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• How to change key binding for Tmux

  - by Severin

  I want to change the key binding in Tmux so I can use Ctrl + Alt instead of Ctrl + b This is my (unfortunately) not working try to do so. unbind C-b set -g prefix M-C What's wrong with this? Thought I followed the documentation for the keys.

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• SQL SERVER Create Primary Key with Specific Name when Creating Table

  It is interesting how sometimes the documentation of simple concepts is not available online. I had received email from one of the reader where he has asked how to create Primary key with a specific name when creating the table itself. He said, he knows the method where he can create the table and then [...]...Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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• 3 Tips on Keyword Research - Keywords Are the Key to Driving Traffic

  Whatever methods of marketing that you choose to use there will always be at least one thing in common with all of them. In order for people to see your advertising, you've got to know how to use key words and phrasing.

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• SEO is Still the Key to Your Internet Marketing Success

  After many years, Search Engine Optimization (SEO) is still the key to creating long term brand awareness, online visibility and attracting increased traffic to your website, enabling you to appear higher in the organic search rankings for a set of targeted, high value keyword phrases. SEO is a set of activities that will provide your website with higher Relevance and Authority, which the search engines use to determine how high your website should rank when a specific keyword phrase is searched for.

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• Keyboard problem with ubuntu 13.10 when holding key down

  - by Lachezar Raychev

  I have the fallowing problem after isntalling the new Ubuntu OS. When i press and hold a key, for example "p" it writes one time "p", and while i am holding it the other "pppp" that are written come with a huge delay between each other - like 1 second or more. If i want to hold down backspace to delete a string of 5 letters it takes me like 10 seconds to do it. Is this a reported problem or does anybody has a solution to this problem?

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• Update app.config key value at run time in WPF

  In this article I am going to show how we can update a key value in app.config file in WPF. For this add reference of System.Configuration in your application.

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• host key verification failed from "connect to server" utility

  - by rambo

  I am able to use SSH from terminal but I am not able to use it from "connect to Server.." utility. it is showing the error in the dialog box as below: Cannot display location "sftp://[email protected]:PORT/ "Host key verification failed" why so? from terminal using below command I am able to access the server: ubuntu# ssh -p 2222 [email protected] Description: Ubuntu 10.04.4 LTS Release: 10.04 Codename: lucid any help please. thank you in advance.

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• The Ultimate Key to Success in Search Engine Optimization

  Most of us seem to be under the impression that the only key to success in search engine optimization is building tons of inbound links. And there is no denying that inbound links do play a very important role in determining how well your pages ultimately rank. In fact, most search-engines won't index a page that doesn't have a link from a page they have already indexed.

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• ssh key error - Permission denied (publickey,gssapi-keyex,gssapi-with-mic)

  - by user1963938

  Amazon Ec2 :: Redhat 6. 64 Bit I'm trying to follow the socks5 guidelines (http://www.catonmat.net/blog/linux-socks5-proxy/ ) to open a socks on one of our servers but unfortunately I got suck at step 1 . ssh -N -D 0.0.0.0:1080 localhost I get error Permission denied (publickey,gssapi-keyex,gssapi-with-mic). How do I fix it ? More debug info ssh -v -f -N -D 0.0.0.0:1080 localhost OpenSSH_5.3p1, OpenSSL 1.0.0-fips 29 Mar 2010 debug1: Reading configuration data /etc/ssh/ssh_config debug1: Applying options for * debug1: Connecting to localhost [127.0.0.1] port 22. debug1: Connection established. debug1: permanently_set_uid: 0/0 debug1: identity file /root/.ssh/identity type -1 debug1: identity file /root/.ssh/id_rsa type -1 debug1: identity file /root/.ssh/id_dsa type -1 debug1: Remote protocol version 2.0, remote software version OpenSSH_5.3 debug1: match: OpenSSH_5.3 pat OpenSSH* debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_5.3 debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug1: kex: server->client aes128-ctr hmac-md5 none debug1: kex: client->server aes128-ctr hmac-md5 none debug1: SSH2_MSG_KEX_DH_GEX_REQUEST(1024<1024<8192) sent debug1: expecting SSH2_MSG_KEX_DH_GEX_GROUP debug1: SSH2_MSG_KEX_DH_GEX_INIT sent debug1: expecting SSH2_MSG_KEX_DH_GEX_REPLY debug1: Host 'localhost' is known and matches the RSA host key. debug1: Found key in /root/.ssh/known_hosts:1 debug1: ssh_rsa_verify: signature correct debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: SSH2_MSG_NEWKEYS received debug1: SSH2_MSG_SERVICE_REQUEST sent debug1: SSH2_MSG_SERVICE_ACCEPT received debug1: Authentications that can continue: publickey,gssapi-keyex,gssapi-with-mic debug1: Next authentication method: gssapi-keyex debug1: No valid Key exchange context debug1: Next authentication method: gssapi-with-mic debug1: Unspecified GSS failure. Minor code may provide more information Credentials cache file '/tmp/krb5cc_0' not found debug1: Unspecified GSS failure. Minor code may provide more information Credentials cache file '/tmp/krb5cc_0' not found debug1: Unspecified GSS failure. Minor code may provide more information debug1: Unspecified GSS failure. Minor code may provide more information debug1: Next authentication method: publickey debug1: Trying private key: /root/.ssh/identity debug1: Trying private key: /root/.ssh/id_rsa debug1: Trying private key: /root/.ssh/id_dsa debug1: No more authentication methods to try. Permission denied (publickey,gssapi-keyex,gssapi-with-mic).

