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  • Cobbler 2.2.2 problems

    - by Peter
    I have setup a dedicated LAN for Cobbler tests. My setup is: Cobbler server: openSUSE 12.3, cobbler 2.2.2 (from openSUSE repos) Imported distros: Centos 6.5, Red Hat 6.5, Red Hat 7.0, openSUSE 13.1 Target Machine: VMs in a Windows 7 Virtualbox Systems provisioning works OK, but I have some problems. The first one is that cobbler does not honor the "pxe_just_once: 1" setting. When the setup of the target OS is finished, after the reboot the target systems continues to PXE boot! The second problem is that the target server is not correctly configured! See my setup: cobbler system report --name=test Name : test TFTP Boot Files : {} Comment : Fetchable Files : {} Gateway : 192.168.0.1 Hostname : testcob1.example.com Image : IPv6 Autoconfiguration : False IPv6 Default Device : Kernel Options : {} Kernel Options (Post Install) : {} Kickstart : <<inherit>> Kickstart Metadata : {} LDAP Enabled : False LDAP Management Type : authconfig Management Classes : [] Management Parameters : <<inherit>> Monit Enabled : False Name Servers : ['192.168.0.1', '8.8.8.8'] Name Servers Search Path : [] Netboot Enabled : False Owners : ['admin'] Power Management Address : Power ID : Power Password : Power Management Type : ipmitool Power Username : Profile : RHEL-6.5-x86_64 Proxy : <<inherit>> Red Hat Management Key : <<inherit>> Red Hat Management Server : <<inherit>> Repos Enabled : False Server Override : <<inherit>> Status : testing Template Files : {} Virt Auto Boot : <<inherit>> Virt CPUs : <<inherit>> Virt Disk Driver Type : <<inherit>> Virt File Size(GB) : <<inherit>> Virt Path : <<inherit>> Virt RAM (MB) : <<inherit>> Virt Type : <<inherit>> Interface ===== : eth0 Bonding Opts : Bridge Opts : DHCP Tag : DNS Name : Master Interface : Interface Type : IP Address : 192.168.0.200 IPv6 Address : IPv6 Default Gateway : IPv6 MTU : IPv6 Secondaries : [] IPv6 Static Routes : [] MAC Address : Management Interface : True MTU : Subnet Mask : 255.255.255.0 Static : True Static Routes : [] Virt Bridge : So, although I have setup the hostname and the network interface of the target system, after the setup, the hostname is set to localhost.localdomain and eth0 is configured as a DHCP not static! How can I find the problem and fix it? Note that I have synced and restarted cobbler a couple of times, but the problems persists.

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  • How to poll the username, when having the UID?

    - by JMW
    we're using ldap with sssd for the usermanagement, so our users are not in the "/etc/passwd" Unfortunately, ps just shows the UIDs: [root@xyz ~]# id jmw uid=1582(jmw) gid=1582(jmw) groups=1582(jmw), 1000(admins) [root@xyz ~]# ps aux [..cutting some output..] 1582 26794 25.0 0.4 190420 38508 ? S 12:15 0:00 /usr/bin/php-cgi -c php.ini [..cutting some output..] How can i poll the username, that belongs to a UID? ( a grep ':1582:' /etc/passwd doesn't work ;-) )

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  • Windows Server 2008 R2, IIS 7.5, Windows Authentication

    - by nick
    Ive a 7.5 IIS installed on my server with windows authentication enabled cause i need it for NTLM / SSO on intranet pages. when windows authentication is activated, iis cant authenticate himself on his own webserver.. thats the error i got in the iis log: 2011-11-24 08:47:10 10.50.2.91 POST /ldap.php - 80 - 10.50.2.91 SWIFT_LoginShare 401 2 5 0 so.. how can i make sure, using windows authentication, that iis authenticates himself? thx for your help

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  • Where are AD registry settings

    - by erdogany
    As you might already know AD does not let you change the password through LDAP if you don't have a secure connection to AD. I heard that there is a registry setting to change this behavior but so far couldn't find where it is... Anyone of you know how can I can change this behavior?

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  • How can I put together services bettwen differents servers?

