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• Elmah on a WebServer

  - by Subbarao

  I Setup Elmah to work on a website. It works fine on my local machine but when I moved it to a web server I get this exception Configuration Error Description: An error occurred during the processing of a configuration file required to service this request. Please review the specific error details below and modify your configuration file appropriately. Parser Error Message: Unrecognized configuration section elmah/security. Source Error: Line 110: </connectionStrings> Line 111: <elmah> Line 112: <security allowRemoteAccess="1" /> Line 113: <errorLog type="Elmah.SqlErrorLog, Elmah" connectionStringName="CadaretGrantConnectionString"/> Line 114: <!-- Don't log 404 --> It shows me a error at Line 112. What should be done in order to get Elmah to work with remote access? Below is my configuration <elmah> <security allowRemoteAccess="1" /> <errorLog type="Elmah.SqlErrorLog, Elmah" connectionStringName="ConnectionString"/> <!-- Don't log 404 --> <errorFilter> <test> <equal binding="HttpStatusCode" value="404" valueType="Int32"/> </test> </errorFilter> </elmah>

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• Adapting Machine Learning Algorithms to my Problem

  - by Berkay

  i'm working on a project and need your ideas, advices. First of all, let me tell my problem. There is power button and some other keys of a machine and there is only one user has authentication to use this machine.There are no other authentication methods, the machine is in public area in a company. the machine is working with the combination of pressing power button and some other keys. The order of pressing keys is secret but we don't trust it, anybody can learn the password and can access the machine. i have the capability of managing the key hold time and also some other metrics to measure the time differences between the key such as horizantal or vertical key press times (differences). and also i can measure the hold time etc. These all means i have some inputs, Now i'm trying to get a user profile by analysing these inputs. My idea is to get the authenticated user to press the password n times and create a threshold or something similar to that. This method also can be said BIOMETRICS, anyone else who knows the machine button combination, can try the password but if he is out of this range can not get access it. How can i adapt these into my algorithms? where should i start ? i don't want to delve deep into machine learning, and also i can see the in my first try i can get false positive and false negative values really high, but i can manage it by changing my inputs. thanks.

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• Check for modification failure in content Integration using VisualSVN Server and Cruisecontrol.net

  - by harun123

  I am using CruiseControl.net for continous integration. I've created a repository for my project using VisualSvn server (uses Windows Authentication). Both the servers are hosted in the same system (Os-Microsoft Windows Server 2003 sp2). When i force build the project using CruiseControl.net "Failed task(s): Svn: CheckForModifications" is shown as the message. When i checked the build report, it says as follows: BUILD EXCEPTION Error Message: ThoughtWorks.CruiseControl.Core.CruiseControlException: Source control operation failed: svn: OPTIONS of 'https://sp-ci.sbsnetwork.local:8443/svn/IntranetPortal/Source': **Server certificate verification failed: issuer is not trusted** (https://sp-ci.sbsnetwork.local:8443). Process command: C:\Program Files\VisualSVN Server\bin\svn.exe log **sameUrlAbove** -r "{2010-04-29T08:35:26Z}:{2010-04-29T09:04:02Z}" --verbose --xml --username ccnetadmin --password cruise --non-interactive --no-auth-cache at ThoughtWorks.CruiseControl.Core.Sourcecontrol.ProcessSourceControl.Execute(ProcessInfo processInfo) at ThoughtWorks.CruiseControl.Core.Sourcecontrol.Svn.GetModifications (IIntegrationResult from, IIntegrationResult to) at ThoughtWorks.CruiseControl.Core.Sourcecontrol.QuietPeriod.GetModifications(ISourceControl sourceControl, IIntegrationResult lastBuild, IIntegrationResult thisBuild) at ThoughtWorks.CruiseControl.Core.IntegrationRunner.GetModifications(IIntegrationResult from, IIntegrationResult to) at ThoughtWorks.CruiseControl.Core.IntegrationRunner.Integrate(IntegrationRequest request) My SourceControl node in the ccnet.config is as shown below: <sourcecontrol type="svn"> <executable>C:\Program Files\VisualSVN Server\bin\svn.exe</executable> <trunkUrl> check out url </trunkUrl> <workingDirectory> C:\ProjectWorkingDirectories\IntranetPortal\Source </workingDirectory> <username> ccnetadmin </username> <password> cruise </password> </sourcecontrol> Can any one suggest how to avoid this error?