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• Installing PHP4 on a Debian (lenny) 7 32bit box

  - by Asim

  I am trying to install PHP4 on a Debian 7 32bit box but I ran into the following root@php4:~# apt-get update Get:1 http://snapshot.debian.org lenny Release.gpg [189 B] Hit http://snapshot.debian.org lenny Release Ign http://snapshot.debian.org lenny Release Hit http://snapshot.debian.org lenny/main Sources/DiffIndex Hit http://snapshot.debian.org lenny/main i386 Packages/DiffIndex Hit http://ftp.us.debian.org wheezy Release.gpg Hit http://security.debian.org wheezy/updates Release.gpg Ign http://snapshot.debian.org lenny/main Translation-en_US Ign http://snapshot.debian.org lenny/main Translation-en Hit http://ftp.us.debian.org wheezy Release Hit http://security.debian.org wheezy/updates Release Hit http://ftp.us.debian.org wheezy/main i386 Packages Hit http://ftp.us.debian.org wheezy/main Translation-en Hit http://security.debian.org wheezy/updates/main i386 Packages Hit http://security.debian.org wheezy/updates/main Translation-en Fetched 189 B in 0s (229 B/s) Reading package lists... Done W: GPG error: http://snapshot.debian.org lenny Release: The following signatures couldn't be verified because the public key is not available: NO_PUBKEY A70DAF536070D3A1 I did the following to fix it gpg --keyserver hkp://subkeys.pgp.net --recv-keys A70DAF536070D3A1 gpg --export --armor A70DAF536070D3A1 | sudo apt-key add - Now I get the following KEYEXPIRED error and unsure how to fix. Even Google does not help root@php4:~# apt-get update Get:1 http://snapshot.debian.org lenny Release.gpg [189 B] Hit http://snapshot.debian.org lenny Release Ign http://snapshot.debian.org lenny Release Hit http://snapshot.debian.org lenny/main Sources/DiffIndex Hit http://security.debian.org wheezy/updates Release.gpg Hit http://snapshot.debian.org lenny/main i386 Packages/DiffIndex Hit http://ftp.us.debian.org wheezy Release.gpg Hit http://security.debian.org wheezy/updates Release Ign http://snapshot.debian.org lenny/main Translation-en_US Hit http://ftp.us.debian.org wheezy Release Hit http://security.debian.org wheezy/updates/main i386 Packages Ign http://snapshot.debian.org lenny/main Translation-en Hit http://ftp.us.debian.org wheezy/main i386 Packages Hit http://security.debian.org wheezy/updates/main Translation-en Hit http://ftp.us.debian.org wheezy/main Translation-en Fetched 189 B in 0s (275 B/s) Reading package lists... Done W: GPG error: http://snapshot.debian.org lenny Release: The following signatures were invalid: KEYEXPIRED 1246455239 Any help?

