Search Results

Search found 14709 results on 589 pages for 'root permission'.

Page 10/589 | < Previous Page | 6 7 8 9 10 11 12 13 14 15 16 17  | Next Page >

• Grails default root path

  - by srinath

  Hi, Is there a variable where we can find out the root directory of my Grails application? I tried request.getSession().getServletContext().getRealPath("/") But shows tmp/App-Test-0.1/ . My app is located in tomcat "/home/srinath/work/projects/tomcat-6.0.18/webapps/App-Test-0.1" could any one help me . thanks in advance, sri..

  Read the article

• how to run python scripts from other script and have thier root in my root

  - by Ahmad Dwaik

  hello all i have a module "B", i want to run it from a script "C", and i want to call global variables in "B", as they wer in "C" root, another problem that is if i imported sys in "B" when i run "C" it doesnt see sys # NameError: global name 'sys' is not defined # what shall i do?

  Read the article

• Execute an external application as root - problem

  - by user598011

  Good morning: I'm trying to run an external application that needs to be executed as root. I have to read the lines from exit after the execution of this application but it says "permission denied", as if the its not been done correctly. I've been thinking over a time and I can not move forward. The code is as follows: process = Runtime.getRuntime().exec("su"); String[] command = {external application command}; process = Runtime.getRuntime().exec(comando); InputStream inputStream = process.getInputStream(); BufferedReader bufferedReader = null; try { bufferedReader = new BufferedReader(new InputStreamReader(inputStream),8192); String line = null; while ((line = bufferedReader.readLine()) != null) { System.out.println("read line:"+line ); } } catch (IOException ioe) { ioe.printStackTrace(); } process.waitFor(); Does anyone know why not let me run the command? Thanks.

  Read the article

• Why can a local root turn into any LDAP user?

  - by Daniel Gollás

  I know this has been asked here before, but I am not satisfied with the answers and don't know if it's ok to revive and hijack an older question. We have workstations that authenticate users on an LDAP server. However, the local root user can su into any LDAP user without needing a password. From my perspective this sounds like a huge security problem that I would hope could be avoided at the server level. I can imagine the following scenario where a user can impersonate another and don't know how to prevent it: UserA has limited permissions, but can log into a company workstation using their LDAP password. They can cat /etc/ldap.conf and figure out the LDAP server's address and can ifconfig to check out their own IP address. (This is just an example of how to get the LDAP address, I don't think that is usually a secret and obscurity is not hard to overcome) UserA takes out their own personal laptop, configures authentication and network interfaces to match the company workstation and plugs in the network cable from the workstation to their laptop, boots and logs in as local root (it's his laptop, so he has local root) As root, they su into any other user on LDAP that may or may not have more permissions (without needing a password!), but at the very least, they can impersonate that user without any problem. The other answers on here say that this is normal UNIX behavior, but it sounds really insecure. Can the impersonated user act as that user on an NFS mount for example? (the laptop even has the same IP address). I know they won't be able to act as root on a remote machine, but they can still be any other user they want! There must be a way to prevent this on the LDAP server level right? Or maybe at the NFS server level? Is there some part of the process that I'm missing that actually prevents this? Thanks!!

  Read the article

• How to make a non-root user to use chown for any user group files?

  - by user1877716

  I would like to make a user super powerful, with almost all root rights but unable to touch a the root user (to change the password of the root). My goal is to user "B" to manage my web server. The problem is user B need to able to run the chown and chmod commands on some files belonging to other users. I tried to put B in root group or use visudo, but it's not enough. I'm working an Centos 6 system. If some body have ideas!

  Read the article

• What's the safest way to kick off a root-level process via cgi on an Apache server?