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• check status application pool iis7 with csharp (access-denied)

  - by jack

  I need to monitor the status of an application in the applications pool of IIS 7 from an other machine on the same domain. My monitoring application must be in C# and running as a Windows service. On my server, I create a user with administration rights and I execute the command aspnet_regiis -ga machine\username wich worked succesfully. My problem is when I try to access the application pool i still get COMExcepttion "Access denied". What did i do wrong or wich step did i miss? I used code from http://patelshailesh.com/index.php/create-a-website-application-pool-programmatically-using-csharp as example. int status = 0; string ipAddress = "10.20.2.13"; string username = "username"; string password = "password"; try { DirectoryEntry de = new DirectoryEntry(string.Format("IIS://{0}/W3SVC/AppPools/MyAppPoolName", ipAddress), username, password); //the exception is thron here. status = (int)de.InvokeGet("AppPoolState"); switch (status) { case 2: //Runnig break; case 4: //Stopped break; default: break; } } catch (Exception ex) { }

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• Django - Override admin site's login form

  - by TrojanCentaur

  I'm currently trying to override the default form used in Django 1.4 when logging in to the admin site (my site uses an additional 'token' field required for users who opt in to Two Factor Authentication, and is mandatory for site staff). Django's default form does not support what I need. Currently, I've got a file in my templates/ directory called templates/admin/login.html, which seems to be correctly overriding the template used with the one I use throughout the rest of my site. The contents of the file are simply as below: # admin/login.html: {% extends "login.html" %} The actual login form is as below: # login.html: {% load url from future %}<!DOCTYPE html> <html> <head> <title>Please log in</title> </head> <body> <div id="loginform"> <form method="post" action="{% url 'id.views.auth' %}"> {% csrf_token %} <input type="hidden" name="next" value="{{ next }}" /> {{ form.username.label_tag }}<br/> {{ form.username }}<br/> {{ form.password.label_tag }}<br/> {{ form.password }}<br/> {{ form.token.label_tag }}<br/> {{ form.token }}<br/> <input type="submit" value="Log In" /> </form> </div> </body> </html> My issue is that the form provided works perfectly fine when accessed using my normal login URLs because I supply my own AuthenticationForm as the form to display, but through the Django Admin login route, Django likes to supply it's own form to this template and thus only the username and password fields render. Is there any way I can make this work, or is this something I am just better off 'hard coding' the HTML fields into the form for?

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• Learning AES: the KeyBytes

  - by Tom Brito

  I got the following example from here: import java.security.Security; import javax.crypto.Cipher; import javax.crypto.spec.SecretKeySpec; public class MainClass { public static void main(String[] args) throws Exception { Security.addProvider(new org.bouncycastle.jce.provider.BouncyCastleProvider()); byte[] input = "www.java2s.com".getBytes(); byte[] keyBytes = new byte[] { 0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, 0x08, 0x09, 0x0a, 0x0b, 0x0c, 0x0d, 0x0e, 0x0f, 0x10, 0x11, 0x12, 0x13, 0x14, 0x15, 0x16, 0x17 }; SecretKeySpec key = new SecretKeySpec(keyBytes, "AES"); Cipher cipher = Cipher.getInstance("AES/ECB/PKCS7Padding", "BC"); System.out.println(new String(input)); // encryption pass cipher.init(Cipher.ENCRYPT_MODE, key); byte[] cipherText = new byte[cipher.getOutputSize(input.length)]; int ctLength = cipher.update(input, 0, input.length, cipherText, 0); ctLength += cipher.doFinal(cipherText, ctLength); System.out.println(new String(cipherText)); System.out.println(ctLength); // decryption pass cipher.init(Cipher.DECRYPT_MODE, key); byte[] plainText = new byte[cipher.getOutputSize(ctLength)]; int ptLength = cipher.update(cipherText, 0, ctLength, plainText, 0); ptLength += cipher.doFinal(plainText, ptLength); System.out.println(new String(plainText)); System.out.println(ptLength); } } I imagine that the byte[] keyBytes should be random generated, so I gone to test the max size before do it. When adding one more byte 0x18 to the array, the exception raised: InvalidKeyException: Key length not 128/192/256 bits. But the original 18 bytes (from 0 to 17) are not multiple of nither 128, 192 or 256. I would like to understand the math here.. can anyone explain me? Thanks!

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• Executing bat file and returning the prompt

