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  • ssh login - execute command - exit

    - by renton
    hi folks, is there a way to just execute a command (or script) on a user (ssh) login and then exit? some kind of replacing the default shell with a custom script. i want a user only be able to restart a service, but not to have regular shell-access thanks

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  • Run script on login with ssh

    - by user912447
    I have a feeling this is quite easy to do but every solution found on google has to do with adding a script to be run whenever someone logs into the machine. What I am looking for is a way to run a script when only I log into the machine. I ssh into a shared computer and need to have it load a couple modules for me and I imagined the easiest way to do this would to just run a script on login. Is there a simple way?

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  • How do I get my Intel HD graphics to work alongside my HD7850, as my second(HDMI out) monitor?

    - by AlexTes
    Title says it all. Further info: Motherboard: http://www.asrock.com/mb/Intel/Z77%20Pro3/ Processor: http://ark.intel.com/products/65520/Intel-Core-i5-3570K-Processor-%286M-Cache-up-to-3_80-GHz%29 So currently my main screen is running on my HD7850. Got drivers from the amd website. I have looked through dozens of questions here. I'm about to try booting Ubuntu from a stick and seeing if the xorg-edgers drivers might help. When booting, all action goes down on the very screen I'm trying to get to work.*EDIT never mind this. Seems to be special boot magic. As the screen only displays whiteline errors once the gui of ubuntu has kicked in and everything graphic is happening through my graphics card again. Connected through HDMI(motherboard)-DVI. So unless having multiple displays is a huge deal the solution hopefully isn't that complicated. I just feel I'm missing something simple. If this really is complicated, I should probably just hook up the display to my graphics card. My CPU is usually the one chilling out though so I'd like to try to get that to work. Also just because I don't want to buy an extra cable and this set up makes me feel warm and fuzzy inside. Tell me what to try or look up, I'll be most appreciative. Thank you! **UPDATE The x-swat ppa installed some intel stuff. Booting with one monitor plugged into the motherboard gives nothing. Doing it with the pc already on gives the purple "Ubuntu" with 5 dots boot/shutdown screen.

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  • How to specify Multiple Secure Webpages with .htaccess RewriteCond

    - by Patrick Ndille
    I have 3 pages that I want to make secure on my website using .htaccess -login.php -checkout.php -account.php I know how to make just one work page at a time using .htaccess RewriteEngine On RewriteCond %{HTTPS} off RewriteCond %{REQUEST_URI} /login.php RewriteRule (.*) https://%{HTTP_HOST}%{REQUEST_URI} [L] I and trying to figure out how to include the other 2 specific pages to make them also secure and used the expression below but it didn't work RewriteEngine On RewriteCond %{HTTPS} off RewriteCond %{REQUEST_URI} /login.php RewriteCond %{REQUEST_URI} /checkout.php RewriteCond %{REQUEST_URI} /account.php RewriteRule (.*) https://%{HTTP_HOST}%{REQUEST_URI} [L] Can someone help me the right expression that will work with multiple pages? The second part of the code is that, if https is already on and a user move to a page that Is not any of the pages i specified about, I want that it should get back to http. how should I write the statement for it to redirect back to http if its not any of the pages above? I have my statement like this but its not working RewriteCond %{HTTPS} on RewriteRule !(checkout|login|account|payment)\.php http://%{HTTP_HOST}%{REQUEST_URI} [L,R] Any thoughts?

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  • SQL RDBMS : one query or multiple calls

    - by None None
    After looking around the internet, I decided to create DAOs that returned objects (POJOs) to the calling business logic function/method. For example: a Customer object with a Address reference would be split in the RDBMS into two tables; Customer and ADDRESS. The CustomerDAO would be in charge of joining the data from the two tables and create both an Address POJO and Customer POJO adding the address to the customer object. Finally return the fulll Customer POJO. Simple, however, now i am at a point where i need to join three or four tables and each representing an attribute or list of attributes for the resulting POJO. The sql will include a group by but i will still result with multiple rows for the same pojo, because some of the tables are joining a one to many relationship. My app code will now have to loop through all the rows trying to figure out if the rows are the same with different attributes or if the record should be a new POJO. Should I continue to create my daos using this technique or break up my Pojo creation into multiple db calls to make the code easier to understand and maintain?

