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• Django website on Apache with wsgi failing

  - by notagain

  I have a website I've built in django that I'm trying to get working on our corporate Apache server (on debian) for our intranet at my workplace. Unfortunately, Apache keeps returning server errors whenever I try to navigate to my site. Although I can navigate to the statics folder. My Apache config and wsgi script look like the following... lbirdf.wsgi import os import sys sys.path.append('/home/lbi/rdfweb/web') sys.path.append('/home/lbi/rdfweb/lbirdf') os.environ['DJANGO_SETTINGS_MODULE'] = 'lbirdf.settings_production' import django.core.handlers.wsgi application = django.core.handlers.wsgi.WSGIHandler() Apache config Listen 8080 <VirtualHost *:8080> ServerName server1 WSGIScriptAlias /rdfweb /home/lbi/rdfweb/web/lbirdf/apache/lbirdf.wsgi Alias /statics /home/lbi/rdfweb/web/lbirdf/statics Alias /admin_media /home/lbi/rdfweb/web/lbirdf/admin_media <Directory /home/lbi/rdfweb/web/lbirdf/apache> Order allow,deny Allow from all </Directory> <Directory /home/lbi/rdfweb/web/lbirdf/admin_media> Order allow,deny Allow from all </Directory> </VirtualHost> Any ideas on where I might be going wrong?

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• When send a meeting invite to a specific user, you get a undeliverable response back from another us

  - by jherlitz

  We have a user, where it doesn't matter who sends it, but if you send her a meeting invite through outlook (exchange 2007), you get a non-deliverable message back from another user. Same one all the time though. I checked her shared calendar properties and removed the user from the list. I checked and made sure she wasn't using cache mode. I checked the manage full permissions and send permissions within exchange and the user is not listed. What am I missing? Notes: Using Outlook 2007 with Exchange 2007 on a XP box and Server 2008.

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• Howto create dynamic named folders with Maven2 Archetype?

  - by chrsk

  Hey, i try to build a structure like this +---src +---main +---webapp +---WEB-INF +---cfg ¦ +---log4j ¦ +---resources ¦ +---extensions ¦ +---${artifactIdKey}-business ¦ +---${artifactIdKey}-layout +---lib I added a required property to my archetype-metadata.xml to have an short-name for the project, which is used among other things for generating unique folders. <requiredProperties> <requiredProperty key="artifactIdKey" /> </requiredProperties> The property artifactIdKey i use in a fileSet: <fileSet> <directory>[..]/resources/extensions/${artifactIdKey}-business</directory> </fileSet> <fileSet> <directory>[..]/resources/extensions/${artifactIdKey}-layout</directory> </fileSet> The command to generate the archetype: mvn archetype:generate -B \ -DgroupId=com.stackoverflow \ -DartifactId=stackoverflow-question -DarchetypeGroupId=com.stackoverflow \ -DarchetypeArtifactId=stackoverflow-archetype -DarchetypeVersion=1.0 \ -DartifactIdKey=soq I assume to get the following folder structure, but i get the ${artifactIdKey} property as folder. Without beeing replaces. ..\ +---soq-business +---soq-layout ..\ +---${artifactIdKey}-business +---${artifactIdKey}-layout How can i achieve this? And how can i place files below soq-business? Without knowing the folder name at this time?

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• Cannot install new certificate in IIS 7 on Windows Server 2008 R2

  - by Alex B.

  We are trying to renew our existing web site certificate on our IIS 7 site under Windows Server 2008 R2, but we continue to get the "Access is denied" error that others have posted. However, when we have gone to implement the common fix of making sure the Administrator group has full access to all folders and subfolders on the C:\ProgramData\Microsoft\Crypto\RSA folder, we get an "Access is Denied" error on changing those permissions. Yes, we are logged in as Administrator user - it just seems to not allow us to modify the group permissions to this folder. Help! We need to renew our certificate before March 2011!