  - by Lieven Cardoen

  I have a problem with cruisecontrol where an ant scripts executes a bat file that doesn't give me the prompt back. As a result, the project in cruisecontrol keeps on bulding forever until I restart cruisecontrol. How can I resolve this? It's a startup.bat from wowza (Streaming Server) that I'm executing: @echo off call setenv.bat if not %WMSENVOK% == "true" goto end set _WINDOWNAME="Wowza Media Server 2" set _EXESERVER= if "%1"=="newwindow" ( set _EXESERVER=start %_WINDOWNAME% shift ) set CLASSPATH="%WMSAPP_HOME%\bin\wms-bootstrap.jar" rem cacls jmxremote.password /P username:R rem cacls jmxremote.access /P username:R rem NOTE: Here you can configure the JVM's built in JMX interface. rem See the "Server Management Console and Monitoring" chapter rem of the "User's Guide" for more information on how to configure the rem remote JMX interface in the [install-dir]/conf/Server.xml file. set JMXOPTIONS=-Dcom.sun.management.jmxremote=true rem set JMXOPTIONS=%JMXOPTIONS% -Djava.rmi.server.hostname=192.168.1.7 rem set JMXOPTIONS=%JMXOPTIONS% -Dcom.sun.management.jmxremote.port=1099 rem set JMXOPTIONS=%JMXOPTIONS% -Dcom.sun.management.jmxremote.authenticate=false rem set JMXOPTIONS=%JMXOPTIONS% -Dcom.sun.management.jmxremote.ssl=false rem set JMXOPTIONS=%JMXOPTIONS% -Dcom.sun.management.jmxremote.password.file= "%WMSCONFIG_HOME%/conf/jmxremote.password" rem set JMXOPTIONS=%JMXOPTIONS% -Dcom.sun.management.jmxremote.access.file= "%WMSCONFIG_HOME%/conf/jmxremote.access" rem log interceptor com.wowza.wms.logging.LogNotify - see Javadocs for ILogNotify %_EXESERVER% "%_EXECJAVA%" %JAVA_OPTS% %JMXOPTIONS% -Dcom.wowza.wms.AppHome="%WMSAPP_HOME%" -Dcom.wowza.wms.ConfigURL="%WMSCONFIG_URL%" -Dcom.wowza.wms.ConfigHome="%WMSCONFIG_HOME%" -cp %CLASSPATH% com.wowza.wms.bootstrap.Bootstrap start :end

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• Hibernate one-to-one mapping

  - by Andrey Yaskulsky

  I have one-to-one hibernate mapping between class Student and class Points: @Entity @Table(name = "Users") public class Student implements IUser { @Id @Column(name = "id") private int id; @Column(name = "name") private String name; @Column(name = "password") private String password; @OneToOne(fetch = FetchType.EAGER, mappedBy = "student") private Points points; @Column(name = "type") private int type = getType(); //gets and sets... @Entity @Table(name = "Points") public class Points { @GenericGenerator(name = "generator", strategy = "foreign", parameters = @Parameter(name = "property", value = "student")) @Id @GeneratedValue(generator = "generator") @Column(name = "id", unique = true, nullable = false) private int Id; @OneToOne @PrimaryKeyJoinColumn private Student student; //gets and sets And then i do: Student student = new Student(); student.setId(1); student.setName("Andrew"); student.setPassword("Password"); Points points = new Points(); points.setPoints(0.99); student.setPoints(points); points.setStudent(student); Session session = HibernateUtil.getSessionFactory().getCurrentSession(); session.beginTransaction(); session.save(student); session.getTransaction().commit(); And hibernate saves student in the table but not saves corresponding points. Is it OK? Should i save points separately?

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• .post inside jQuery.validator.addMethod always returns false :(

  - by abdullah kahraman

  Hello! I am very new to jQuery and javascript programming. I have a program below that checks whether username is taken or not. For now, the PHP script always returns if(isset($_POST["username"]) )//&& isset($_POST["checking"])) { $xml="<register><message>Available</message></register>"; echo $xml; } Login function works, but username checking doesn't. Any ideas? Here is all of my code: $(document).ready(function() { jQuery.validator.addMethod("checkAvailability",function(value,element){ $.post( "login.php" , {username:"test", checking:"yes"}, function(xml){ if($("message", xml).text() == "Available") return true; else return false; }); },"Sorry, this user name is not available"); $("#loginForm").validate({ rules: { username: { required: true, minlength: 4, checkAvailability: true }, password:{ required: true, minlength: 5 } }, messages: { username:{ required: "You need to enter a username." , minlength: jQuery.format("Your username should be at least {0} characters long.") } }, highlight: function(element, errorClass) { $(element).fadeOut("fast",function() { $(element).fadeIn("slow"); }) }, success: function(x){ x.text("OK!") }, submitHandler: function(form){send()} }); function send(){ $("#message").hide("fast"); $.post( "login.php" , {username:$("#username").val(), password:$("#password").val()}, function(xml){ $("#message").html( $("message", xml).text() ); if($("message", xml).text() == "You are successfully logged in.") { $("#message").css({ "color": "green" }); $("#message").fadeIn("slow", function(){location.reload(true);}); } else { $("#message").css({ "color": "red" }); $("#message").fadeIn("slow"); } }); } $("#newUser").click(function(){ return false; }); });

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• Help securing files access with htaccess and php?

