct_logon", False whr.setRequestHeader "User-Agent", "Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.0)" whr.setRequestHeader "Connection", "Keep-Alive" whr.Send whr.WaitForResponse postData = "user=paschom1&password=change01" 'Or the following?' postData = "user=paschom1&password=change01&orig_url=&pageId=custom&auth_mode=custom&ct_orig_uri=/BC/dispatcher&lpSp=&lpCookie=off" whr.Send postData whr.WaitForResponse Debug.Print whr.responseText End Sub It just refuses to work though. Not sure if I need to use more setRequestHeader with Content-Form or something similar, and if I do, not sure what exactly I am supposed to pass it. If anyone has any advice regarding this, it would be hugely appreciated. I could probably use a perl module to do it, but I would rather keep it all in VBA if possible. Thanks, Marc.

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  • Help With Proxy Username & Pass with GeckoFX??!!

    - by John
    Hello, I am trying to set the proxy username and password. I saw this posting (http://geckofx.org/viewtopic.php?id=832) and I thought it might be a similar setting for the username/password, such as : Skybound.Gecko.GeckoPreferences.User["network.proxy.user"] = (user); Skybound.Gecko.GeckoPreferences.User["network.proxy.password"] = (password); But, nothing has worked so far. Can anyone help? I would really appreciate it!!! I am using VB.net if that helps. Thanks!!

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  • How to create datetime string in soapui using groovy

    - by Arunkumar
    Hi am using Soapui for testing web services. i need to create a customer record with email address and password. create customer record service contains emailid and password, wen i click the run(submit request) button in create customer record in soapui, i should get the emailid appended with current time of creation and any password. how to do this with groovy?

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  • Set Proxy Credential in Web Browser Control

    - by rockacola
    I am working on a legacy code where an application uses AxSHDocVw.AxWebBrowser (NOT System.Windows.Forms.Control) to open up web pages and am extending it to take proxy into considerations. I have following example on http://www.pinvoke.net/default.aspx/wininet/internetsetoption.html to use InternetSetOption() to go through specified proxy and tested that it works. Now the hurdle is I tried everything but failed to pass username and password with following code: //-- Set Proxy Username bool resultF = InternetSetOption(IntPtr.Zero, INTERNET_OPTION_PROXY_USERNAME, username, username.Length+1); var errorF = Marshal.GetLastWin32Error(); //-- Set Proxy Password bool resultG = InternetSetOption(IntPtr.Zero, INTERNET_OPTION_PROXY_PASSWORD, password, password.Length+1); var errorG = Marshal.GetLastWin32Error(); Both resultF and resultG return true and has no errors but it still working. Any hint on what may be happening here? and what method do I have to debug this? Thanks in advance.

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  • org.apache.struts2.dispatcher.Dispatcher: Could not find action or result Error

    - by peterwkc
    i tried to code the following simple struts but encounter this error during run time. [org.apache.struts2.dispatcher.Dispatcher] Could not find action or result: No result defined for action com.peter.action.LoginAction and result success index.jsp <%@ page language="java" contentType="text/html; charset=ISO-8859-1" pageEncoding="ISO-8859-1"%> <%@ taglib prefix="s" uri="/struts-tags" %> <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1"> <title>Struts Tutorial</title> </head> <body> <h2>Hello Struts</h2> <s:form action="login" > <s:textfield name="username" label="Username:" /> <s:password name="password" label="Password:"/> <s:submit /> </s:form> </body> </html> LoginAction.java /** * */ package com.peter.action; //import org.apache.struts2.convention.annotation.Namespace; import org.apache.struts2.convention.annotation.ResultPath; import org.apache.struts2.convention.annotation.Result; import org.apache.struts2.convention.annotation.Action; import com.opensymphony.xwork2.ActionSupport; /** * @author nicholas_tse * */ //@Namespace("/") To define URL namespace @ResultPath("/") // To instruct Struts where to search result page(jsp) public class LoginAction extends ActionSupport { private String username, password; /** * */ private static final long serialVersionUID = -8992836566328180883L; /** * */ public LoginAction() { // TODO Auto-generated constructor stub } public String getUsername() { return username; } public void setUsername(String username) { this.username = username; } public String getPassword() { return password; } public void setPassword(String password) { this.password = password; } @Override @Action(value = "login", results = {@Result(name="success", location="welcome.jsp")}) public String execute() { return SUCCESS; } } /* Remove * struts2-gxp-plugin * struts2-portlet-plugin * struts2-jsf-plugin * struts2-osgi-plugin and its related osgi-plugin * struts-rest-plugin * * Add * velocity-tools-view * * */ web.xml <?xml version="1.0" encoding="UTF-8"?> <web-app id="WebApp_ID" version="3.0" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"> <display-name>Struts</display-name> <!-- org.apache.struts2.dispatcher.ng.filter.StrutsPrepareAndExecuteFilter org.apache.struts2.dispatcher.FilterDispatcher --> <filter> <filter-name>Struts_Filter</filter-name> <filter-class>org.apache.struts2.dispatcher.FilterDispatcher</filter-class> <init-param> <param-name>actionPackages</param-name> <param-value>com.peter.action</param-value> </init-param> </filter> <filter-mapping> <filter-name>Struts_Filter</filter-name> <url-pattern>/*</url-pattern> </filter-mapping> <listener> <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> </listener> <welcome-file-list> <welcome-file>index.jsp</welcome-file> </welcome-file-list> </web-app> Besides the runtime error, there is deployment error which is ERROR [com.opensymphony.xwork2.util.finder.ClassFinder] (MSC service thread 1-2) Unable to read class [WEB-INF.classes.com.peter.action.LoginAction]: Could not load WEB-INF/classes/com/peter/action/LoginAction.class - [unknown location] at com.opensymphony.xwork2.util.finder.ClassFinder.readClassDef(ClassFinder.java:785) [xwork-core-2.3.1.2.jar:2.3.1.2] AFAIK, the scanning methodology of struts will scan the default packages named struts2 for any annotated class but i have instructed struts2 to scan in com.peter.action using init-param but still unable to find the class. It is pretty weird. Please help. Thanks.