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• How to highlight the new created file in JTree

  - by newbie123

  I want to make it like when I click a button, it will create a new file. Then the jTree will highlight the new file. Below are my code. Currently I create new file, i will show the new file but no highlight the file. class FileTreeModel implements TreeModel { private FileNode root; public FileTreeModel(String directory) { root = new FileNode(directory); } public Object getRoot() { return root; } public Object getChild(Object parent, int index) { FileNode parentNode = (FileNode) parent; return new FileNode(parentNode, parentNode.listFiles()[index].getName()); } public int getChildCount(Object parent) { FileNode parentNode = (FileNode) parent; if (parent == null || !parentNode.isDirectory() || parentNode.listFiles() == null) { return 0; } return parentNode.listFiles().length; } public boolean isLeaf(Object node) { return (getChildCount(node) == 0); } public int getIndexOfChild(Object parent, Object child) { FileNode parentNode = (FileNode) parent; FileNode childNode = (FileNode) child; return Arrays.asList(parentNode.list()).indexOf(childNode.getName()); } public void valueForPathChanged(TreePath path, Object newValue) { } public void addTreeModelListener(TreeModelListener l) { } public void removeTreeModelListener(TreeModelListener l) { } } class FileNode extends java.io.File { public FileNode(String directory) { super(directory); } public FileNode(FileNode parent, String child) { super(parent, child); } @Override public String toString() { return getName(); } } jTree = new JTree(); jTree.setBounds(new Rectangle(164, 66, 180, 421)); jTree.setBackground(SystemColor.inactiveCaptionBorder); jTree.setBorder(BorderFactory.createTitledBorder(null, "", TitledBorder.LEADING, TitledBorder.TOP, new Font("Arial", Font.BOLD, 12), new Color(0, 0, 0))); FileTreeModel model = new FileTreeModel(root); jTree.setRootVisible(false); jTree.setModel(model); expandAll(jTree); public void expandAll(JTree tree) { int row = 0; while (row < tree.getRowCount()) { tree.expandRow(row); row++; } }

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• Using terminal vs KDE in linux?

  - by Ke

  Im used to using nautilus within centos but have recently just got a VPS and quickly realising that using a KDE is unacceptable in this environment. Although I do find it so much quicker doing things like folder permissions in KDE rather than typing it all out in the terminal? Everyone I speak to says, use the terminal and I should learn this way as opposed to using the KDE, but theres certain things I just dont get How is it possible to make quick changes to scripts and viewing them in a browser etc , without a mouse or using KDE? and only using a terminal?? I am wondering how to develop websites just using the terminal??? How can it be quicker to type out/view permissions etc in the terminal when its instant and just a few clicks in the KDE? Any thoughts are much appreciated. I would love to understand the benefits but just cant seem to see them right now. Cheers Ke.

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• how to remotely open an URL in Firefox in a specific profile?

  - by miernik

  I have several instances of Firefox with several different profiles running. Among them profiles with the names "software" and "test". I am trying to open an URL from a bash script to have it open in profile "test", like this: firefox -P "test" http://www.example.org/ However that opens it in profile "software" anyway. Any ideas? Mozilla/5.0 (X11; U; Linux x86_64; en-US; rv:1.9.1.8) Gecko/20100308 Iceweasel/3.5.8 (like Firefox/3.5.8) No, it is not a permissions problem, all my profile directories are perfectly under my permissions: root@przehyba:~/.mozilla# ls -ld firefox/ drwx------ 13 miernik miernik 4096 Mar 11 09:15 firefox/ root@przehyba:~/.mozilla# ls -ld firefox/* drwxr-xr-x 9 miernik miernik 4096 Mar 12 11:29 firefox/info -rw-r--r-- 1 miernik miernik 560 Mar 11 09:15 firefox/profiles.ini drwxr-xr-x 10 miernik miernik 4096 Mar 16 11:51 firefox/software drwxr-xr-x 9 miernik miernik 4096 Mar 11 09:14 firefox/tech drwxr-xr-x 11 miernik miernik 4096 Mar 15 22:48 firefox/test root@przehyba:~/.mozilla#

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• icacls, Network Service, and setting ACLs on Windows Server 2008