  - by MartyMacGyver

  The problem: I have a script that runs periodically via a cron job as root, but I want to give people a way to kick it off asynchronously too, via a webpage. (The script will be written to ensure it doesn't run overlapping instances or such.) I don't need the users to log in or have an account, they simply click a button and if the script is ready to be run it'll run. The users may select arguments for the script (heavily filtered as inputs) but for simplicity we'll say they just have the button to choose to press. As a simple test, I've created a Python script in cgi-bin. chown-ing it to root:root and then applying "chmod ug+" to it didn't have the desired results: it still thinks it has the effective group of the web server account... from what I can tell this isn't allowed. I read that wrapping it with a compiled cgi program would do the job, so I created a C wrapper that calls my script (its permissions restored to normal) and gave the executable the root permissions and setuid bit. That worked... the script ran as if root ran it. My main question is, is this normal (the need for the binary wrapper to get the job done) and is this the secure way to do this? It's not world-facing but still, I'd like to learn best practices. More broadly, I often wonder why a compiled binary is more "trusted" than a script in practice? I'd think you'd trust a file that was human-readable over a cryptic binaryy. If an attacker can edit a file then you're already in trouble, more so if it's one you can't easily examine. In short, I'd expect it to be the other way 'round on that basis. Your thoughts?

  Read the article

• How can I start the desktop without having to "startx"?

  - by gtldsp

  I dont want to start every time startx is there any way to get GUI Direct login screen. my files are root@ubuntu:~# locate org.conf /usr/share/X11/xorg.conf.d /usr/share/X11/xorg.conf.d/10-evdev.conf /usr/share/X11/xorg.conf.d/11-evdev-quirks.conf /usr/share/X11/xorg.conf.d/11-evdev-trackpoint.conf /usr/share/X11/xorg.conf.d/50-synaptics.conf /usr/share/X11/xorg.conf.d/50-vmmouse.conf /usr/share/X11/xorg.conf.d/50-wacom.conf /usr/share/X11/xorg.conf.d/51-synaptics-quirks.conf /usr/share/man/man5/xorg.conf.5.gz /usr/share/man/man5/xorg.conf.d.5.gz root@ubuntu:~# cd /usr/share/X11/xorg.conf.d root@ubuntu:/usr/share/X11/xorg.conf.d# ll total 36 drwxr-xr-x 2 root root 4096 Apr 23 04:38 ./ drwxr-xr-x 5 root root 4096 Apr 23 04:38 ../ -rw-r--r-- 1 root root 1099 Apr 4 17:04 10-evdev.conf -rw-r--r-- 1 root root 590 Mar 15 08:52 11-evdev-quirks.conf -rw-r--r-- 1 root root 364 Mar 15 08:52 11-evdev-trackpoint.conf -rw-r--r-- 1 root root 956 Apr 13 06:00 50-synaptics.conf -rw-r--r-- 1 root root 115 Mar 22 09:54 50-vmmouse.conf -rw-r--r-- 1 root root 842 Mar 30 03:13 50-wacom.conf -rw-r--r-- 1 root root 590 Apr 13 05:59 51-synaptics-quirks.conf root@ubuntu:/usr/share/X11/xorg.conf.d# Please provide me step by step details.

  Read the article

• Can I distribute a software with the following permission notice

  - by Parham

  I've recently written a piece of software (without any other contributors) for a company which I part own. I was wondering if I could distribute it with the following permission notice, which is a modified version of the MIT License. Are there any obvious risks if I do distribute with this licence and does it give me the right to reuse the code in other projects? Permission is hereby granted, to any person within CompanyName (the "Company") obtaining a copy of this software and associated documentation files, excluding any third party libraries (the "Software"), to deal with the Software, with limitations restricted to use, copy, modify and merge, the Software may not be published, distributed, sublicensed and/or sold without the explicit permission from AuthorName (the "Author"). This notice doesn't apply to sections of the Software where copyright is held by any persons other than the Author. The Author remains the owner of the Software and may deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software. The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software. THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.

  Read the article

• What kind of permission is this? (Groups+Roles)

  - by Jorge

  I'm starting to need an access control for roles in my app. I don't know much of this, but I understand how vBulletin works: I create groups, then give permissions to groups. I think that what I need is the Role Bases Access Control (RBAC) , but i'm not sure, because I need groups to give permissions instead of single users (Maybe it's not that complicated to achieve). Example of what I'm thinking: Given a post: Editor's Group has permission to view it before it's published. Editor's Group has permission to edit its content. Public Group (Default) has not permission to view it before it's published. Admin Group has permission to delete the post. So basically I wan't orientation about if RBAC is what I need. And also, how would it be good to store group membership in a user, for example, would be good to have: ID NAME PASSWORD GROUPS (1, MyName, MyPassword, 1/2/3/4/5) and explode it via PHP or one registry for every Group membership in a table named permissions, example: USERID, USERGROUP values (1, 1), (1, 2) Maybe should be the second way because of the formal norms but I didn't study yet Databases 1 at college.