    - by poz2k4444
    For a schoolar project, I have to run differents services in a lab enviroment where I'll have 6 computers working as servers, what services can I put together, and what cannot be, in order to prevent security risks, and considereiting that if one service goes down, affects less possible the function of the server farm, the services are: MySql Http for intranet Https DHCP IPP SMTP LDAP VPN SSH NTP DNS NFS I'll use linux

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  • Enterprise Wireless Authentication without Active Directory

    - by ank
    We are in the process of redoing our wireless access network and would like to know if there is any method to get Windows clients/users access to the network using 802.1x WITHOUT having an Active Directory server for authentication and WITHOUT installing additional software on each and every client. Note that we already use Radius servers, LDAP servers (all on CentOS). Users employ a variety of clients including Windows, Mac, Linux, Android, iOS.

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  • How to I create a user that is allowed to only add/remove users to distribution lists in Active Directory?

    - by Sorin Sbarnea
    I do have a third party product (Jira) that has Active Directory integration via LDAP. I want to enable Jira administrators to edit group memberships and have them syncronized inside Active Directory. This currently works but I needed to use a Domain Administrator service account in order to do this. The question is how can I do this without giving the entire Domain Administrator permission to the service account.

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  • How to find a user's (or mine) access rights on Windows Server 2008?

    - by Faiz
    I was given access to a Windows Server 2008 box and I need to check what all permissions I have on that box (if possible in the entire domain). I don't have access to domain controller and I don't want to write LDAP queries but just some GUI option or some command line stuff. Is there anyway? PS: I am not in to network administration, I am a BI developer. Pardon me if asked a stupid question.

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  • Does anyone have a valid and working example of OpenLDAP meta backend?

    - by QXT
    I have been Google'ing my fingers off and simply can not find a working example of how to merge/proxy a OpenLDAP server and windows AD server. Have anyone worked with this before? Any suggestions would be appreciated. The idea is simple: openldap.mydomain.local ---- Linux LDAP Server winad.mydomain.local ---- Windows AD Server Some users are one Linux and some on WinAD. OpenLDAP should search both on login. A working example would be appreciated.

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  • Access a remote active directory

    - by theXs
    I'm currently trying to query a remote Active Directory on a Windows Server 2008 R2. However, I'm not able to query the directory if I enter the following string in the cmd line: dsquery user -name m* -s ip:389 -u -p Furthermore, I tried to access the directory with: ldap://: but it didn't work either. I received the following error message: The server is not operational. Is there an option with which I can enable the remote access of an Active Directory?

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  • Google chrome proxy authentication dialogue timeout

    - by Nihar Sarangi
    I am on a network that uses LDAP proxy for authentication based on a username and password. Whenever I start Google Chrome, it pops up with a proxy authentication dialogue, but the dialogue disappears automatically after variable amount of time (sometimes it stays for 5 seconds some times less than 1 second). I have found the same issue with Chromium also. Is there any configuration I can set to control this timeout, or say, auto-authenticate with my authentication details from the shell or DE (Gnome3 on Arch)?

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  • sqlite3 permission denied android

    - by Sephy
    Hi, I'm trying to access the database of the application I'm developping directly on my Nexus, but I get a "permission denied" when I tried to execute the "sqlite3" command. I also tried to start the adb in root mod, but again, permission denied on the device... I guess I will have to do that with the emulator but I have a lot of data to load and it would have been 10 times faster with the phone on Wifi than the emulator... Unless someone has any idea? thanks

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  • [Java] RSA BadPaddingException : data must start with zero