  - by Ted

  Setting ACLs on Windows Server 2008 via the command line is giving me some problems. As per http://web2.minasi.com/forum/topic.asp?TOPIC%5FID=26907 I've tried all sorts of variations: C:\Windows\system32icacls "D:\Websites\site.com\Web\bin*" /grant 'NT A uthority\NETWORK SERVICE: (OI) (CI)M' C:\Windows\system32icacls "D:\Websites\site.com\Web\bin*" /grant "NETWORK SERVICE": (OI) (CI)M And all variations in between. However, each try leads to i.e. "Invalid parameter "'NETWORK'"" depending on the variation above. As per http://technet.microsoft.com/en-us/library/cc753525%28WS.10%29.aspx (see in comments), it appears that others have experienced the same issue where the same command works on Windows 7/Vista/etc., but not on Windows Server 2008. What's the best way to apply permissions to Network Service account on a directory and/or files via the command line in Windows Server 2008? Especially as there's no way to do multiple file permissions at once via the GUI (see http://serverfault.com/questions/30991/windows-server-2008-change-security-settings-for-multiple-files-at-once).

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• wsgi django not working

  - by MaKo

  im installing django, the test for wsgi is ok, but when i point my default file to the django test, it doesnt work, this is the test that works fine: default: /etc/apache2/sites-available/default <VirtualHost *:80> ServerName www.example.com ServerAlias example.com ServerAdmin [email protected] DocumentRoot /var/www <Directory /var/www/documents> Order allow,deny Allow from all </Directory> WSGIScriptAlias / /home/ubuntu/djangoProj/micopiloto/application.wsgi <Directory /home/ubuntu/djangoProj/mysitio/wsgi_handler.py> Order allow,deny Allow from all </Directory> </VirtualHost> application.wsgi:: ~/djangoProj/micopiloto import os import sys sys.path.append('/srv/www/cucus/application') os.environ['PYTHON_EGG_CACHE'] = '/srv/www/cucus/.python-egg' def application(environ, start_response): status = '200 OK' output = 'Hello World!MK SS9 tkt kkk' response_headers = [('Content-type', 'text/plain'), ('Content-Length', str(len(output)))] start_response(status, response_headers) return [output] but if I change the default to point to application_sa.wsgi the django test, it doesnt work :( application_sa.wsgi import os, sys sys.path.append('/home/ubuntu/djangoProj/micopiloto') os.environ['DJANGO_SETTINGS_MODULE'] = 'micopiloto.settings' import django.core.handlers.wsgi application = django.core.handlers.wsgi.WSGIHandler() I restart the apache server every time i change the wsgi to test, so what im i missing? thanks a lot!

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• How to use dirent.h correctly.

  - by Nick

  Hello, I am new to C++ and I am experimenting with the dirent.h header to manipulate directory entries. The following little app compiles but pukes after you supple a directory name. Can someone give me a hint? The int quit is there to provide a while loop. I removed the loop in an attempt to isolate my problem. thanks! #include <iostream> #include <dirent.h> using namespace std; int main() { char *dirname = 0; DIR *pd = 0; struct dirent *pdirent = 0; int quit = 1; cout<< "Enter a directory path to open (leave blank to quit):\n"; cin >> dirname; if(dirname == NULL) { quit = 0; } pd = opendir(dirname); if(pd == NULL) { cout << "ERROR: Please provide a valid directory path.\n"; } return 0; }

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• My QFileSystemModel doesn't work as expected in PyQt

  - by Skilldrick

  I'm learning the Qt Model/View architecture at the moment, and I've found something that doesn't work as I'd expect it to. I've got the following code (adapted from Qt Model Classes): from PyQt4 import QtCore, QtGui model = QtGui.QFileSystemModel() parentIndex = model.index(QtCore.QDir.currentPath()) print model.isDir(parentIndex) #prints True print model.data(parentIndex).toString() #prints name of current directory childIndex = model.index(0, 0, parentIndex) print model.data(childIndex).toString() rows = model.rowCount(parentIndex) print rows #prints 0 (even though the current directory has directory and file children) The question: Is this a problem with PyQt, have I just done something wrong, or am I completely misunderstanding QFileSystemModel? According to the documentation, model.rowCount(parentIndex) should return the number of children in the current directory. The QFileSystemModel docs say that it needs an instance of a Gui application, so I've also placed the above code in a QWidget as follows, but with the same result: import sys from PyQt4 import QtCore, QtGui class Widget(QtGui.QWidget): def __init__(self, parent=None): QtGui.QWidget.__init__(self, parent) model = QtGui.QFileSystemModel() parentIndex = model.index(QtCore.QDir.currentPath()) print model.isDir(parentIndex) print model.data(parentIndex).toString() childIndex = model.index(0, 0, parentIndex) print model.data(childIndex).toString() rows = model.rowCount(parentIndex) print rows def main(): app = QtGui.QApplication(sys.argv) widget = Widget() widget.show() sys.exit(app.exec_()) if __name__ == '__main__': main()