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• Windows Vista claims wireless key is the wrong length

  - by humble coffee

  A family member of mine is house sitting and has been given the details of their wifi. The access point is an Airport Express, it has WEP encryption (I think) and they've been given a passphrase to use. I know it's a passphrase and not the encrypted key as it's an English word. The passphrase is 10 characters long. The problem is that Vista complains that it's not a valid key as it must be a 5 or 13 character non-hex key or a 10 or 26 character hex key. (From what I've read this suggests the encryption is WEP?) I've found a couple of suggested solutions, but I'm not actually at the house at the moment so I wanted to make sure I have a good chance of getting it to work when I'm there but have no internets to ask. Solution 1: Vista needs to be told explicitly what kind of encryption and key is being used. Specify in the connection settings that you are using WEP and that it is a "shared key". Solution2: Try converting the passphrase to hexadecimal using an ASCII-hex converter and entering that.

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• Can SSH into remote server but can't SCP?

  - by ArtfulDodger2012

  I can SSH into remote server just fine using private key authentication with prompt for passphrase. However I'm getting permission denied when I try to SCP a file using the same passphrase. Here's my output: $ scp -v [file] [user]@[remoteserver.com]:/home/[my dir] Executing: program /usr/bin/ssh host [remoteserver.com], user [user], command scp -v -t /home/[my dir] OpenSSH_5.3p1 Debian-3ubuntu7, OpenSSL 0.9.8k 25 Mar 2009 debug1: Reading configuration data /home/[my dir].ssh/config debug1: Applying options for [remoteserver.com] debug1: Reading configuration data /etc/ssh/ssh_config debug1: Applying options for * debug1: Connecting to [remoteserver.com] [[remoteserver.com]] port 22. debug1: Connection established. debug1: identity file /home/[user]/.ssh/aws_corp type -1 debug1: Remote protocol version 2.0, remote software version OpenSSH_5.3p1 Debian-3ubuntu7 debug1: match: OpenSSH_5.3p1 Debian-3ubuntu7 pat OpenSSH* debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_5.3p1 Debian-3ubuntu7 debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug1: kex: server->client aes128-ctr hmac-md5 none debug1: kex: client->server aes128-ctr hmac-md5 none debug1: SSH2_MSG_KEX_DH_GEX_REQUEST(1024<1024<8192) sent debug1: expecting SSH2_MSG_KEX_DH_GEX_GROUP debug1: SSH2_MSG_KEX_DH_GEX_INIT sent debug1: expecting SSH2_MSG_KEX_DH_GEX_REPLY debug1: Host '[remoteserver.com]' is known and matches the RSA host key. debug1: Found key in /home/[my dir]/.ssh/known_hosts:12 debug1: ssh_rsa_verify: signature correct debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: SSH2_MSG_NEWKEYS received debug1: SSH2_MSG_SERVICE_REQUEST sent debug1: SSH2_MSG_SERVICE_ACCEPT received debug1: Authentications that can continue: publickey debug1: Next authentication method: publickey debug1: Trying private key: /home/[my dir]/.ssh/aws_corp debug1: PEM_read_PrivateKey failed debug1: read PEM private key done: type <unknown> Enter passphrase for key '/home/[my dir]/.ssh/aws_corp': debug1: read PEM private key done: type RSA Connection closed by [remote server] lost connection I've searched for answers but can't find quite the same problem or am just being thick. Either way any help is much appreciated. Cheers!

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• Can't upload project to PPA using Quickly

  - by RobinJ

  I can't get Quickly to upload my project into my PPA. I've set up my PGP key and used it so sign the code of conduct, and the PPA exists. I don't know what other usefull information I can supply. robin@RobinJ:~/Ubuntu One/Python/gtkreddit$ quickly share --ppa robinj/gtkredditGet Launchpad Settings Launchpad connection is ok gpg: WARNING: unsafe permissions on configuration file `/home/robin/.gnupg/gpg.conf' gpg: WARNING: unsafe enclosing directory permissions on configuration file `/home/robin/.gnupg/gpg.conf' gpg: WARNING: unsafe permissions on configuration file `/home/robin/.gnupg/gpg.conf' gpg: WARNING: unsafe enclosing directory permissions on configuration file `/home/robin/.gnupg/gpg.conf' Traceback (most recent call last): File "/usr/share/quickly/templates/ubuntu-application/share.py", line 138, in <module> license.licensing() File "/usr/share/quickly/templates/ubuntu-application/license.py", line 284, in licensing {'translatable': 'yes'}) File "/usr/share/quickly/templates/ubuntu-application/internal/quicklyutils.py", line 166, in change_xml_elem xml_tree.find(parent_node).insert(0, new_node) AttributeError: 'NoneType' object has no attribute 'insert' ERROR: share command failed Aborting I reported this as a bug on Launchpad, because I assume that it is a bug. If you know a quick workaround, please let me know. https://bugs.launchpad.net/ubuntu/+source/quickly/+bug/1018138