  Read the article

• SVN best practice - checking out root folder

  - by Stephen Dolier

  Hi all, quick question about svn checkout best practice. Once the structure of a repository is set up, ie trunk, branches, tags, is it normal to have the root checked out to our local machines. Or should you only check out the trunk if that's what you are working on or a branch if we so choose to create one. The reason i ask is that every time someone creates a branch or tag we all get a copy when we do an update. btw, we're recently migrated from vss.

  Read the article

• What Belongs to the Aggregate Root

  - by jlembke

  This is a practical Domain Driven Design question: Conceptually, I think I get Aggregate roots until I go to define one. I have an Employee entity, which has surfaced as an Aggregate root. In the Business, some employees can have work-related Violations logged against them: Employee-----*Violations Since not all Employees are subject to this, I would think that Violations would not be a part of the Employee Aggregate, correct? So when I want to work with Employees and their related violations, is this two separate Repository interactions by some Service? Lastly, when I add a Violation, is that method on the Employee Entity? Thanks for the help!

  Read the article

• Arrays- Square root of an Array and printing the result JAVA

  - by roger34

  Hello, The title says it all, really. I'm trying to get an array of (9) numbers squared then printed but I keep coming back with only one result - the number of numbers in the array squared- obviously not what I want. Thanks for any help. Ok, here is my terrible code so far. Trying to pass it to a method as well. public static void main ( String args[] ) { double[] nums = {126, 12.939, 795, 320.16, 110, 34.7676, 7773, 67, 567, 323}; System.out.println ("Square root is " +square); square(nums); } public static double square (double [] array) { double result; for( double i = 0; i < array.length ; i++ ) result = Math.sqrt(array[i]); return result; } }

  Read the article

• sudo make install: permission denied

  - by Wojtek Rzepala

  I have a most annoying question about building from sources. I've searched for an answer for a long while and nowhere can I find one... I've compiled software from sources before and this just baffled me. So I am trying to install python 2.7.2 from sources. I can do a successful ./configure, also make seems to run fine. But when I do sudo make install, I get a variety of errors... First, sudo make install gives me this: make: stat: GNUmakefile: Permission denied make: stat: makefile: Permission denied make: stat: Makefile: Permission denied make: stat: install: Permission denied make: *** No rule to make target `install'. Stop. So I did chmod +rx Makefile*. To no avail. Then, sudo ls . says ls: cannot access .: Permission denied Then ls -d . says the permissions are drwxr-x--- Then, as a desperate measure, chmod +rx .. That gave me: make: stat: Modules/config.c.in: Permission denied make: *** No rule to make target `Modules/config.c.in', needed by `Makefile'. Stop. So some progress... What is happening here? It looks like some sort of permission problem. I presumed that sudo would be the solution but clearly there is something else going on here... I tried sudo -s but I get those permission problems all over again... I am using Ubuntu 10.04LTS.

  Read the article

• XSLT: Add namespace to root element

  - by Ingrid

  I need to change namespaces in the root element as follows: input document: <?xml version="1.0" encoding="UTF-8" standalone="yes"?> <foo xsi:schemaLocation="urn:isbn:1-931666-22-9 http://www.loc.gov/ead/ead.xsd" xmlns:ns2="http://www.w3.org/1999/xlink" xmlns="urn:isbn:1-931666-22-9" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> desired output: <foo audience="external" xsi:schemaLocation="urn:isbn:1-931666-22-9 http://www.loc.gov/ead/ead.xsd" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="urn:isbn:1-931666-22-9"> I was trying to do it as I copy over the whole document and before I give any other transformation instructions, but the following doesn't work: <xsl:template match="* | processing-instruction() | comment()"> <xsl:copy copy-namespaces="no"> <xsl:for-each select="."> <xsl:attribute name="audience" select="'external'"/> <xsl:namespace name="xlink" select="'http://www.w3.org/1999/xlink'"/> </xsl:for-each> <xsl:apply-templates/> <xsl:copy-of select="@*"/> <xsl:apply-templates/> </xsl:copy> </xsl:template> Thanks for any advice!