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• Reenabling the Spotlight Menubar item in Mac OS X 10.6

  - by Tim Visher

  I believe I followed the instructions here to disable Spotlight indexing and remove the menubar item. I reenabled indexing just fine, but when I changed the permissions back to 744, the spotlight search position came back (as in the space it would normally occupy), but the actual icon and search box will not show up. If I click that portion of the screen I get a blue box, but I can't type anything in to anything. Currently, permissions look like this: [~]$ ll /System/Library/CoreServices/Search.bundle.bak/Contents/MacOS/ total 648 -rwxr-xr-x 1 root wheel 835K Sep 17 14:48 Search* ll is an alias mapped to the following alias ll='${LS_PREAMBLE} -hl' with $LS_PREAMBLE [~]$ echo $LS_PREAMBLE ls -GF (Ignore the .bak extension. I decided that until I found a way to fully restore it, I would just remove it entirely following the directions here) That looks right to me and obviously something is launching, but the UI elements aren't there. So how can I restore it? Thanks in advance!

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• Should I impersonate PHP via FastCGI?

  - by AKeller

  I am installing the latest version of PHP onto IIS 7.5 via FastCGI, and all of the instructions say that FastCGI should impersonate the calling client by setting fastcgi.impersonate = 1 If my website will have this configuration dedicated application pool application pool identity of ApplicationPoolIdentity anonymous authentication only (as IUSR) why do I want to impersonate? I come from an ASP.NET background, where the IUSR gets read-only permissions and the application pool identity gets any write permissions. Giving write access to the IUSR usually opens the door for WebDAV vulnerabilities. So I hesitate to let PHP run as the IUSR. I can't find many people asking this question (1 | 2) so I think I must be missing something. Can someone clarify this for me?

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• Ldap access lists users even if user has no rights...

  - by Patkos Csaba

  I am trying to set up a more complex Active Directory structure for some testing purposes. What I did so far: set up 2 windows (one 2008 and one 2003) to control the same domain set up an Organizational Unit (ou): Developers set up 2 child OUs: "one" and "two" each OU has it's admin: adminOne and adminTwo I denied all access to OU "two" by removing on the Security tab all the groups I don't want to access it. now, when I log in as adminOne and I try to click on OU "two" it says I don't have permissions to see the users and properties of "two" - this is perfect, it's what I want Here comes my problem: I do a LDAP query with the adminOne user on the "Developers" What I expect to happen: I expect to retrieve the users from Developer - One I expect to NOT be able to retrieve the users from Developers - Two What actually happens: ldap shows all the users, both from Developers - One and Developers - Two, even if the user should not have permissions to Developers - Two And now my question: is there any specific settings on Windows 2003 or 2008 Active Directory servers which allow or deny access over LDAP? I could not find any.

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• Rename files and directories using substitution and variables

  - by rednectar

  I have found several similar questions that have solutions, except they don't involve variables. I have a particular pattern in a tree of files and directories - the pattern is the word TEMPLATE. I want a script file to rename all of the files and directories by replacing the word TEMPLATE with some other name that is contained in the variable ${newName} If I knew that the value of ${newName} was say "Fred lives here", then the command find . -name '*TEMPLATE*' -exec bash -c 'mv "$0" "${0/TEMPLATE/Fred lives here}"' {} \; will do the job However, if my script is: newName="Fred lives here" find . -name '*TEMPLATE*' -exec bash -c 'mv "$0" "${0/TEMPLATE/${newName}}"' {} \; then the word TEMPLATE is replaced by null rather than "Fred lives here" I need the "" around $0 because there are spaces in the path name, so I can't do something like: find . -name '*TEMPLATE*' -exec bash -c 'mv "$0" "${0/TEMPLATE/"${newName}"}"' {} \; Can anyone help me get this script to work so that all files and directories that contain the word TEMPLATE have TEMPLATE replaced by whatever the value of ${newName} is eg, if newName="A different name" and a I had directory of /foo/bar/some TEMPLATE directory/with files then the directory would be renamed to /foo/bar/some A different name directory/with files and a file called some TEMPLATE file would be renamed to some A different name file