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• Updating Banshee to 2.4

  - by Lucasguy11

  I have banshee 2.2.1 with Ubuntu 11.10 I have been trying to update banshee to 2.4 (released yesterday) but it just isnt working, I have been using sudo add-apt-repository ppa:banshee-team/ppa in terminal, from the Banshee.fm website. but after running through terminal it says this: sudo add-apt-repository ppa:banshee-team/ppa You are about to add the following PPA to your system: PPA for Banshee Team This PPA contains the latest stable debs of Banshee for Ubuntu. To install Banshee, you must first enable the PPA on your system: 1. Open Software Sources (System->Administration->Software Sources) 2. Navigate to the "Third Party Sources" tab. 3. Click "Add" 4. Enter the APT line below that corresponds to your Ubuntu version that starts with "deb". 5. Click "Add Source" 6. Click "Close" 7. It will prompt you to reload your software cache. Click "Reload". 8. Now install the package "banshee" from Synaptic, or using the command below: sudo apt-get install banshee For those who wish to compile from trunk, add the deb-src line and then run "sudo apt-get build-dep" to install all required dependencies before starting to compile. Unstable (version which have odd minor version numbers) debs of Banshee can be found here: https://launchpad.net/~banshee-team/+archive/banshee-unstable More info: https://launchpad.net/~banshee-team/+archive/ppa Press [ENTER] to continue or ctrl-c to cancel adding it Executing: gpg --ignore-time-conflict --no-options --no-default-keyring --secret-keyring /tmp/tmp.OPAjxemDQr --trustdb-name /etc/apt/trustdb.gpg --keyring /etc/apt/trusted.gpg --primary-keyring /etc/apt/trusted.gpg --keyserver hkp://keyserver.ubuntu.com:80/ --recv 9D2C2E0A3C88DD807EC787D74874D3686E80C6B7 gpg: requesting key 6E80C6B7 from hkp server keyserver.ubuntu.com gpg: key 6E80C6B7: "Launchpad PPA for Banshee Team" not changed gpg: Total number processed: 1 gpg: unchanged: 1 I believe I have the ppa but, im not sure. I need a step by step process to get this, ive been trying to figure it out for quite a while now...

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• Problems after bumblebee installation

  - by Samuel

  I tried to install bumblebee on Ubuntu 12.04 LTS by following steps on ubuntuwiki site. But when i used this code: sudo add-apt-repository ppa:bumblebee/stable && sudo apt-get update this output came out: Executing: gpg --ignore-time-conflict --no-options --no-default-keyring --secret-keyring /tmp/tmp.q0zzLiXVT3 --trustdb-name /etc/apt/trustdb.gpg --keyring /etc/apt/trusted.gpg --primary-keyring /etc/apt/trusted.gpg --keyserver hkp://keyserver.ubuntu.com:80/ --recv 46C0364A882F14F899448FFCB22A95F88110A93A gpg: requesting key 8110A93A from hkp server keyserver.ubuntu.com gpg: key 8110A93A: "Launchpad PPA for Bumlebee Project" not changed gpg: Total number processed: 1 gpg: unchanged: 1 E: Type 'ain' is not known on line 3 in source list /etc/apt/sources.list.d/bumblebee-stable-precise.list E: The list of sources could not be read. There´s also the same problem message when I try to run the update center. ´E:Type´ain´ is not known on line 3 in source list /etc/apt/sources.list.d/bumblebee-stable-precise.list, E:The list of sources could not be read., E:The package lists or status file could not be parsed or opened.´ I don´t know what to do since I´m a newbie at Linux. Thanks in advance, Samuel.

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• How to make the internal subwoofer work on an Asus G73JW?