  Read the article

• Cannot find root device after latest kernel upgrade

  - by DisgruntledGoat

  I'm running Ubuntu 13.04. Yesterday I tried to install updates but there was an error, and it suggested running apt-get -f install which I did. Now when I try to boot, I get an error "Gave up waiting for root device". The text is almost identical to the text shown in this and this question. However, the "built-in shell" simply doesn't work! Nothing I type shows up on the screen or does anything. I tried adding a rootdelay to grub but it just waits longer and shows the same screen. Loading the previous kernel works (although there are a few graphics glitches) but as far as I can tell, it should be booting the exact same stuff. The new kernel is 3.8.0-31-generic and the previous working one is 3.8.0-25-generic. Here is my entire /boot/grub/menu.lst file, comments removed: default 0 timeout 3 title Ubuntu 13.04, kernel 3.8.0-31-generic uuid c690c1e6-beb9-46e7-85c2-145cd07d44ac kernel /boot/vmlinuz-3.8.0-31-generic root=UUID=c690c1e6-beb9-46e7-85c2-145cd07d44ac rootdelay=120 ro quiet splash initrd /boot/initrd.img-3.8.0-31-generic quiet title Ubuntu 13.04, kernel 3.8.0-31-generic (recovery mode) uuid c690c1e6-beb9-46e7-85c2-145cd07d44ac kernel /boot/vmlinuz-3.8.0-31-generic root=UUID=c690c1e6-beb9-46e7-85c2-145cd07d44ac ro single initrd /boot/initrd.img-3.8.0-31-generic title Ubuntu 13.04, kernel 3.8.0-25-generic uuid c690c1e6-beb9-46e7-85c2-145cd07d44ac kernel /boot/vmlinuz-3.8.0-25-generic root=UUID=c690c1e6-beb9-46e7-85c2-145cd07d44ac ro quiet splash initrd /boot/initrd.img-3.8.0-25-generic quiet title Ubuntu 13.04, kernel 3.8.0-25-generic (recovery mode) uuid c690c1e6-beb9-46e7-85c2-145cd07d44ac kernel /boot/vmlinuz-3.8.0-25-generic root=UUID=c690c1e6-beb9-46e7-85c2-145cd07d44ac ro single initrd /boot/initrd.img-3.8.0-25-generic title Ubuntu 13.04, kernel 3.8.0-23-generic uuid c690c1e6-beb9-46e7-85c2-145cd07d44ac kernel /boot/vmlinuz-3.8.0-23-generic root=UUID=c690c1e6-beb9-46e7-85c2-145cd07d44ac ro quiet splash initrd /boot/initrd.img-3.8.0-23-generic quiet title Ubuntu 13.04, kernel 3.8.0-23-generic (recovery mode) uuid c690c1e6-beb9-46e7-85c2-145cd07d44ac kernel /boot/vmlinuz-3.8.0-23-generic root=UUID=c690c1e6-beb9-46e7-85c2-145cd07d44ac ro single initrd /boot/initrd.img-3.8.0-23-generic title Ubuntu 13.04, memtest86+ uuid c690c1e6-beb9-46e7-85c2-145cd07d44ac kernel /boot/memtest86+.bin quiet title -------------------------------- root title Windows Vista rootnoverify (hd0,2) savedefault makeactive chainloader +1 As you can see the UUID is the same for all kernels. Why am I getting this problem, and what can I do to fix it?

  Read the article

• Adobe flash plug-in crashed/needs permission in Chromium

  - by sally

  A couple of days ago I started to get these messages when trying to watch videos on Youtube and a couple of other sites. (Using Chromium browser) Normally starts by just seeing a black screen with Missing Plug-in faintly visible, but will also ask me to give permission to run both at top (yellow bar), and if I tell it to 'allow' it follows up with the white puzzle piece on black background requesting permission. Occasionally I have seen the message telling me the flash-player has crashed, too. Giving permission does not fix the problem. Have also navigated to 'about:plugins' and checked that Adobe flash is enabled and permitted. I have just done a fresh installation of Lubuntu 11.10 as a last resort but problem still not fixed.