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• GVFS Locations Not Available In ~/.gvfs

  - by Aaron Copley

  So, I can mount GVFS locations correctly (specifically CIFS) either from the Gnome "Places" menu, or via the command line gvfs-mount, but the filesystem is not mounted in the expected location; ~/.gvfs. In fact, running the mount command does not list any GVFS filesystem at all. This is reproducible for non-root users while the root user behaves as expected. Strace reveals a permissions error for the user mounting the filesystem for the path /home/username/.gvfs. Ownership and permissions are correct and there are no extended attributes for the path as revealed by lsattr. Also, /root/.gvfs and /home/username/.gvfs are on the same filesystem. All packages are current. Any ideas?

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• How to troubleshoot git "unable to set permission" on adding project?

  - by Brian Knoblauch

  Finally decided to move from Subversion to Git, but am having problems with my first project. Did my "git init" and am trying to do a "git add" of my project, but it's failing with: $ git add . error: unable to set permission to '.git/objects/6b/6018c1c76dc5ec159d5cb65bab72 fa300d52f6' error: build.xml: failed to insert into database error: unable to index file build.xml fatal: adding files failed I have full permissions to the directories in question. The only odd thing about it is that it's a drive mounted (and mapped) from a server over CIFS. No problems creating/editing files/permissions with other applications. The host is Windows Vista x64 and I'm running git under Cygwin. Server is Windows 2008. Any other ideas on what I might be doing wrong?

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• Drag Drop copy file

  - by Graham Warrender

  I've perhaps done something marginally stupid, but can't see what it is!! string pegasusKey = @"HKEY_LOCAL_MACHINE\SOFTWARE\Pegasus\"; string opera2ServerPath = @"Server VFP\"; string opera3ServerPath = @"O3 Client VFP\"; string opera2InstallationPath = null; string opera3InstallationPath = null; //Gets the opera Installtion paths and reads to the string opera*InstallationPath opera2InstallationPath = (string)Registry.GetValue(pegasusKey + opera2ServerPath + "System", "PathToServerDynamic", null); opera3InstallationPath = (string)Registry.GetValue(pegasusKey + opera3ServerPath + "System", "PathToServerDynamic", null); string Filesource = null; string[] FileList = (string[])e.Data.GetData(DataFormats.FileDrop, false); foreach (string File in FileList) Filesource = File; label.Text = Filesource; if (System.IO.Directory.Exists(opera3InstallationPath)) { System.IO.File.Copy(Filesource, opera3InstallationPath); MessageBox.Show("File Copied from" + Filesource + "\n to" + opera3InstallationPath); } else { MessageBox.Show("Directory Doesn't Exist"); } The user drags the file onto the window, I then get the installation path of an application which is then used as the destination for the source file.. When the application is runs, it throws the error directory not found. But surely if the directory doesn't exists is should step into the else statement? a simple application that is becoming a headache!!

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• how to lucene serch in android