    - by Robin Monjo
    Hello everyone. I try to implement an RSA algorithm in a Java program. I am facing the "BadPaddingException : data must start with zero". Here are the methods used to encrypt and decrypt my data : public byte[] encrypt(byte[] input) throws Exception { Cipher cipher = Cipher.getInstance("RSA/ECB/PKCS1Padding");// cipher.init(Cipher.ENCRYPT_MODE, this.publicKey); return cipher.doFinal(input); } public byte[] decrypt(byte[] input) throws Exception { Cipher cipher = Cipher.getInstance("RSA/ECB/PKCS1Padding");/// cipher.init(Cipher.DECRYPT_MODE, this.privateKey); return cipher.doFinal(input); } privateKey and publicKey attributes are read from files this way : public PrivateKey readPrivKeyFromFile(String keyFileName) throws IOException { PrivateKey key = null; try { FileInputStream fin = new FileInputStream(keyFileName); ObjectInputStream ois = new ObjectInputStream(fin); BigInteger m = (BigInteger) ois.readObject(); BigInteger e = (BigInteger) ois.readObject(); RSAPrivateKeySpec keySpec = new RSAPrivateKeySpec(m, e); KeyFactory fact = KeyFactory.getInstance("RSA"); key = fact.generatePrivate(keySpec); ois.close(); } catch (Exception e) { e.printStackTrace(); } return key; } Private key and Public key are created this way : public void Initialize() throws Exception { KeyPairGenerator keygen = KeyPairGenerator.getInstance("RSA"); keygen.initialize(2048); keyPair = keygen.generateKeyPair(); KeyFactory fact = KeyFactory.getInstance("RSA"); RSAPublicKeySpec pub = fact.getKeySpec(keyPair.getPublic(), RSAPublicKeySpec.class); RSAPrivateKeySpec priv = fact.getKeySpec(keyPair.getPrivate(), RSAPrivateKeySpec.class); saveToFile("public.key", pub.getModulus(), pub.getPublicExponent()); saveToFile("private.key", priv.getModulus(), priv.getPrivateExponent()); } and then saved in files : public void saveToFile(String fileName, BigInteger mod, BigInteger exp) throws IOException { FileOutputStream f = new FileOutputStream(fileName); ObjectOutputStream oos = new ObjectOutputStream(f); oos.writeObject(mod); oos.writeObject(exp); oos.close(); } I can't figured out how the problem come from. Any help would be appreciate ! Thanks in advance.

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  • how does linear probing handle this?

    - by Weadadada Awda
    • the hash function: h(x) = | 2x + 5 | mod M • a bucket array of capacity N • a set of objects with keys: 12, 44, 13, 88, 23, 94, 11, 39, 20, 16, 5 (to input from left to right) 4.a [5 pts] Write the hash table where M=N=11 and collisions are handled using linear probing. So I got up to here x x x x x 44 88 12 23 13 94 but the next variable should go after the 94 now, (the 11) but does it start from the beggining or what? thx

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  • Resultant of a polynomial with x^n–1

    - by devin.omalley
    Resultant of a polynomial with x^n–1 (mod p) I am implementing the NTRUSign algorithm as described in http://grouper.ieee.org/groups/1363/lattPK/submissions/EESS1v2.pdf , section 2.2.7.1 which involves computing the resultant of a polynomial. I keep getting a zero vector for the resultant which is obviously incorrect. private static CompResResult compResMod(IntegerPolynomial f, int p) { int N = f.coeffs.length; IntegerPolynomial a = new IntegerPolynomial(N); a.coeffs[0] = -1; a.coeffs[N-1] = 1; IntegerPolynomial b = new IntegerPolynomial(f.coeffs); IntegerPolynomial v1 = new IntegerPolynomial(N); IntegerPolynomial v2 = new IntegerPolynomial(N); v2.coeffs[0] = 1; int da = a.degree(); int db = b.degree(); int ta = da; int c = 0; int r = 1; while (db > 0) { c = invert(b.coeffs[db], p); c = (c * a.coeffs[da]) % p; IntegerPolynomial cb = b.clone(); cb.mult(c); cb.shift(da - db); a.sub(cb, p); IntegerPolynomial v2c = v2.clone(); v2c.mult(c); v2c.shift(da - db); v1.sub(v2c, p); if (a.degree() < db) { r *= (int)Math.pow(b.coeffs[db], ta-a.degree()); r %= p; if (ta%2==1 && db%2==1) r = (-r) % p; IntegerPolynomial temp = a; a = b; b = temp; temp = v1; v1 = v2; v2 = temp; ta = db; } da = a.degree(); db = b.degree(); } r *= (int)Math.pow(b.coeffs[0], da); r %= p; c = invert(b.coeffs[0], p); v2.mult(c); v2.mult(r); v2.mod(p); return new CompResResult(v2, r); } There is pseudocode in http://www.crypto.rub.de/imperia/md/content/texte/theses/da_driessen.pdf which looks very similar. Why is my code not working? Are there any intermediate results I can check? I am not posting the IntegerPolynomial code because it isn't too interesting and I have unit tests for it that pass. CompResResult is just a simple "Java struct".