  - by bschaeffer

  I'm working on a site that allows users to purchase digital content and have implemented a method that attempts to serve secure downloads. I'm using CodeIgniter to force downloads like this: $file = file_get_contents($path); force_download("my_file_name.zip", $file); Of course, I make sure the user has access to the file using a database before serving the download, but I was wondering if there was a way to make these files more secure. I'm using a some 8-10 letter keys to create the file paths so urls to the files aren't exactly easy to figure out... something like http://mysite.com/as67Hgr/asdo0980/uth89.zip in lieu of http://mysite.com/downloads/my_file.zip. Also, I'm using .htaccess to deny directory browsing like so: Options All -Indexes. Other than that... I have no idea what steps to take. I've seen articles suggesting username and password methods using .htaccess, but I don't understand how to bypass the password prompt that would occur using that method. I was hoping there might be a method where I could send a username and password combination using headers and cUrl (or something similar), but I wouldn't know where to start. Any help would be hugely appreciated. Thanks in advance!

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• MySQL Connection Error in PHP

  - by user309381

  I have set the password for root and grant all privileges for root. Why does it say it is denied? ****mysql_query() [function.mysql-query]: Access denied for user 'SYSTEM'@'localhost' (using password: NO) in C:\wamp\www\photo_gallery\includes\database.php on line 56 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\wamp\www\photo_gallery\includes\database.php on line 56 The Query has problemAccess denied for user 'SYSTEM'@'localhost' (using password: NO) Code as follows: <?php include("DB_Info.php"); class MySQLDatabase { public $connection; function _construct() { $this->open_connection(); } public function open_connection() { /* $DB_SERVER = "localhost"; $DB_USER = "root"; $DB_PASS = ""; $DB_NAME = "photo_gallery";*/ $this->connection = mysql_connect($DBSERVER,$DBUSER,$DBPASS); if(!$this->connection) { die("Database Connection Failed" . mysql_error()); } else { $db_select = mysql_select_db($DBNAME,$this->connection); if(!$db_select) { die("Database Selection Failed" . mysql_error()); } } } function mysql_prep($value) { if (get_magic_quotes_gpc()) { $value = stripslashes($value); } // Quote if not a number if (!is_numeric($value)) { $value = "'" . mysql_real_escape_string($value) . "'"; } return $value; } public function close_connection() { if(isset($this->connection)) { mysql_close($this->connection); unset($this->connection); } } public function query($sql) { //$sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //$found_user = mysql_fetch_assoc($result); //echo $found_user; return $found_user; } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); ?>

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• How to check name of Domino Server to which lotus notes is configured? using C#

  - by Preeti

  Hi, I am trying to login Domino server. For that i am taking Lotus Notes Password and Domino Server Name from user. if (notesPassword == "" && serverName == "") { MessageBox.Show("Please enter the server name !! "); return; } else { if (connectToDomino(notesPassword, serverName)) { MessageBox.Show("Connection Established Succesfully!!.."); } else { MessageBox.Show("Connection Fail.Please Login Again To Begin"); } }//else and in public bool connectToDomino(string NotesPassword, string strDominoServerName) { try { if (_lotesNotesSession == null) { NotesSession notesSession = new Domino.NotesSessionClass(); notesSession.Initialize(NotesPassword); } return true; } catch(Exception ex) { return false; } } Here i am initializing notes password.So in this case it is just verifying Notes Password. So even if user enters invalid entry of server name above function will return true. I tried : string serverName = notesSession.ServerName; But it is showing null value. :( Regards, Preeti

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• Invalid argument supplied for foreach() using adldap

  - by Brad

  I am using adldap http://adldap.sourceforge.net/ And I am passing the session from page to page, and checking to make sure the username within the session is a member of a certain member group, for this example, it is the STAFF group. <?php ini_set('display_errors',1); error_reporting(E_ALL); require_once('/web/ee_web/include/adLDAP.php'); $adldap = new adLDAP(); session_start(); $group = "STAFF"; //$authUser = $adldap->authenticate($username, $password); $result=$adldap->user_groups($_SESSION['user_session']); foreach($result as $key=>$value) { switch($value) { case $group: print '<h3>'.$group.'</h3>'; break; default: print '<h3>Did not find specific value: '.$value.'</h3>'; } if($value == $group) { print 'for loop broke'; break; } } ?> It gives me the error: Warning: Invalid argument supplied for foreach() on line 15, which is this line of code: foreach($result as $key=$value) { When I uncomment the code $authUser = $adldap-authenticate($username, $password); and enter in the appropriate username and password, it works fine, but I shouldn't have to, since the session is valid, I just want to see if the username stored within the valid_session is apart of the STAFF group. Why would it be giving me that problem?

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• Sample twitter application.