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  • How do I move the login stuff from Wordpress (login, register etc.) to its own subdomain [migrated]

    - by surferconor425
    Title says it all, hope this is possible as I want to start a network of sites for different stuff but want to use one account system with that subdomain using ssl with extended verification, thanks. EDIT: Ok, it has been closed because I am not being clear enough so I will narrow it down a bit more. I am wondering how to move all registration files and login files into a sub-domain instead of just the normal domain so I can put an SSL on it to make it more secure but leave the rest of the blog on the normal domain.

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  • Login Facebook using Web-Harvest

    - by parin
    I tried to login Facebook using Web-Harvest. I used the following xml code to login < ?xml version="1.0" encoding="UTF-8"? < config charset="ISO-8859-1" < file action="write" path="homepage.xml" charset="UTF-8" < html-to-xml < http method="post" url="http://www.facebook.com/login.php" cookie-policy="browser" < http-param name="email"myemail < http-param name="pass"mypassword < /http < /html-to-xml < /file The homepage.xml (output) file contains the xml code for the login page of facebook along with the following lines: < h2 class="main_message" id="standard_error"Cookies Required< /h2< p class="sub_message" id="standard_explanation"Cookies are not enabled on your browser. Please adjust this in your security preferences before continuing.< /p I tried all the allowed values for cookie-policy in the http processor of the xml code but was unsuccessful.

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  • HttpServletRequest#login() not working in Java.

    - by Nitesh Panchal
    Hello, j_security_check just doesn't seem enough for me to perform login process. So, instead of submitting the form to j_security_check i created my own servlet and in that i am programmatically trying to do login. This works but i am not able to redirect to my restricted resource. Can anybody tell me what can be the problem? This is processRequest method of my servlet :- protected void processRequest(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { response.setContentType("text/html;charset=UTF-8"); PrintWriter out = response.getWriter(); try { String strUsername = request.getParameter("txtusername"); String strPassword = request.getParameter("txtpassword"); if(strUsername == null || strPassword == null || strUsername.equals("") || strPassword.equals("")) throw new Exception("Username and/or password missing."); request.login(strUsername, strPassword); System.out.println("Login succeeded!!"); if(request.isUserInRole(ROLES.ADMIN.getValue())){//enum System.out.println("Found in Admin Role"); response.sendRedirect("/Admin/home.jsf"); } else if (request.isUserInRole(ROLES.GENERAL.getValue())) response.sendRedirect("/Common/index.jsf"); else //guard throw new Exception("No role for user " + request.getRemoteUser()); }catch(Exception ex){ //patch work why there needs to be blogger here? System.out.println("Invalid username and/or password!!"); response.sendRedirect("/Common/index.jsf"); }finally { out.close(); } } Everything works fine and i can even see message "Found in Admin Role" but problem is even after authenticating i am not able to redirect my request to some other page. Please help geeks.

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  • Asp.net Login Status Question: It Aint Working

    - by contactmatt
    I'm starting to use Role Management in my website, and I'm current following along on the tutorial from http://www.asp.net/Learn/Security/tutorial-02-vb.aspx . I'm having a problem with the asp:LoginStatus control. It is not telling me that I am currently logged in after a successful login. This can't be true because after successfully logging in, my LoggedInTemplate is shown. The username and passwords are simply stored in a array. Heres the Login.aspx page code. Protected Sub btnLogin_Click(ByVal sender As Object, ByVal e As System.EventArgs) _ Handles btnLogin.Click ' Three valid username/password pairs: Scott/password, Jisun/password, and Sam/password. Dim users() As String = {"Scott", "Jisun", "Sam"} Dim passwords() As String = {"password", "password", "password"} For i As Integer = 0 To users.Length - 1 Dim validUsername As Boolean = (String.Compare(txtUserName.Text, users(i), True) = 0) Dim validPassword As Boolean = (String.Compare(txtPassword.Text, passwords(i), False) = 0) If validUsername AndAlso validPassword Then FormsAuthentication.RedirectFromLoginPage(txtUserName.Text, chkRemember.Checked) End If Next ' If we reach here, the user's credentials were invalid lblInvalid.Visible = True End Sub Here is the content place holder on the master page specifically designed to hold Login Information. On successfull login, the page is redirected to '/Default.aspx', and the LoggedIn Template below is shown...but the status says Log In. <asp:ContentPlaceHolder Id="LoginContent" runat="server"> <asp:LoginView ID="LoginView1" runat="server"> <LoggedInTemplate> Welcome back, <asp:LoginName ID="LoginName1" runat="server" />. </LoggedInTemplate> <AnonymousTemplate> Hello, stranger. </AnonymousTemplate> </asp:LoginView> <br /> <asp:LoginStatus ID="LoginStatus1" runat="server" LogoutAction="Redirect" LogoutPageUrl="~/Logout.aspx" /> </asp:ContentPlaceHolder> Forms authentication is enabled. I'm not sure what to do about this :o.