  Read the article

• Error while running Jetty Server on port 80 as non root user

  - by user75016

  All, I was trying to setup jetty on port 80 but its giving exception saying permission denied as below. I have setup jetty to work with setuid and configured start.ini as follows: OPTIONS=Server,jsp,jmx,resources,websocket,ext,plus,annotations,jta,jdbc,setuid (below as first configuration file in start.ini) etc/jetty-setuid.xml and jetty-setuid.xml file with username and group name of non root user. 2012-07-03 15:29:02.411:INFO:oejdp.ScanningAppProvider:Deployment monitor /opt/jetty-hightide-8.1.3.v20120416/contexts at interval 1 2012-07-03 15:29:02.454:WARN:oejuc.AbstractLifeCycle:FAILED [email protected]:80: java.net.SocketException: Permission denied java.net.SocketException: Permission denied at sun.nio.ch.Net.bind(Native Method) at sun.nio.ch.ServerSocketChannelImpl.bind(ServerSocketChannelImpl.java:126) at sun.nio.ch.ServerSocketAdaptor.bind(ServerSocketAdaptor.java:59) at org.eclipse.jetty.server.nio.SelectChannelConnector.open(SelectChannelConnector.java:182) at org.eclipse.jetty.server.AbstractConnector.doStart(AbstractConnector.java:311) at org.eclipse.jetty.server.nio.SelectChannelConnector.doStart(SelectChannelConnector.java:260) at org.eclipse.jetty.util.component.AbstractLifeCycle.start(AbstractLifeCycle.java:59) at org.eclipse.jetty.server.Server.doStart(Server.java:273) at org.eclipse.jetty.util.component.AbstractLifeCycle.start(AbstractLifeCycle.java:59) at org.eclipse.jetty.xml.XmlConfiguration$1.run(XmlConfiguration.java:1215) at java.security.AccessController.doPrivileged(Native Method) at org.eclipse.jetty.xml.XmlConfiguration.main(XmlConfiguration.java:1138) 2012-07-03 15:29:02.455:WARN:oejuc.AbstractLifeCycle:FAILED org.eclipse.jetty.server.Server@66da9ea4: java.net.SocketException: Permission denied java.net.SocketException: Permission denied

  Read the article

• Restrict access to apache2 web root but allow it to subfolders

  - by razor7

  I need to restrict access by password to my web root apache test server (ie http://localhost) but allow access to subfolders (ie: http://localhost/testsite) I did create the .htpasswd and .htaccess, and put the .htaccess to web root (http://localhost) so when trying to access web root, it asks for user and pass, but so does in subfolders (ie: trying to access http://localhost/testite) I want to be asked for password on web root, but not on subfolders. Is that possible?

  Read the article

• Change permission of raw device by UUID

  - by Omid Kosari

  I am using "Apache Traffic Server" which may and it is better to use raw device like /dev/sdb without mounting it as a directory . So i should change the permission of /dev/sdb from root to user "trafficserver" to allow it formatting whole device as it likes . The problem starts after each reboot because /dev/sdb changes randomize so i should use UUID . But i can not change the permission of /dev/disk/by-uuid/48c2fac6-06eb-42c7-8b8a-3b83b781f28c . Is there a way to permanently change permission of disk drive by UUID ?

  Read the article

• Setting folder permission so it isn't deletable, but everything else is allowed

  - by user10324

  I want to set permissions to a folder such that, when I'm normally logged in (meaning not as root), this folder isn't deletable, but I can still change the name of this folder and move in around in my system ? IS this possible and if yes, how to set the permission ? I already tried different combinations for the permissions but couldn't figure out how to do it. Also (side question), if some folder hast some permission set and I copy this folder, along with its contents (assuming I am allowed to read it) to a memory, are these permission preserved ? I suppose not, since under Windows I could delete the folder on the stick, even if under Ubuntu it isn't allowed to delete it...