  - by CodyLoco

  I have an Asus G73JW laptop which has an internal subwoofer built-in. Currently, the system detects the internal speakers as a 2.0 system (or I can change do 4.0 is the only other option). I found a bug report here: https://bugs.launchpad.net/ubuntu/+source/alsa-driver/+bug/673051 which discusses the bug and according to them a fix was sent upstream back at the end of 2010. I would have thought this would have made it into 12.04 but I guess not? I tried following the link given at the very bottom to install the latest ALSA drivers, here: https://wiki.ubuntu.com/Audio/InstallingLinuxAlsaDriverModules however I keep running into an error when trying to install: sudo apt-get install linux-alsa-driver-modules-$(uname -r) Reading package lists... Done Building dependency tree Reading state information... Done E: Unable to locate package linux-alsa-driver-modules-3.2.0-24-generic E: Couldn't find any package by regex 'linux-alsa-driver-modules-3.2.0-24-generic' I believe I have added the repository correctly: sudo add-apt-repository ppa:ubuntu-audio-dev/ppa [sudo] password for codyloco: You are about to add the following PPA to your system: This PPA will be used to provide testing versions of packages for supported Ubuntu releases. More info: https://launchpad.net/~ubuntu-audio-dev/+archive/ppa Press [ENTER] to continue or ctrl-c to cancel adding it Executing: gpg --ignore-time-conflict --no-options --no-default-keyring --secret-keyring /tmp/tmp.7apgZoNrqK --trustdb-name /etc/apt/trustdb.gpg --keyring /etc/apt/trusted.gpg --primary-keyring /etc/apt/trusted.gpg --keyserver hkp://keyserver.ubuntu.com:80/ --recv 4E9F485BF943EF0EABA10B5BD225991A72B194E5 gpg: requesting key 72B194E5 from hkp server keyserver.ubuntu.com gpg: key 72B194E5: public key "Launchpad Ubuntu Audio Dev team PPA" imported gpg: Total number processed: 1 gpg: imported: 1 (RSA: 1) And I also ran an update as well (followed the instructions on the fix above). Any ideas?

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• How to disable the alt-hotkey behavior on gnome terminal?

  - by farleyknight

  This is not just gnome terminal, but pretty much all gnome windows: When you hold the "alt" key, you can press the first letter of one of the menu items. This will let you scroll that menu without clicking on it directly. This is okay on any other window, like say Firefox, but on gnome terminal, it steals the keys I use for emacs!! There is very little chance of me learning a new set key combinations if I can avoid. If I can't isolate this just to gnome terminal, I'm fine with that.

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• iPhone OS: Why is my managedModelObject not complying with Key Value Coding?

  - by nickthedude

  Ok so I'm trying to build this stat tracker for my app and I have built a data model object called statTracker that keeps track of all the stuff I want it to. I can set and retrieve values using the selectors, but if I try and use KVC (ie setValue: forKey: ) everything goes bad and says my StatTracker class is not KVC compliant: valueForUndefinedKey:]: the entity StatTracker is not key value coding-compliant for the key "timesLauched".' 2010-05-18 15:55:08.573 here's the code that is triggering it: NSArray *statTrackerArray = [[NSArray alloc] init]; statTrackerArray = [[CoreDataSingleton sharedCoreDataSingleton] getStatTracker]; NSNumber *number1 = [[NSNumber alloc] init]; number1 = [NSNumber numberWithInt:(1 + [[(StatTracker *)[statTrackerArray objectAtIndex:0] valueForKey:@"timesLauched"] intValue])]; [(StatTracker *)[statTrackerArray objectAtIndex:0] setValue:number1 forKey:@"timesLaunched" ]; NSError *error; if (![[[CoreDataSingleton sharedCoreDataSingleton] managedObjectContext] save:&error]) { NSLog(@"error writing to db"); } Not sure if this is enough code for you folks let me know what you need if you do need more. This would be so sweet if I could use KVC because I could then abstract all this stat tracking stuff into a single method call with a string argument for the value in question. At least that is what I hope to accomplish here. I'm actually now understanding the power of KVC but now I'm just trying to figure out how to make it work. Thanks! Nick

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• Unlock all private keys on Ubuntu, entering password only once at login