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  • django: cannot import settings, cannot login to admin, cannot change admin password

    - by xpanta
    Hi, It seems that I am completely lost here. Yesterday I noticed that I cannot login to the admin panel (don't use it much, so it's been some weeks since last login). I thought that I might have changed the admin password and now I can't remember it (though I doubt it). I tried django-admin.py changepassword (using django 1.2.1) but it said that 'changepassword' is unknown command (I have all the necessary imports in my settings.py. Admin interface used to work ok). Then I gave a django-admin.py validate. Then the hell begun. django-admin.py validate gave me this error: Error: Settings cannot be imported, because environment variable DJANGO_SETTINGS_MODULE is undefined. I then gave a set DJANGO_SETTINGS_MODULE=myproject.settings and then again a django-admin.py validate This is what I get now: Error: Could not import settings 'myproject.settings' (Is it on sys.path? Does it have syntax errors?): No module named myproject.settings and now I am lost. I tried django console and sys.path.append('c:\workspace') or sys.append('c:\workspace\myproject') but still get the same errors. I use windows 7 and my project dir is c:\workspace. I don't use a PYTHONPATH variable (although I tried setting it temporarily to C:\workspace but I still get the same error). I don't use Apache, just the django development server. What am I doing wrong? My web page works fine. I think that the fact that I can't login as admin is related to the previous import error, no? PS: I also tried this: http://coderseye.com/2007/howto-reset-the-admin-password-in-django.html but still I couldn't change admin password for some reason. Although I could create another admin user (with which I couldn't login).

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  • Have to login twice. PHP sessions and login troubles with Chrome and Opera.

    - by Robert
    The problem I am encountering is that for my login form I have to login twice for the session to register properly, but only in Chrome (my version is 4.0.249.89) and Opera (my version is 10.10). Here is the stripped down code that I am testing on: Login Page: session_start(); $_SESSION['user_id'] = 8; $_SESSION['user_name'] = 'Jim'; session_write_close(); header('Location: http://www.my-domain-name.com/'); exit(); Home Page: session_start(); if ( isset($_SESSION['user_id']) ) { echo "You are logged in!"; } else { echo "You are NOT logged in!"; } Logout Page: session_start(); session_unset(); session_destroy(); header('Location: http://www.my-domain-name.com/'); exit(); Currently, under a fresh load with no cookies, if I go to my-domain-name.com/login/ it will redirect to the home page and say "You are NOT logged in!" but if I go there again it will say "You are logged in!". Any ideas? Thanks for your help.

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  • php curl login not work

    - by Massimo Zampieri
    Hi i have a problem with the curl. I watched an old post Remote Login not Working With Curl, but it not work. I followed baba's advice but the code enter in the if statement. Sorry for my bad english. Can anyone help me? This is the code: $url="http://hipfile.com/"; $urllog="http://hipfile.com/login.html"; $postdata = "login=bnnoor&password=########&op=login"; $ch = curl_init(); curl_setopt ($ch, CURLOPT_URL, $url); curl_setopt ($ch, CURLOPT_SSL_VERIFYPEER, FALSE); curl_setopt ($ch, CURLOPT_USERAGENT, "Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.6) Gecko/20070725 Firefox/2.0.0.6"); curl_setopt ($ch, CURLOPT_TIMEOUT, 60); curl_setopt ($ch, CURLOPT_FOLLOWLOCATION, 1); curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt ($ch, CURLOPT_REFERER, $urllog); curl_setopt ($ch, CURLOPT_POSTFIELDS, $postdata); curl_setopt ($ch, CURLOPT_POST, 1); $result = curl_exec ($ch); if (!$result) { $http_code = curl_getinfo($ch, CURLINFO_HTTP_CODE); curl_close($ch); // make sure we closeany current curl sessions die($http_code.' Unable to connect to server. Please come back later.'); } echo $result; curl_close($ch);

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  • Cannot open database "DataDir" requested by the login. The login failed