  Read the article

• Hardening non-root standalone Linux Tomcat install

  - by NoozNooz42

  I want to know if you have any tips as to how to strengthen the security of a non-root install of Tomcat in standalone mode once Tomcat is already installed in a non-root account, in standalone mode. I precise this because, for example, I'm not at all interested by the answers given here (because both Java and Tomcat requires root priviledges there to be installed and I've got zero interest in running jsvc): http://serverfault.com/questions/43765 So far, here's what I've done for my non-root standalone Tomcat 6 install: download and install the JRE .bin provided by Oracle/Sun (no need to be root here) (no need for a full JDK anymore right seen that Jasper [Tomcat's JSP engine] has its own compiler now right?) download and tar -xzf tomcat 6 (no need to be root here) set up transparent port-forwarding (must be root here) Note that my distribution is a Debian one and I have exactly zero interest in downloading Debian package / backports / whatever... Because, once again, I DO NOT want to need to be root to install Java & Tomcat. The only moment I needed to be root was to configure the firewall to transparently do the port forwarding 80 <-- 8080 and 443 <-- 8443. I then deleted all the default webapps but one: cd ~/apache-tomcat-6.0.26/webapps rm -rf docs rm -rf examples/ rm -rf manager/ rm -rf ROOT/ What about the directory ~/apache-tomcat-6.0.26/webapps/host-manager, do I need it or can I delete it? So, once I've installed Tomcat standalone in a non-root account (and taken into account that I don't want to enter the root password anymore and that I don't plan to install the whole Apache shebang), what more can I do? Are there connectors I can disable? (how?)

  Read the article

• Display all images from outside web root folder using PHP

  - by micmola

  Hello, I want to display all images that are stored outside my web root folder. Please help me. I am only able to display one image repeatedly. For example, if I have 5 images in my folder, only one image is displayed on my browser 5 times. Please help me on this. I've been working on this problem for over a month now. I'm a newbie. Help. Thank you. Here is the code I'm using. images.php <?php // Get our database connector require("includes/copta.php"); // Grab the data from our people table $sql = "select * from people"; $result = mysql_query($sql) or die ("Could not access DB: " . mysql_error()); $imgLocation = " /uploadfile/"; while ($row = mysql_fetch_array($result)) { $imgName = $row["filename"]; $imgPath = $imgLocation . $imgName; echo "<img src=\"call_images.php?imgPath=" . $imgName . "\" alt=\"\"><br/>"; echo $row['id'] . " " . $imgName. "<br />"; } ?> call_images.php <?php // Get our database connector require("includes/copta.php"); $imgLocation = '/ uploadz/'; $sql = "select * from people"; $result = mysql_query($sql) or die ("Could not access DB: " . mysql_error()); while ($row = mysql_fetch_array($result)) { $imgName = $row["filename"]; $imgPath = $imgLocation . $imgName; // Make sure the file exists if(!file_exists($imgPath) || !is_file($imgPath)) { header('HTTP/1.0 404 Not Found'); die('The file does not exist'); } // Make sure the file is an image $imgData = getimagesize($imgPath); if(!$imgData) { header('HTTP/1.0 403 Forbidden'); die('The file you requested is not an image.'); } // Set the appropriate content-type // and provide the content-length. header("Pragma: public"); header("Expires: 0"); header("Cache-Control: must-revalidate, post-check=0, pre-check=0"); header("Content-Type: image/jpg"); header("Content-length: " . filesize($imgPath)); // Print the image data readfile($imgPath); exit(); } ?>

  Read the article

• Anonymous user permission issue in SharePoint Server

  - by George2

  Hello everyone, I am using SharePoint Server 2007 x64 and Windows Server 2008 x64. I have setup a site with template publishing portal. I have grant anonymous access to all the site. My question is, if I create a new page, how to grant permission to anonymous user to access (read permission) the page? I got this confusion because, for the permission setting of a page, a permission is set according to a user name (e.g. read permission for user "foo" of a page). Since anonymous user does not have a related "user name", how could I grant read permission to anonymous user? BTW: I use Windows Forms authentication and Windows NTLM authentication for my sites. Thanks in advance, George