  - by Jack

  <?php function updateTwitter($status) { // Twitter login information $username = 'xxxxx'; $password = 'xxxxxx'; // The url of the update function $url = 'http://twitter.com/statuses/update.xml'; // Arguments we are posting to Twitter $postargs = 'status='.urlencode($status); // Will store the response we get from Twitter $responseInfo=array(); // Initialize CURL $ch = curl_init($url); // Tell CURL we are doing a POST curl_setopt ($ch, CURLOPT_POST, true); // Give CURL the arguments in the POST curl_setopt ($ch, CURLOPT_POSTFIELDS, $postargs); // Set the username and password in the CURL call curl_setopt($ch, CURLOPT_USERPWD, $username.':'.$password); // Set some cur flags (not too important) curl_setopt($ch, CURLOPT_VERBOSE, 1); curl_setopt($ch, CURLOPT_NOBODY, 0); curl_setopt($ch, CURLOPT_HEADER, 0); curl_setopt($ch, CURLOPT_FOLLOWLOCATION,1); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); // execute the CURL call $response = curl_exec($ch); // Get information about the response $responseInfo=curl_getinfo($ch); // Close the CURL connection curl_close($ch); // Make sure we received a response from Twitter if(intval($responseInfo['http_code'])==200){ // Display the response from Twitter echo $response; }else{ // Something went wrong echo "Error: " . $responseInfo['http_code']; } } updateTwitter("Just finished a sweet tutorial on http://brandontreb.com"); ?> I get the following output Error: 0 Please help.

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• How to find that Mutex in C# is acquired?

  - by TN

  How can I find from mutex handle in C# that a mutex is acquired? When mutex.WaitOne(timeout) timeouts, it returns false. However, how can I find that from the mutex handle? (Maybe using p/invoke.) UPDATE: public class InterProcessLock : IDisposable { readonly Mutex mutex; public bool IsAcquired { get; private set; } public InterProcessLock(string name, TimeSpan timeout) { bool created; var security = new MutexSecurity(); security.AddAccessRule(new MutexAccessRule(new SecurityIdentifier(WellKnownSidType.WorldSid, null), MutexRights.Synchronize | MutexRights.Modify, AccessControlType.Allow)); mutex = new Mutex(false, name, out created, security); IsAcquired = mutex.WaitOne(timeout); } #region IDisposable Members public void Dispose() { if (IsAcquired) mutex.ReleaseMutex(); } #endregion } Currently, I am using my own property IsAcquired to determine whether I should release a mutex. Not essential but clearer, would be not to use a secondary copy of the information represented by IsAcquired property, but rather to ask directly the mutex whether it is acquired by me. Since calling mutex.ReleaseMutex() throws an exception if it is not acquired by me. (By acquired state I mean that the mutex is in not-signaled state when I am owning the mutex.)

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• running scala apps with java -jar

  - by paintcan

  Yo dawgs, I got some problems with the java. Check it out. sebastian@sebastian-desktop:~/scaaaaaaaaala$ java -cp /home/sebastian/.m2/repository/org/scala-lang/scala-library/2.8.0.RC3/scala-library-2.8.0.RC3.jar:target/scaaaaaaaaala-1.0.jar scaaalaaa.App Hello World! That's cool, right, but how bout this: sebastian@sebastian-desktop:~/scaaaaaaaaala$ java -cp /home/sebastian/.m2/repository/org/scala-lang/scala-library/2.8.0.RC3/scala-library-2.8.0.RC3.jar -jar target/scaaaaaaaaala-1.0.jar Exception in thread "main" java.lang.NoClassDefFoundError: scala/Application at java.lang.ClassLoader.defineClass1(Native Method) at java.lang.ClassLoader.defineClassCond(ClassLoader.java:632) at java.lang.ClassLoader.defineClass(ClassLoader.java:616) at java.security.SecureClassLoader.defineClass(SecureClassLoader.java:141) at java.net.URLClassLoader.defineClass(URLClassLoader.java:283) at java.net.URLClassLoader.access$000(URLClassLoader.java:58) at java.net.URLClassLoader$1.run(URLClassLoader.java:197) at java.security.AccessController.doPrivileged(Native Method) at java.net.URLClassLoader.findClass(URLClassLoader.java:190) at java.lang.ClassLoader.loadClass(ClassLoader.java:307) at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:301) at java.lang.ClassLoader.loadClass(ClassLoader.java:248) at scaaalaaa.App.main(App.scala) Caused by: java.lang.ClassNotFoundException: scala.Application at java.net.URLClassLoader$1.run(URLClassLoader.java:202) at java.security.AccessController.doPrivileged(Native Method) at java.net.URLClassLoader.findClass(URLClassLoader.java:190) at java.lang.ClassLoader.loadClass(ClassLoader.java:307) at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:301) at java.lang.ClassLoader.loadClass(ClassLoader.java:248) ... 13 more Wat the heck? Any idea why the first works and not the 2nd? How do I -jar my scala?? Thanks in advance bro.