  - by xyz Sad

  Lucen with android logic ..??? public class TestAndroidLuceneActivity extends Activity { @Override public void onCreate(Bundle icicle) { super.onCreate(icicle); setContentView(R.layout.main); try { Directory directory = new RAMDirectory(); Analyzer analyzer = new StandardAnalyzer(); Document doc = new Document(); doc.add(new Field("header", "ABC", Field.Store.YES,Field.Index.TOKENIZED)); indexWriter.addDocument(doc); doc.add(new Field("header", "DEF", Field.Store.YES,Field.Index.TOKENIZED)); indexWriter.addDocument(doc); doc.add(new Field("header", "GHI", Field.Store.YES,Field.Index.TOKENIZED)); indexWriter.addDocument(doc); doc.add(new Field("header", "JKL", Field.Store.YES,Field.Index.TOKENIZED)); indexWriter.addDocument(doc); indexWriter.optimize(); indexWriter.close(); IndexSearcher indexSearcher = new IndexSearcher(directory); QueryParser parser = new QueryParser("header", analyzer); // Query query = parser.parse("(" + "Anil" + ")"); Query query = parser.parse("(" + "ABC" + ")"); Hits hits = indexSearcher.search(query); for (int i = 0; i < hits.length(); i++) { Document hitDoc = hits.doc(i); Log.i("TestAndroidLuceneActivity", "Lucene: " +hitDoc.get("header")); // Toast.makeText(this, hitDoc.get("header"),Toast.LENGTH_LONG).show(); } indexSearcher.close(); directory.close(); } catch (Exception ex) { System.out.println(ex.getMessage()); } } } i have this code but i m not able to understnd plz send me related or modifed and set it main.xml show me some out put plzz..its does not serch after "ABC" plz tell me wat is the problem in logic any thing missing???..

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• How to catch non exist requested URL in Java servlet ?

  - by Frank

  My objects are stored online in two different places : <1 On my nmjava.com site, where I can put them in a directory called "Dir_My_App/Dir_ABC/" <2 On Google App Engine datastore When my Java app runs it checks both places for the objects, I designed the app so that it tries to get an object from a Url, it doesn't care whether it's an object in a directory or an object returned by a servlet. My_Object Get_Object(String Site_Url,String Object_Path) { ... get object by the name of Object_Path from the Site_Url ... } Now the request Url for my web site nmjava.com might look like this : http://nmjava.com/Dir_My_App/Dir_ABC/My_Obj_123 [ In a directory ] Or in the case of Google App Engine servlet : http://nm-java.appspot.com/Check_License/Dir_My_App/Dir_ABC/My_Obj_123 [ Non exist ] The "Object_Path" was generated by my app automatically. It can now get the object from my site by the above method like this : My_Object Get_Object("http://nmjava.com","/Dir_My_App/Dir_ABC/My_Obj_123"); In the Google App Engine, my servlet is running and ready to serve the object, if the request comes in correctly, but since I don't want to design my app to know whether the object is in one site's directory or in other site's datastore, I need to design the servlet to catch the non exist Url, such as the one above, and be able to make a call : My_Object Get_Object("http://nm-java.appspot.com/Check_License","/Dir_My_App/Dir_ABC/My_Obj_123"); So my question is : When a request comes into the servlet with a non exist Url, how should it catch it and analyze the url in order to respond properly, in my case it should know that : http://nm-java.appspot.com/Check_License/Dir_My_App/Dir_ABC/My_Obj_123 is asking for the object "My_Obj_123" [ ignore the dirs ] and return the object from the datastore. Now I'm getting this : Error: Not Found The requested URL /Check_License/Dir_My_App/Dir_ABC/My_Obj_123 was not found on this server. Where in my servlet and how do I detect the request for this non exist Url ?

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• JTree TreePath casting problem

  - by newbie123

  I got this casting problem java.lang.String cannot be cast to java.io.File, when I trying to do this TreePath p = new TreePath(new Object[] {"src","file","My Diary" }); This is my jtree file model class FileTreeModel implements TreeModel { private FileNode root; public FileTreeModel(String directory) { root = new FileNode(directory); } public Object getRoot() { return root; } public Object getChild(Object parent, int index) { FileNode parentNode = (FileNode) parent; return new FileNode(parentNode, parentNode.listFiles()[index].getName()); } public int getChildCount(Object parent) { FileNode parentNode = (FileNode) parent; if (parent == null || !parentNode.isDirectory() || parentNode.listFiles() == null) { return 0; } return parentNode.listFiles().length; } public boolean isLeaf(Object node) { return (getChildCount(node) == 0); } public int getIndexOfChild(Object parent, Object child) { FileNode parentNode = (FileNode) parent; FileNode childNode = (FileNode) child; return Arrays.asList(parentNode.list()).indexOf(childNode.getName()); } public void valueForPathChanged(TreePath path, Object newValue) { } public void addTreeModelListener(TreeModelListener l) { } public void removeTreeModelListener(TreeModelListener l) { } } class FileNode extends java.io.File { public FileNode(String directory) { super(directory); } public FileNode(FileNode parent, String child) { super(parent, child); } @Override public String toString() { return getName(); } }