  - by conradlee

  I login to Ubuntu 12.04 using a password. Later on, when I use my browser(Chrome), I'm asked for a password to unlock the keychain so that the browser can access my saved credentials for various websites (it's the same password). Also, whenever I use SSH to connect to other computers using my private key, I am prompted for the same password to unlock my private key. How can I make it so that I am asked for my password exactly once per login (given that my login password is the same as the one I use for all my private keys)? Probably someone will try to label this question as a duplicate of this question, this question, or this question. While these questions are similar, none of them explicitly say that there still needs to be a password entered on login, as I am demanding here. As a result, the accepted solutions just say "set your passwords to blank"--I don't want that, it's dangerous! So I am aware of the similar questions, but none of them has received the correct answer yet, because they are slightly different.

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• retrieve value from hashtable with clone of key; C#

  - by Johnny

  I would like to know if there is any possible way to retrieve an item from a hashtable using a key that is identical to the actual key, but a different object. I understand why it is probably not possible, but I would like to see if there is any tricky way to do it. My problem arises from the fact that, being as stupid as I am, I created hashtables with int[] as the keys, with the integer arrays containing indices representing spatial position. I somehow knew that I needed to create a new int[] every time I wanted to add a new entry, but neglected to think that when I generated spatial coordinate arrays later they would be worthless in retrieving the values from my hashtables. Now I am trying to decide whether to rearrange things so that I can store my values in ArrayLists, or whether to search through the list of keys in the Hashtable for the one I need every time I want to get a value, neither of the options being very cool. Unless of course there is a way to get //1 to work like //2! Thanks in advance. static void Main(string[] args) { Hashtable dog = new Hashtable(); //1 int[] man = new int[] { 5 }; dog.Add(man, "hello"); int[] cat = new int[] { 5 }; Console.WriteLine(dog.ContainsKey(cat)); //false //2 int boy = 5; dog.Add(boy, "wtf"); int kitten = 5; Console.WriteLine(dog.ContainsKey(kitten)); //true; }

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• How to figure out which key was pressed on a BlackBerry

  - by Skrud

  What I want: To know when the user has pressed the button that has the number '2' on it, for example. I don't care whether "Alt" or "Shift" has been pressed. The user has pressed a button, and I want to evaluate whether this button has '2' printed on it. Naturally, if I switch devices this key will change. On a Bold 9700/9500 this is the 'E' key. On a Pearl, this is the 'T'/'Y' key. I've managed to get this working in what appears to be a roundabout way, by looking up the keycode of the '2' character with the ALT button enabled and using Keypad.key() to get the actual button: // figure out which key the '2' is on: final int BUTTON_2_KEY = Keypad.key(KeypadUtil.getKeyCode('2', KeypadListener.STATUS_ALT, KeypadUtil.MODE_EN_LOCALE)); protected boolean keyDown(int keycode, int time) { int key = Keypad.key(keycode); if ( key == BUTTON_2_KEY ) { // do something return true; } return super.keyDown(keycode,time); } I can't help but wonder if there is a better way to do this. I've looked at the constants defined in KeypadListener and Keypad but I can't find any constants mapped to the actual buttons on the device. Would any more experienced BlackBerry devs care to lend a helping hand? Thanks!

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• How do you assign commands to keys in Terminal?

  - by NES

  Is there a solution to assign special key combinations to words in terminal use. For example the less command is very usefull and i use i a lot to pipe the output of another process through it. The idea would be to set up special key combinations that are only active in terminal use assigned to write different commands? So pressing CTRL + l in terminal window could write | less or CTRL + G could stand for | grep Note: i just mean adding the letters to commandline not execute the finally. A similar way what's tabcompletion but more specific.