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• Getting content from PHP: Trouble with POST and query.

  - by vgm64

  Apologies for my longest question on SO ever. I'm trying to interface with a php frontend for a mysql database in ROOT (a CERN framework in C++ for high energy physics analysis). To start off with, I tried to get this php interface to play nice with wget and curl first because I'm more familiar with them. The following command works: wget --post-data "hostname=localhost:3306&un=joeuser&pw=psswd&myquery=show_spazio_databases;" http://some.host.edu/log/log_query_matlab.php The results are: database1 database2 That's good. If I leave out the --post-data then I get the result: Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'admin'@'localhost' (using password: NO) in /log/log_query_matlab.php on line 6 i'm dead! Access denied for user 'admin'@'localhost' (using password: NO) Warning: mysql_query() [function.mysql-query]: Access denied for user 'admin'@'localhost' (using password: NO) in /log/log_query_matlab.php on line 29 Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /log/log_query_matlab.php on line 29 I have access to the php script (read only), but the error itself isn't too important. What matters it that using ROOT, I use a function called as socket.SendRaw(message, message.Length()) (socket is a TSocket) and this gives me the same "error" as wget without the post data switch if my "message" is "POST http://some.host.edu/log/log_query_matlab.php?hostname=localhost:3306&un=joeuser&pw=psswd&myquery=show_spazio_databases" This may be in vain, but does someone knows a way I should format the "message" that includes something that is equivalent to the --post-data switch. Or, is there a standard way to format POST requests in a single line (I've seen multi-line stuff. Is that right?) Sorry I'm clueless! PS. The mysql query is show databases but the space has been replaced with _spazio_, Italian for space. The author of the db and php interface requires it (and various replacements for symbols), but has anyone seen this before? Trying to troubleshoot that was terrible!

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• Problem running java code through command line

  - by kunjaan

  I have a simple Class package chapter10; public class CompilationTest { public static void main(String[] args) { System.out.println("HELLO WORLD"); } } The path is Test\src\chapter10\CompilationTest.java I successfully compiled the code into the same folder and now I have Test\src\chapter10\CompilationTest.class However when I try to run from the same folder it I get this error >java CompilationTest Exception in thread "main" java.lang.NoClassDefFoundError: CompilationTest (wrong name: chapter10/CompilationTest) at java.lang.ClassLoader.defineClass1(Native Method) at java.lang.ClassLoader.defineClassCond(Unknown Source) at java.lang.ClassLoader.defineClass(Unknown Source) at java.security.SecureClassLoader.defineClass(Unknown Source) at java.net.URLClassLoader.defineClass(Unknown Source) at java.net.URLClassLoader.access$000(Unknown Source) at java.net.URLClassLoader$1.run(Unknown Source) at java.security.AccessController.doPrivileged(Native Method) at java.net.URLClassLoader.findClass(Unknown Source) at java.lang.ClassLoader.loadClass(Unknown Source) at sun.misc.Launcher$AppClassLoader.loadClass(Unknown Source) at java.lang.ClassLoader.loadClass(Unknown Source) Could not find the main class: CompilationTest. Program will exit. When I run using >java chapter10/PropertiesTest Exception in thread "main" java.lang.NoClassDefFoundError: chapter10/PropertiesTest Caused by: java.lang.ClassNotFoundException: chapter10.PropertiesTest at java.net.URLClassLoader$1.run(Unknown Source) at java.security.AccessController.doPrivileged(Native Method) at java.net.URLClassLoader.findClass(Unknown Source) at java.lang.ClassLoader.loadClass(Unknown Source) at sun.misc.Launcher$AppClassLoader.loadClass(Unknown Source) at java.lang.ClassLoader.loadClass(Unknown Source) Could not find the main class: chapter10/PropertiesTest. Program will exit.

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• Emailing smtp with Python error

  - by jakecar

  I can't figure out why this isn't working. I'm trying to send an email from my school email address with this code I got online. The same code works for sending from my GMail address. Does anyone know what this error means? The error occurs after waiting for about one and a half minutes. import smtplib FROMADDR = "FROM_EMAIL" LOGIN = "USERNAME" PASSWORD = "PASSWORD" TOADDRS = ["TO_EMAIL"] SUBJECT = "Test" msg = ("From: %s\r\nTo: %s\r\nSubject: %s\r\n\r\n" % (FROMADDR, ", ".join(TOADDRS), SUBJECT) ) msg += "some text\r\n" server = smtplib.SMTP('OUTGOING_SMTP', 587) server.set_debuglevel(1) server.ehlo() server.starttls() server.login(LOGIN, PASSWORD) server.sendmail(FROMADDR, TOADDRS, msg) server.quit() And here's the error I get: Traceback (most recent call last): File "emailer.py", line 13, in server = smtplib.SMTP('OUTGOING_SMTP', 587) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/smtplib.py", line 239, in init (code, msg) = self.connect(host, port) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/smtplib.py", line 295, in connect self.sock = self._get_socket(host, port, self.timeout) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/smtplib.py", line 273, in _get_socket return socket.create_connection((port, host), timeout) File "/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/socket.py", line 514, in create_connection raise error, msg socket.error: [Errno 60] Operation timed out