    - by Turkan Caglar
    Hi, I have a single user software that uses SQL Server database 2005. My computer was crushed while my software was on. I saved a copy of my database after I rescued my computer. However my software gives my following message. ( I don't know anything about SQL) However, I guess my database has the last login information. Since software was not shot down properly, it still thinks that I am logged in doesn't allow me to login again. How can I solve the problem? Can you help me? Or do you know any source that I can get help? Cannot open database "DataDir" requested by the login. The login failed ComputerName = GURELS User ID=sa;Initial Catalog=DataDir;Data Source=GURELS\CSS;Application Name=ChefTec User ID=sa;Initial Catalog=CTDir;Data Source=GURELS\CSS;Use Procedure for Prepare=1;Auto Translate=True;Packet Size=4096;Application Name=ChefTec;Workstation ID=GURELS;Use Encryption for Data=False;Tag with column collation when possible=False Thank you very much Turkan e-mail" [email protected] phone:617 259 6644

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  • YQL + PHP : how to make a facebook login

    - by Jonathan
    Hi! I was reading some stuff about the YQL api that Yahoo! has provided, I am not sure, but it appears to be a collection of lots of third party api into one common language, right? what I don't get is how to make the facebook login through it so I can get the user profile data... My project is to add a facebook(and other social networks) form login, because the website won't have his own login, people will have to use a social network to link in. Then I thought the YQL would help me out with this task so I wouldn't have to develop lots of functions to each one of the networks. Reading this http://developer.yahoo.com/yql/guide/yql-code-examples.html#sdk_yql, I understood how to make a Yahoo login so I can access some private data, but couldn't find how I could do it with facebook and others So my question... Can YQL help me with this? Can you give me a simple example of a facebook session using it within PHP? Are there alternatives to aid me in this task? thanks, Jonathan

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  • Xcode: Display Login View in applicationDidBecomeActive

    - by Patrick
    In my app I would like to show a login screen - which will be displayed when the app starts and when the app becomes active. For reference, I am using storyboards, ARC and it is a tabbed bar application. First off, I have this method which returns the topViewController. - (UIViewController *)topViewController:(UIViewController *)rootViewController { if (rootViewController.presentedViewController == nil) { return rootViewController; } if ([rootViewController.presentedViewController isMemberOfClass:[UINavigationController class]]) { UINavigationController *navigationController = (UINavigationController *)rootViewController.presentedViewController; UIViewController *lastViewController = [[navigationController viewControllers] lastObject]; return [self topViewController:lastViewController]; } UIViewController *presentedViewController = (UIViewController *)rootViewController.presentedViewController; return [self topViewController:presentedViewController]; } And I call this method here: - (void)applicationDidBecomeActive:(UIApplication *)application { if ( ... ) { // if the user needs to login PasswordViewController *passwordView = [[PasswordViewController alloc] init]; UIViewController *myView = [self topViewController:self.window.rootViewController]; [myView presentModalViewController:passwordView animated:NO]; } } To an extent this does work - I can call a method in viewDidAppear which shows an alert view to allow the user to log in. However, this is undesirable and I would like to have a login text box and other ui elements. If I do not call my login method, nothing happens and the screen stays black, even though I have put a label and other elements on the view. Does anyone know a way to resolve this? My passcode view is embedded in a Navigation Controller, but is detached from the main storyboard.

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  • Codeigniter Session Data not available in other pages after login

    - by jswat
    So, I have set up a login page that verifies the user's credentials, and then sets codeigniter session data 'email' and 'is_logged_in' and a few other items. The first page after the login, the data is accessible. After that page, I can no longer access the session data. In fact, if I try reloading that first page, the session data is gone. I have tried storing it in the database, storing it unencrypted (bad idea I know, but it was for troubleshooting), and storing it encrypted. I have autoloaded the session library in config.php. Here's an example of the code I'm using to set the session data: $data = array( 'email' => $this->input->post('username'), 'is_logged_in' => true ); $this->session->set_userdata($data); And to retrieve it, I'm using : $this->session->userdata('email'); Or $this->session->userdata('is_logged_in'); I've done lots of work with PHP and cookies, and sessions before, but this is my first project with Codeigniter and I'm perplexed. Could it have something to do with directory issues? I have the login page and process controlled by a 'login' controller, and then it redirects to a 'site' controller. Thanks for your help, and please let me know if I need to clarify anything.