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  • NSData-AES Class Encryption/Decryption in Cocoa

    - by David Schiefer
    hi, I am attempting to encrypt/decrypt a plain text file in my text editor. encrypting seems to work fine, but the decrypting does not work, the text comes up encrypted. I am certain i've decrypted the text using the word i encrypted it with - could someone look through the snippet below and help me out? Thanks :) Encrypting: NSAlert *alert = [NSAlert alertWithMessageText:@"Encryption" defaultButton:@"Set" alternateButton:@"Cancel" otherButton:nil informativeTextWithFormat:@"Please enter a password to encrypt your file with:"]; [alert setIcon:[NSImage imageNamed:@"License.png"]]; NSSecureTextField *input = [[NSSecureTextField alloc] initWithFrame:NSMakeRect(0, 0, 300, 24)]; [alert setAccessoryView:input]; NSInteger button = [alert runModal]; if (button == NSAlertDefaultReturn) { [[NSUserDefaults standardUserDefaults] setObject:[input stringValue] forKey:@"password"]; NSData *data; [self setString:[textView textStorage]]; NSMutableDictionary *dict = [NSDictionary dictionaryWithObject:NSPlainTextDocumentType forKey:NSDocumentTypeDocumentAttribute]; [textView breakUndoCoalescing]; data = [[self string] dataFromRange:NSMakeRange(0, [[self string] length]) documentAttributes:dict error:outError]; NSData*encrypt = [data AESEncryptWithPassphrase:[input stringValue]]; [encrypt writeToFile:[absoluteURL path] atomically:YES]; Decrypting: NSAlert *alert = [NSAlert alertWithMessageText:@"Decryption" defaultButton:@"Open" alternateButton:@"Cancel" otherButton:nil informativeTextWithFormat:@"This file has been protected with a password.To view its contents,enter the password below:"]; [alert setIcon:[NSImage imageNamed:@"License.png"]]; NSSecureTextField *input = [[NSSecureTextField alloc] initWithFrame:NSMakeRect(0, 0, 300, 24)]; [alert setAccessoryView:input]; NSInteger button = [alert runModal]; if (button == NSAlertDefaultReturn) { NSLog(@"Entered Password - attempting to decrypt."); NSMutableDictionary *dict = [NSDictionary dictionaryWithObject:NSPlainTextDocumentType forKey:NSDocumentTypeDocumentOption]; NSData*decrypted = [[NSData dataWithContentsOfFile:[self fileName]] AESDecryptWithPassphrase:[input stringValue]]; mString = [[NSAttributedString alloc] initWithData:decrypted options:dict documentAttributes:NULL error:outError];

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  • Subsonic 3, MySql, won't update record.

    - by Warspawn
    [WebMethod] public string GetAuthToken(string username, string password) { var db = new LogicDB(); //var results = from u in db.Users // where u.Username == username && u.Password == password // select u; User u = db.Select .From<User>() .Where(UsersTable.UsernameColumn).IsEqualTo(username) .And(UsersTable.PasswordColumn).IsEqualTo(password) .ExecuteSingle<User>(); if (u == null) { return "{'success': false, 'reason': 'Invalid username and/or password.'}"; } else { // really there should only be one match... Guid code = Guid.NewGuid(); u.Securitycode = code.ToString(); u.Securityexp = System.DateTime.Now.AddHours(24); //u.Save(db.Provider); return "{'id':'" + u.Id.ToString() + "', 'code':'" + code.ToString() + "', 'exp':'" + u.Securityexp.ToString() + "'}" + "\n\n<br/><br/>" + u.GetDirtyColumns().ToArray().ToString(); } } When I run that, I keep getting: System.Collections.Generic.KeyNotFoundException: The given key was not present in the dictionary. This is when u.Save(db.Provider); is uncommented. And happens even with just u.Save(); or using the linq query above results instead.

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  • Brute force characters into a textbox in c#

    - by Fred Dunly
    Hey everyone, I am VERY new to programming and the only language I know is C# So I will have to stick with that... I want to make a program that "test passwords" to see how long they would take to break with a basic brute force attack. So what I did was make 2 text boxes. (textbox1 and textbox2) and wrote the program so if the text boxes had the input, a "correct password" label would appear, but i want to write the program so that textbox2 will run a brute force algorithm in it, and when it comes across the correct password, it will stop. I REALLY need help, and if you could just post my attached code with the correct additives in it that would be great. The program so far is extremely simple, but I am very new to this, so. Thanks in advance. private void textBox2_TextChanged(object sender, EventArgs e) { } private void button1_Click(object sender, EventArgs e) { if (textBox2.Text == textBox1.Text) { label1.Text = "Password Correct"; } else { label1.Text = "Password Wrong"; } } private void label1_Click(object sender, EventArgs e) { } } } `

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