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• Send mail in asp.net

  - by Zerotoinfinite

  Hi Experts, I am using asp.net 3.5 and C#. I want to send mail from asp.net, for that I have got some details from my hosting provider which are these: mail.MySite.net UserName Password But I am unable to send mail through these details, I have done the following changes in my web.config file: <system.net> <mailSettings> <smtp> <network host="mail.MySite.net" port="8080" userName="UserName" password="Password" /> </smtp> </mailSettings> </system.net> Also, at the code behind I am writing this function: MailMessage mail = new MailMessage("[email protected]", "[email protected]"); mail.Subject = "Hi"; mail.Body = "Test Mail from ASP.NET"; mail.IsBodyHtml = false; SmtpClient smp = new SmtpClient(); smp.Send(mail); but I am getting error message as message sending failed. Please let me know what I am doing wrong and what I have to do to make it work fine. Thanks in advance.

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• Setup SSL (self signed cert) with tomcat

  - by Danny

  I am mostly following this page: http://tomcat.apache.org/tomcat-6.0-doc/ssl-howto.html I used this command to create the keystore keytool -genkey -alias tomcat -keyalg RSA -keystore /etc/tomcat6/keystore and answered the prompts Then i edited my server.xml file and uncommented/edited this line <Connector port="8443" protocol="HTTP/1.1" SSLEnabled="true" maxThreads="150" scheme="https" secure="true" clientAuth="false" sslProtocol="TLS" keystoreFile="/etc/tomcat6/keystore" keystorePass="tomcat" /> then I go to the web.xml file for my project and add this into the file <security-constraint> <web-resource-collection> <web-resource-name>Security</web-resource-name> <url-pattern>/*</url-pattern> </web-resource-collection> <user-data-constraint> <transport-guarantee>CONFIDENTIAL</transport-guarantee> </user-data-constraint> </security-constraint> When I try to run my webapp I am met with this: Unable to connect Firefox can't establish a connection to the server at localhost:8443. * The site could be temporarily unavailable or too busy. Try again in a few moments. * If you are unable to load any pages, check your computer's network connection. If I comment out the lines I've added to my web.xml file, the webapp works fine. My log file in /var/lib/tomcat6/logs says nothing. I can't figure out if this is a problem with my keystore file, my server.xml file or my web.xml file.... Any assistance is appreciated I am using tomcat 6 on ubuntu.

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• How to combine twill and python into one code that could be run on "Google App Engine"?

  - by brilliant

  Hello everybody!!! I have installed twill on my computer (having previously installed Python 2.5) and have been using it recently. Python is installed on disk C on my computer: C:\Python25 And the twill folder (“twill-0.9”) is located here: E:\tmp\twill-0.9 Here is a code that I’ve been using in twill: go “some website’s sign-in page URL” formvalue 2 userid “my login” formvalue 2 pass “my password” submit go “URL of some other page from that website” save_html result.txt This code helps me to log in to one website, in which I have an account, record the HTML code of some other page of that website (that I can access only after logging in), and store it in a file named “result.txt” (of course, before using this code I firstly need to replace “my login” with my real login, “my password” with my real password, “some website’s sign-in page URL” and “URL of some other page from that website” with real URLs of that website, and number 2 with the number of the form on that website that is used as a sign-in form on that website’s log-in page) This code I store in “test.twill” file that is located in my “twill-0.9” folder: E:\tmp\twill-0.9\test.twill I run this file from my command prompt: python twill-sh test.twill Now, I also have installed “Google App Engine SDK” from “Google App Engine” and have also been using it for awhile. For example, I’ve been using this code: import hashlib m = hashlib.md5() m.update("Nobody inspects") m.update(" the spammish repetition ") print m.hexdigest() This code helps me transform the phrase “Nobody inspects the spammish repetition” into md5 digest. Now, how can I put these two pieces of code together into one python script that I could run on “Google App Engine”? Let’s say, I want my code to log in to a website from “Google App Engine”, go to another page on that website, record its HTML code (that’s what my twill code does) and than transform this HTML code into its md5 digest (that’s what my second code does). So, how can I combine those two codes into one python code? I guess, it should be done somehow by importing twill, but how can it be done? Can a python code - the one that is being run by “Google App Engine” - import twill from somewhere on the internet? Or, perhaps, twill is already installed on “Google App Engine”?