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  • Ruby on Rails user login form in main layout

    - by Jimmy
    Hey guys I have a simple ror application for some demo stuff. I am running into a problem with trying to move my login form from the users controller and just have it displayed in the main navigation so that a user can easily log in from anywhere. The problem is the form doesn't generate the correct action for the html form. Ruby code: <% form_for(url_for(:action => 'login'), :method => 'post') do |f| %> <li><%= f.text_field("username") %></li> <li><%= f.password_field("password") %></li> <li><%= submit_tag("Login")%></li> <% end %> The problem is depending on the controller I am currently in this generates HTML actions like <form action="/home" method="post">...</form> when it should be generating HTML like so <form action="/login" method="post">...</form> I know I could simply do an HTML form here but I want to keep things as easy to maintain as possible. Any help?

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  • Login to website and use cookie to get source for another page

    - by Stu
    I am trying to login to the TV Rage website and get the source code of the My Shows page. I am successfully logging in (I have checked the response from my post request) but then when I try to perform a get request on the My Shows page, I am re-directed to the login page. This is the code I am using to login: private string LoginToTvRage() { string loginUrl = "http://www.tvrage.com/login.php"; string formParams = string.Format("login_name={0}&login_pass={1}", "xxx", "xxxx"); string cookieHeader; WebRequest req = WebRequest.Create(loginUrl); req.ContentType = "application/x-www-form-urlencoded"; req.Method = "POST"; byte[] bytes = Encoding.ASCII.GetBytes(formParams); req.ContentLength = bytes.Length; using (Stream os = req.GetRequestStream()) { os.Write(bytes, 0, bytes.Length); } WebResponse resp = req.GetResponse(); cookieHeader = resp.Headers["Set-cookie"]; String responseStream; using (StreamReader sr = new StreamReader(resp.GetResponseStream())) { responseStream = sr.ReadToEnd(); } return cookieHeader; } I then pass the cookieHeader into this method which should be getting the source of the My Shows page: private string GetSourceForMyShowsPage(string cookieHeader) { string pageSource; string getUrl = "http://www.tvrage.com/mytvrage.php?page=myshows"; WebRequest getRequest = WebRequest.Create(getUrl); getRequest.Headers.Add("Cookie", cookieHeader); WebResponse getResponse = getRequest.GetResponse(); using (StreamReader sr = new StreamReader(getResponse.GetResponseStream())) { pageSource = sr.ReadToEnd(); } return pageSource; } I have been using this previous question as a guide but I'm at a loss as to why my code isn't working.

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  • how to programatically compare permissions of login/user in sql server 2005

    - by titanium
    There's a login/user in SQL Server who is having a problem importing accounts in production server. I don't have an idea what method he is doing this. According to the one importing, this import is working fine in development server. But when he did the same import in production it is giving him errors. Below are the errors he is getting for each accounts. 2009-06-05 18:01:05.8254 ERROR [engine-1038] Task [1038:00001 - Members]: Step 1.0 [<Insert step description>]: Task.RunStep(): StoreRow has failed 2009-06-05 18:01:05.9035 ERROR [engine-1038] Task [1038:00001 - Members]: Step 1.0 [<Insert step description>]: Task.RunStep(): StoreRow exception: Exception caught while storing Data. [Microsoft][ODBC SQL Server Driver][SQL Server]'ACCOUNT1' is not a valid login or you do not have permission. Please note that 'ACCOUNT1' is not the real account name. I just changed it for security reason. Using SQL Server Management Studio (SSMS), I viewed/checked the permissions of the user/login who is performing the import from development server and production for comparison. I found no difference. My question is: Is there a way to programmatically query permissions in server and database level of a particular login/user so I can compare/contrast for any differences?

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  • PHP login, getting wrong count value from query / fetch array

    - by Chris
    Hello, *EDIT*Thanks to the comments below it has been figured out that the problem lies with the md5, without everything works as it should. But how do i implent the md5 then? I am having some troubles with the following code below to login. The database and register system are already working. The problem lies that it does not find any result at all in the query. IF the count is 0 it should redirect the user to a secured page. But this only works if i write count = 0, but this should be 0 , only if the user name and password is found he should be directed to the secure (startpage) of the site after login. For example root (username) root (password) already exists but i cannot seem to properly login with it. <?php session_start(); if (!empty($_POST["send"])) { $username = ($_POST["username"]); $password = (md5($_POST["password"])); $count = 0; $con = mysql_connect("localhost" , "root", ""); mysql_select_db("testdb", $con); $result = mysql_query("SELECT name, password FROM user WHERE name = '".$username."' AND password = '".$password."' ") or die("Error select statement"); $count = mysql_num_rows($result); if($count > 0) // always goes the to else, only works with >=0 but then the data is not found in the database, hence incorrect { $row = mysql_fetch_array($result); $_SESSION["username"] = $row["name"]; header("Location: StartPage.php"); } else { echo "Wrong login data, please try again"; } mysql_close($con); } ?>