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  • Simple caesar cipher in java

    - by Max Canlas
    Hey I'm making a simple caesar cipher in Java using the formula [x- (x+shift-1) mod 127 + 1] I want to have my encrypted text to have the ASCII characters except the control characters(i.e from 32-127). How can I avoid the control characters from 0-31 applying in the encrypted text. Thank you.

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  • rotating bitmaps. In code.

    - by Marco van de Voort
    Is there a faster way to rotate a large bitmap by 90 or 270 degrees than simply doing a nested loop with inverted coordinates? The bitmaps are 8bpp and typically 2048*2400*8bpp Currently I do this by simply copying with argument inversion, roughly (pseudo code: for x = 0 to 2048-1 for y = 0 to 2048-1 dest[x][y]=src[y][x]; (In reality I do it with pointers, for a bit more speed, but that is roughly the same magnitude) GDI is quite slow with large images, and GPU load/store times for textures (GF7 cards) are in the same magnitude as the current CPU time. Any tips, pointers? An in-place algorithm would even be better, but speed is more important than being in-place. Target is Delphi, but it is more an algorithmic question. SSE(2) vectorization no problem, it is a big enough problem for me to code it in assembler Duplicates How do you rotate a two dimensional array?. Follow up to Nils' answer Image 2048x2700 - 2700x2048 Compiler Turbo Explorer 2006 with optimization on. Windows: Power scheme set to "Always on". (important!!!!) Machine: Core2 6600 (2.4 GHz) time with old routine: 32ms (step 1) time with stepsize 8 : 12ms time with stepsize 16 : 10ms time with stepsize 32+ : 9ms Meanwhile I also tested on a Athlon 64 X2 (5200+ iirc), and the speed up there was slightly more than a factor four (80 to 19 ms). The speed up is well worth it, thanks. Maybe that during the summer months I'll torture myself with a SSE(2) version. However I already thought about how to tackle that, and I think I'll run out of SSE2 registers for an straight implementation: for n:=0 to 7 do begin load r0, <source+n*rowsize> shift byte from r0 into r1 shift byte from r0 into r2 .. shift byte from r0 into r8 end; store r1, <target> store r2, <target+1*<rowsize> .. store r8, <target+7*<rowsize> So 8x8 needs 9 registers, but 32-bits SSE only has 8. Anyway that is something for the summer months :-) Note that the pointer thing is something that I do out of instinct, but it could be there is actually something to it, if your dimensions are not hardcoded, the compiler can't turn the mul into a shift. While muls an sich are cheap nowadays, they also generate more register pressure afaik. The code (validated by subtracting result from the "naieve" rotate1 implementation): const stepsize = 32; procedure rotatealign(Source: tbw8image; Target:tbw8image); var stepsx,stepsy,restx,resty : Integer; RowPitchSource, RowPitchTarget : Integer; pSource, pTarget,ps1,ps2 : pchar; x,y,i,j: integer; rpstep : integer; begin RowPitchSource := source.RowPitch; // bytes to jump to next line. Can be negative (includes alignment) RowPitchTarget := target.RowPitch; rpstep:=RowPitchTarget*stepsize; stepsx:=source.ImageWidth div stepsize; stepsy:=source.ImageHeight div stepsize; // check if mod 16=0 here for both dimensions, if so -> SSE2. for y := 0 to stepsy - 1 do begin psource:=source.GetImagePointer(0,y*stepsize); // gets pointer to pixel x,y ptarget:=Target.GetImagePointer(target.imagewidth-(y+1)*stepsize,0); for x := 0 to stepsx - 1 do begin for i := 0 to stepsize - 1 do begin ps1:=@psource[rowpitchsource*i]; // ( 0,i) ps2:=@ptarget[stepsize-1-i]; // (maxx-i,0); for j := 0 to stepsize - 1 do begin ps2[0]:=ps1[j]; inc(ps2,RowPitchTarget); end; end; inc(psource,stepsize); inc(ptarget,rpstep); end; end; // 3 more areas to do, with dimensions // - stepsy*stepsize * restx // right most column of restx width // - stepsx*stepsize * resty // bottom row with resty height // - restx*resty // bottom-right rectangle. restx:=source.ImageWidth mod stepsize; // typically zero because width is // typically 1024 or 2048 resty:=source.Imageheight mod stepsize; if restx>0 then begin // one loop less, since we know this fits in one line of "blocks" psource:=source.GetImagePointer(source.ImageWidth-restx,0); // gets pointer to pixel x,y ptarget:=Target.GetImagePointer(Target.imagewidth-stepsize,Target.imageheight-restx); for y := 0 to stepsy - 1 do begin for i := 0 to stepsize - 1 do begin ps1:=@psource[rowpitchsource*i]; // ( 0,i) ps2:=@ptarget[stepsize-1-i]; // (maxx-i,0); for j := 0 to restx - 1 do begin ps2[0]:=ps1[j]; inc(ps2,RowPitchTarget); end; end; inc(psource,stepsize*RowPitchSource); dec(ptarget,stepsize); end; end; if resty>0 then begin // one loop less, since we know this fits in one line of "blocks" psource:=source.GetImagePointer(0,source.ImageHeight-resty); // gets pointer to pixel x,y ptarget:=Target.GetImagePointer(0,0); for x := 0 to stepsx - 1 do begin for i := 0 to resty- 1 do begin ps1:=@psource[rowpitchsource*i]; // ( 0,i) ps2:=@ptarget[resty-1-i]; // (maxx-i,0); for j := 0 to stepsize - 1 do begin ps2[0]:=ps1[j]; inc(ps2,RowPitchTarget); end; end; inc(psource,stepsize); inc(ptarget,rpstep); end; end; if (resty>0) and (restx>0) then begin // another loop less, since only one block psource:=source.GetImagePointer(source.ImageWidth-restx,source.ImageHeight-resty); // gets pointer to pixel x,y ptarget:=Target.GetImagePointer(0,target.ImageHeight-restx); for i := 0 to resty- 1 do begin ps1:=@psource[rowpitchsource*i]; // ( 0,i) ps2:=@ptarget[resty-1-i]; // (maxx-i,0); for j := 0 to restx - 1 do begin ps2[0]:=ps1[j]; inc(ps2,RowPitchTarget); end; end; end; end;