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• android: consume key press, bypassing framework processing

  - by user360024

  What I want android to do: when user presses a single key, have the view respond, but do so without opening a text area and displaying the character associated with the key that was pressed, and without requiring that the Enter key be pressed, and without requiring that the user press Esc to make the text area go away. For example, when user presses "u" (and doesn't press Enter), that means "undo the last action", so the controller and model immediately undo the last action, then the view does an invalidate() and user sees that their last action has been undone. In other words the "u" key press should be silently processed, such that the only visual result is that user's last action has been undone. I've implemented OnKeyListener and provided an onKey() method: the class: public class MyGameView extends View implements OnKeyListener{ in the constructor: //2010jun06, phj: With onKey(), helps let this View consume key presses // before the framework gets a chance to consume the key press. setOnKeyListener((View.OnKeyListener)this); the onKey() method: public boolean onKey(View v, int keyCode, KeyEvent event) { if (keyCode == KeyEvent.KEYCODE_R) { Log.d("BWA", "In onKey received keycode associated with R."); } return true; // meaning the event (key press) has been consumed, so // the framework should not handle this event. } but when user presses "u" key on the emulator keypad, a textarea is opened at the bottom of the screen, the "u" charater is displayed there, and the onKey() method doesn't execute until user presses the Enter key. Is there a way to make android do what I want? Thanks,

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• OpenSSH does not accept public key?

  - by Bob

  I've been trying to solve this for a while, but I'm admittedly quite stumped. I just started up a new server and was setting up OpenSSH to use key-based SSH logins, but I've run into quite a dilemma. All the guides are relatively similar, and I was following them closely (despite having done this once before). I triple checked my work to see if I would notice some obvious screw up - but nothing is apparent. As far as I can tell, I haven't done anything wrong (and I've checked very closely). If it's any help, on my end I'm using Cygwin and the server is running Ubuntu 12.04.1 LTS. Anyways, here is the output (I've removed/censored some parts for privacy (primarily anything with my name, website, or its IP address), but I can assure you that nothing is wrong there): $ ssh user@host -v OpenSSH_5.9p1, OpenSSL 0.9.8r 8 Feb 2011 debug1: Connecting to host [ipaddress] port 22. debug1: Connection established. debug1: identity file /home/user/.ssh/id_rsa type 1 debug1: identity file /home/user/.ssh/id_rsa-cert type -1 debug1: identity file /home/user/.ssh/id_dsa type -1 debug1: identity file /home/user/.ssh/id_dsa-cert type -1 debug1: identity file /home/user/.ssh/id_ecdsa type -1 debug1: identity file /home/user/.ssh/id_ecdsa-cert type -1 debug1: Remote protocol version 2.0, remote software version OpenSSH_5.9p1 Debian-5ubuntu1 debug1: match: OpenSSH_5.9p1 Debian-5ubuntu1 pat OpenSSH* debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_5.9 debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug1: kex: server->client aes128-ctr hmac-md5 none debug1: kex: client->server aes128-ctr hmac-md5 none debug1: sending SSH2_MSG_KEX_ECDH_INIT debug1: expecting SSH2_MSG_KEX_ECDH_REPLY debug1: Server host key: ECDSA 24:68:c3:d8:13:f8:61:94:f2:95:34:d1:e2:6d:e7:d7 debug1: Host 'host' is known and matches the ECDSA host key. debug1: Found key in /home/user/.ssh/known_hosts:2 debug1: ssh_ecdsa_verify: signature correct debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: SSH2_MSG_NEWKEYS received debug1: Roaming not allowed by server debug1: SSH2_MSG_SERVICE_REQUEST sent debug1: SSH2_MSG_SERVICE_ACCEPT received debug1: Authentications that can continue: publickey debug1: Next authentication method: publickey debug1: Offering RSA public key: /home/user/.ssh/id_rsa debug1: Authentications that can continue: publickey debug1: Trying private key: /home/user/.ssh/id_dsa debug1: Trying private key: /home/user/.ssh/id_ecdsa debug1: No more authentication methods to try. Permission denied (publickey). What can I do to resolve my problem?

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• Query for model by key

  - by Jason Hall

  What I'm trying to do is query the datastore for a model where the key is not the key of an object I already have. Here's some code: class User(db.Model): partner = db.SelfReferenceProperty() def text_message(self, msg): user = User.get_or_insert(msg.sender) if not user.partner: # user doesn't have a partner, find them one # BUG: this line returns 'user' himself... :( other = db.Query(User).filter('partner =', None).get() if other: # connect users else: # no one to connect to! The idea is to find another User who doesn't have a partner, that isn't the User we already know. I've tried filter('key !=, user.key()), filter('__key__ !=, user.key()) and a couple others, and nothing returns another User who doesn't have a partner. filter('foo !=, user.key()) also returns nothing, for the record.

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