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  • Importing Multiple Schemas to a Model in Oracle SQL Developer Data Modeler

    - by thatjeffsmith
    Your physical data model might stretch across multiple Oracle schemas. Or maybe you just want a single diagram containing tables, views, etc. spanning more than a single user in the database. The process for importing a data dictionary is the same, regardless if you want to suck in objects from one schema, or many schemas. Let’s take a quick look at how to get started with a data dictionary import. I’m using Oracle SQL Developer in this example. The process is nearly identical in Oracle SQL Developer Data Modeler – the only difference being you’ll use the ‘File’ menu to get started versus the ‘File – Data Modeler’ menu in SQL Developer. Remember, the functionality is exactly the same whether you use SQL Developer or SQL Developer Data Modeler when it comes to the data modeling features – you’ll just have a cleaner user interface in SQL Developer Data Modeler. Importing a Data Dictionary to a Model You’ll want to open or create your model first. You can import objects to an existing or new model. The easiest way to get started is to simply open the ‘Browser’ under the View menu. The Browser allows you to navigate your open designs/models You’ll see an ‘Untitled_1′ model by default. I’ve renamed mine to ‘hr_sh_scott_demo.’ Now go back to the File menu, and expand the ‘Data Modeler’ section, and select ‘Import – Data Dictionary.’ This is a fancy way of saying, ‘suck objects out of the database into my model’ Connect! If you haven’t already defined a connection to the database you want to reverse engineer, you’ll need to do that now. I’m going to assume you already have that connection – so select it, and hit the ‘Next’ button. Select the Schema(s) to be imported Select one or more schemas you want to import The schemas selected on this page of the wizard will dictate the lists of tables, views, synonyms, and everything else you can choose from in the next wizard step to import. For brevity, I have selected ALL tables, views, and synonyms from 3 different schemas: HR SCOTT SH Once I hit the ‘Finish’ button in the wizard, SQL Developer will interrogate the database and add the objects to our model. The Big Model and the 3 Little Models I can now see ALL of the objects I just imported in the ‘hr_sh_scott_demo’ relational model in my design tree, and in my relational diagram. Quick Tip: Oracle SQL Developer calls what most folks think of as a ‘Physical Model’ the ‘Relational Model.’ Same difference, mostly. In SQL Developer, a Physical model allows you to define partitioning schemes, advanced storage parameters, and add your PL/SQL code. You can have multiple physical models per relational models. For example I might have a 4 Node RAC in Production that uses partitioning, but in test/dev, only have a single instance with no partitioning. I can have models for both of those physical implementations. The list of tables in my relational model Wouldn’t it be nice if I could segregate the objects based on their schema? Good news, you can! And it’s done by default Several of you might already know where I’m going with this – SUBVIEWS. You can easily create a ‘SubView’ by selecting one or more objects in your model or diagram and add them to a new SubView. SubViews are just mini-models. They contain a subset of objects from the main model. This is very handy when you want to break your model into smaller, more digestible parts. The model information is identical across the model and subviews, so you don’t have to worry about making a change in one place and not having it propagate across your design. SubViews can be used as filters when you create reports and exports as well. So instead of generating a PDF for everything, just show me what’s in my ‘ABC’ subview. But, I don’t want to do any work! Remember, I’m really lazy. More good news – it’s already done by default! The schemas are automatically used to create default SubViews Auto-Navigate to the Object in the Diagram In the subview tree node, right-click on the object you want to navigate to. You can ask to be taken to the main model view or to the SubView location. If you haven’t already opened the SubView in the diagram, it will be automatically opened for you. The SubView diagram only contains the objects from that SubView Your SubView might still be pretty big, many dozens of objects, so don’t forget about the ‘Navigator‘ either! In summary, use the ‘Import’ feature to add existing database objects to your model. If you import from multiple schemas, take advantage of the default schema based SubViews to help you manage your models! Sometimes less is more!

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