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• VC++ 6 and MS Speech SDK 5.1 fatal error C1083: Cannot open source file: 'files\microsoft': No such

  - by eg123

  Trying to compile an application (flite synthesis sapi) on vc++6. This requires Microsoft Speech SDK 5.1 Have included C:\Program Files\Microsoft Speech SDK 5.1\IDL C:\Program Files\Microsoft Speech SDK 5.1\include using Toolsoptionsdirectories and also on another attempt via ProjectSettings Repeatedly get this error microsoft fatal error C1083: Cannot open source file: 'files\microsoft': No such file or directory speech fatal error C1083: Cannot open source file: 'speech': No such file or directory sdk fatal error C1083: Cannot open source file: 'sdk': No such file or directory idl fatal error C1083: Cannot open source file: '5.1\idl': No such file or directory FliteCMUKalDiphone.idl Thought it may be spaces related so included full path in quotes in relevant .h files. No joy Installed Microsoft Speech SDK 5.1 on another machine in same folder as flite and renamed to mssdk51 (so no spaces in pathname) but same error came up. Tried pasting in contents of each .idl called in file where glitch seems to generate Still same message. I am new to C++ and programming in general. My only guess is that something in the speech sdk is calling the .idl file and I can't find where from. Of course this is probably way wrong!

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• Resolve `strace` file number to a filename

  - by Mike Pennington

  I am debugging a problem where MoinMoin on CentOS is throwing a permissions error, but I can't track down where the problematic file / directory is. I ran strace -vp <pid> on the apache pid; when I have the problem I see this: epoll_wait(10, {{EPOLLIN, {u32=3487534344, u64=140367313734920}}}, 2, 10000) = 1 accept4(6, {sa_family=AF_INET6, sin6_port=htons(52621), inet_pton (AF_INET6, "::ffff:105.193.30.91", &sin6_addr), sin6_flowinfo=0, sin6_scope_id=0}, [28], SOCK_CLOEXEC) = 11 ## Later on... read(7, 0x7fffa658ad7f, 1) = -1 EAGAIN (Resource temporarily unavailable) However, since apache is already running, I see no corresponding open() on the file referred to as 7; thus I see the permissions problem, but I still don't know which file is the problem. I know I could try to catch all the file opens when I respawn apache, but I'm hoping there is a way to map file 7 to a real filename... is there a way to do this?

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• Easily manage vsftpd virtual users?

  - by Phil

  I have a vsftpd server configured with many virtual users. logins are stored in a Berkeley DB file One configuration file exists for each user to define his permissions (read-only or read-write, home directory, etc.). To do that, I use the user_config_dir parameter (set in vsftpd.conf). I am wondering if it would be possible to manage these virtual users from a simple GUI (such as web interface). I have found some tools but they are limited to generic vsftpd configuration, not virtual users management. Otherwise, PAM-MySQL seems to be a good way to manage users efficiently but only username/password and logs can be stored in database, not permissions. Finally, I've found this thread, but the solution is a bit awkward... Is there any way to easily manage the vsftpd users ?

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• Firing through HTTP a Perl script for sending signals to daemons

  - by Eric Fortis

  Hello guys, I'm using apache2 on Ubuntu. I have a Perl script which basically read the files names of a directory, then rewrites a text file, then sends a signal to a daemon. How can this be done, as secure as possible through a web-page? Actually I can run the code below, but not if I remove the comments. I'm looking for advise considering: Using HTTP Requests? How about Apache file permissions on the directory shown in code? Is htaccess enough to enable user/pass access to the cgi? Should I use a database instead of writing to a file and run a cron querying the db with permission granted to write and send the signal? Granting as less permissions as possible to the webserver. Should I set a VPN? #!/usr/bin/perl -wT use strict; use CGI; #@fileList = </home/user/*>; #read a directory listing my $query = CGI->new(); print $query->header( "text/html" ), $query->p( "FirstFileNameInArray" ), #$query->p( $fileList[0] ), #output the first file in directory $query->end_html;

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