  Read the article

• MegaCli newly created disk doesn't appear under /dev/sdX

  - by Henry-Nicolas Tourneur

  After having successfully added 2 new disks in a new RAID virtual drive (background initialization done), I would have exepected it to appear under /dev/sdh but it's not there (so, unusable). The system is running a CentOS 5.2 64 bits, HAL and udev daemons are running, not records of any sdh apparition under the messsage log file or in dmesg, only MegaCli do see that virtual drive. Any idea ? Some data: [root@server ~]# ./MegaCli -LDInfo -LALL -a0 Adapter 0 -- Virtual Drive Information: Virtual Disk: 0 (target id: 0) Name: RAID Level: Primary-1, Secondary-0, RAID Level Qualifier-0 Size:139392MB State: Optimal Stripe Size: 64kB Number Of Drives:2 Span Depth:1 Default Cache Policy: WriteBack, ReadAheadNone, Direct, No Write Cache if Bad BBU Current Cache Policy: WriteBack, ReadAheadNone, Direct, No Write Cache if Bad BBU Access Policy: Read/Write Disk Cache Policy: Disk's Default Virtual Disk: 1 (target id: 1) Name: RAID Level: Primary-1, Secondary-0, RAID Level Qualifier-0 Size:285568MB State: Optimal Stripe Size: 64kB Number Of Drives:2 Span Depth:1 Default Cache Policy: WriteBack, ReadAheadNone, Direct, No Write Cache if Bad BBU Current Cache Policy: WriteBack, ReadAheadNone, Direct, No Write Cache if Bad BBU Access Policy: Read/Write Disk Cache Policy: Disk's Default [root@server ~]# ls -l /dev/disk/by-id/scsi-360* lrwxrwxrwx 1 root root 9 Nov 17 2010 /dev/disk/by-id/scsi-36001ec90f82fe100108ca0a704098d09 -> ../../sda lrwxrwxrwx 1 root root 10 Nov 17 2010 /dev/disk/by-id/scsi-36001ec90f82fe100108ca0a704098d09-part1 -> ../../sda1 lrwxrwxrwx 1 root root 10 Nov 17 2010 /dev/disk/by-id/scsi-36001ec90f82fe100108ca0a704098d09-part2 -> ../../sda2 lrwxrwxrwx 1 root root 9 Nov 17 2010 /dev/disk/by-id/scsi-36090a028e0fe07e78f94940c0000a0ee -> ../../sdf lrwxrwxrwx 1 root root 10 Nov 17 2010 /dev/disk/by-id/scsi-36090a028e0fe07e78f94940c0000a0ee-part1 -> ../../sdf1 lrwxrwxrwx 1 root root 9 Nov 17 2010 /dev/disk/by-id/scsi-36090a028e0fe972a3f91240a0000005f -> ../../sdb lrwxrwxrwx 1 root root 10 Nov 17 2010 /dev/disk/by-id/scsi-36090a028e0fe972a3f91240a0000005f-part1 -> ../../sdb1 lrwxrwxrwx 1 root root 9 Nov 17 2010 /dev/disk/by-id/scsi-36090a028e0fea7e18f94640c000020ec -> ../../sde lrwxrwxrwx 1 root root 10 Nov 17 2010 /dev/disk/by-id/scsi-36090a028e0fea7e18f94640c000020ec-part1 -> ../../sde1 lrwxrwxrwx 1 root root 9 Nov 17 2010 /dev/disk/by-id/scsi-36090a028e0feb7da8f94340c0000203d -> ../../sdd lrwxrwxrwx 1 root root 10 Nov 17 2010 /dev/disk/by-id/scsi-36090a028e0feb7da8f94340c0000203d-part1 -> ../../sdd1 lrwxrwxrwx 1 root root 9 Nov 17 2010 /dev/disk/by-id/scsi-36090a028e0fed7d78f94040c000080b7 -> ../../sdc lrwxrwxrwx 1 root root 10 Nov 17 2010 /dev/disk/by-id/scsi-36090a028e0fed7d78f94040c000080b7-part1 -> ../../sdc1 lrwxrwxrwx 1 root root 9 Nov 17 2010 /dev/disk/by-id/scsi-36090a05830145e58e0b9c479000010a1 -> ../../sdg lrwxrwxrwx 1 root root 10 Nov 17 2010 /dev/disk/by-id/scsi-36090a05830145e58e0b9c479000010a1-part1 -> ../../sdg1

  Read the article

• Can an aggregate root hold references of members of another aggregate root?

  - by Rushino

  Hello, I know outside aggregates cant change anything inside an aggregate without passing by his root. That said i would like to know if an aggregate root can hold references of members (objects insides) of another aggregate root? (fellowing DDD rules) Example : a Calendar contain a list of phases which contain a list of sequences which contain a list of assignations Calendar is root because phases and sequences and assignations only work in context of a calendar. You also have Students and Groups of student (called groups) It is possible (fellowing DDD rules) to make Groups holding references of assignations or it need to pass by the root for accessing groups from assignations ? Thanks.

  Read the article

< Previous Page | 6 7 8 9 10 11 12 13 14 15 16 17  | Next Page >