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  • Server-side web browser for PHP?

    - by Lee
    If you wanted to re-create the DOM server-side and manipulate it in PHP, how would you choose to go about it? I'm looking for a fast, multi-user complete server-side web browser that can interface with PHP and run complete Javascript. Like Jaxer but something I can use with PHP... an extension would be fine, or even an Apache mod. Ideas?

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  • dynamically load PHP code from external file

    - by jera
    code is in a static class in an external file eg. /home/test/public_html/fg2/templatecode/RecordMOD/photoslide.mod how do I load this into my script on demand, and be able to call its functions ? I am a novice at php , so please explain your code. help is appreciated. Jer

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  • Emulator typing "="

    - by Anton
    Hello. I have a problem with Android Emulator. I have created avd and run it. But when it was started I can't work, because it emulate typing many "=" symbol. I launch this avd with -debug-all parameter and debuger write "could not handle (sym=61, mod=0, str=EQUAL) KEY [0x00d, down]". OS - Windows Vista. Last version Java platform, SDK, Eclipse. Thank you very mach if you can hep me =)

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  • Question about C behaviour for unsigned integer underflow

    - by nn
    I have read in many places that integer overflow is well-defined in C unlike the signed counterpart. Is underflow the same? For example: unsigned int x = -1; // Does x == UINT_MAX? Thanks. I can't recall where, but i read somewhere that arithmetic on unsigned integral types is modular, so if that were the case then -1 == UINT_MAX mod (UINT_MAX+1).

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  • How to not write the full url in javascript ?

    - by Axel
    Hi, I did a Mod rewrite for my website so the URLs looks like this : http://www.mydomain.com/health/54856 http://www.mydomain.com/economy/strategy/911025/ http://www.mydomain.com/tags/obama/new So, the problem is that i make AJAX calls to a file here : http://www.mydomain.com/login.php And i don't want to write the FULL url or even use ../ trick because there isn't a fixed number of folders. So, what i want now, is something worked for my code to access the login.php from the root whatever the domain name is : $.ajax({ type: "POST", url: "http://www.mydomain.com/login.php" });

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