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• MySqlDataReader giving error at build

  - by TuxMeister

  Hey there. I have a function in VB.net that authenticates a user towards a MySQL DB before launching the main application. Here's the code of the function: Public Function authConnect() As Boolean Dim dbserver As String Dim dbuser As String Dim dbpass As String dbserver = My.Settings.dbserver.ToString dbuser = My.Settings.dbuser.ToString dbpass = My.Settings.dbpass.ToString conn = New MySqlConnection myConnString = "server=" & dbserver & ";" & "user id=" & dbuser & ";" & "password=" & dbpass & ";" & "database=rtadmin" Dim myCommand As New MySqlCommand Dim myAdapter As New MySqlDataAdapter Dim myData As New DataTable Dim myDataReader As New MySqlDataReader Dim query As String myCommand.Parameters.Add(New MySqlParameter("?Username", login_usr_txt.Text)) myCommand.Parameters.Add(New MySqlParameter("?Password", login_pass_txt.Text)) query = "select * from users where user = ?Username and passwd = ?Password" conn.ConnectionString = myConnString Try conn.Open() Try myCommand.Connection = conn myCommand.CommandText = query myAdapter.SelectCommand = myCommand myDataReader = myCommand.ExecuteReader If myDataReader.HasRows() Then MessageBox.Show("You've been logged in.", "RT Live! Information", MessageBoxButtons.OK, MessageBoxIcon.Information) End If Catch ex As Exception End Try Catch ex As Exception End Try End Function The function is not yet complete, there are a few other things that need to be done before launching the application, since I'm using a MessageBox to display the result of the login attempt. The error that I'm getting is the following: Error 1 'MySql.Data.MySqlClient.MySqlDataReader.Friend Sub New(cmd As MySql.Data.MySqlClient.MySqlCommand, statement As MySql.Data.MySqlClient.PreparableStatement, behavior As System.Data.CommandBehavior)' is not accessible in this context because it is 'Friend'. C:\Users\Mario\documents\visual studio 2010\Projects\Remote Techs Live!\Remote Techs Live!\Login.vb 43 13 Remote Techs Live! Any ideas? Thanks.

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• Send email using GMail SMTP server from PHP page

  - by skb

  I am trying to send an email via GMail's SMTP server from a PHP page but I get this error: authentication failure [SMTP: SMTP server does no support authentication (code: 250, response: mx.google.com at your service, [98.117.99.235] SIZE 35651584 8BITMIME STARTTLS ENHANCEDSTATUSCODES PIPELINING)] Can anyone help? Here is my code: <?php require_once "Mail.php"; $from = "Sandra Sender <[email protected]>"; $to = "Ramona Recipient <[email protected]>"; $subject = "Hi!"; $body = "Hi,\n\nHow are you?"; $host = "smtp.gmail.com"; $port = "587"; $username = "[email protected]"; $password = "testtest"; $headers = array ('From' => $from, 'To' => $to, 'Subject' => $subject); $smtp = Mail::factory('smtp', array ('host' => $host, 'port' => $port, 'auth' => true, 'username' => $username, 'password' => $password)); $mail = $smtp->send($to, $headers, $body); if (PEAR::isError($mail)) { echo("<p>" . $mail->getMessage() . "</p>"); } else { echo("<p>Message successfully sent!</p>"); } ?>

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• Minimum privileges to read SQL Jobs using SQL SMO

  - by Gustavo Cavalcanti

  I wrote an application to use SQL SMO to find all SQL Servers, databases, jobs and job outcomes. This application is executed through a scheduled task using a local service account. This service account is local to the application server only and is not present in any SQL Server to be inspected. I am having problems getting information on job and job outcomes when connecting to the servers using a user with dbReader rights on system tables. If we set the user to be sysadmin on the server it all works fine. My question to you is: What are the minimum privileges a local SQL Server user needs to have in order to connect to the server and inspect jobs/job outcomes using the SQL SMO API? I connect to each SQL Server by doing the following: var conn = new ServerConnection { LoginSecure = false, ApplicationName = "SQL Inspector", ServerInstance = serverInstanceName, ConnectAsUser = false, Login = user, Password = password }; var smoServer = new Server (conn); I read the jobs by reading smoServer.JobServer.Jobs and read the JobSteps property on each of these jobs. The variable server is of type Microsoft.SqlServer.Management.Smo.Server. user/password are of the user found in each SQL Server to be inspected. If "user" is SysAdmin on the SQL Server to be inspected all works ok, as well as if we set ConnectAsUser to true and execute the scheduled task using my own credentials, which grants me SysAdmin privileges on SQL Server per my Active Directory membership. Thanks!

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