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  • Ubuntu Preseed set Norwegian Keyboard?

    - by Vangelis Tasoulas
    It's been a couple of days now that I am trying to make a fully automated unattended installation. I managed to make it work with Ubuntu/Cobbler and a preseed file, but I cannot set the correct keyboard layout which is Norwegian in this case. I am doing the tests on a virtual machine and when I am going with a normal manual installation (no preseed) everything is working fine. When I am using the preseed file, I always end up with an "English (US)" keyboard no matter the many different options I have tried. I can change it manually with the "dpkg-reconfigure keyboard-configuration" command, but that's not the case. It should be handled automatically using the preseed file. I am using DEBCONF_DEBUG=5 when the grub is loading, and as I see in "/var/log/installer/syslog" file after the installation has finished, the preseeding commands are accepted. Can anyone help on this? The preseed file I am using is following: d-i debian-installer/country string NO d-i debian-installer/language string en_US:en d-i debian-installer/locale string en_US.UTF-8 d-i console-setup/ask_detect boolean false d-i keyboard-configuration/layout select Norwegian d-i keyboard-configuration/variant select Norwegian d-i keyboard-configuration/modelcode string pc105 d-i keyboard-configuration/layoutcode string no d-i keyboard-configuration/xkb-keymap select no d-i netcfg/choose_interface select auto d-i netcfg/get_hostname string myhostname d-i netcfg/get_domain string simula.no d-i hw-detect/load_firmware boolean true d-i mirror/country string manual d-i mirror/http/hostname string ftp.uninett.no d-i mirror/http/directory string /ubuntu d-i mirror/http/proxy string http://10.0.1.253:3142/ d-i mirror/codename string precise d-i mirror/suite string precise d-i clock-setup/utc boolean true d-i time/zone string Europe/Oslo d-i clock-setup/ntp boolean true d-i clock-setup/ntp-server string 10.0.1.254 d-i partman-auto/method string lvm partman-auto-lvm partman-auto-lvm/new_vg_name string vg0 d-i partman-auto/purge_lvm_from_device boolean true d-i partman-lvm/device_remove_lvm boolean true d-i partman-md/device_remove_md boolean true d-i partman-lvm/confirm boolean true d-i partman-lvm/confirm_nooverwrite boolean true d-i partman-auto-lvm/guided_size string max d-i partman-auto/choose_recipe select 30atomic d-i partman/default_filesystem string ext4 d-i partman-partitioning/confirm_write_new_label boolean true d-i partman/choose_partition select finish d-i partman/confirm boolean true d-i partman/confirm_nooverwrite boolean true d-i partman/mount_style select uuid d-i passwd/root-login boolean false d-i passwd/make-user boolean true d-i passwd/user-fullname string vangelis d-i passwd/username string vangelis d-i passwd/user-password-crypted password $6$asdafdsdfasdfasdf d-i passwd/user-uid string d-i user-setup/allow-password-weak boolean false d-i passwd/user-default-groups string adm cdrom dialout lpadmin plugdev sambashare d-i user-setup/encrypt-home boolean false d-i apt-setup/restricted boolean true d-i apt-setup/universe boolean true d-i apt-setup/backports boolean true d-i apt-setup/services-select multiselect security d-i apt-setup/security_host string security.ubuntu.com d-i apt-setup/security_path string /ubuntu tasksel tasksel/first multiselect Basic Ubuntu server, OpenSSH server d-i pkgsel/include string build-essential htop vim nmap ntp d-i pkgsel/upgrade select safe-upgrade d-i pkgsel/update-policy select none d-i pkgsel/updatedb boolean true d-i grub-installer/only_debian boolean true d-i grub-installer/with_other_os boolean true d-i finish-install/keep-consoles boolean false d-i finish-install/reboot_in_progress note d-i cdrom-detect/eject boolean true d-i debian-installer/exit/halt boolean false d-i debian-installer/exit/poweroff boolean false

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  • String contains trailing zeroes when converted from decimal [migrated]

    - by Locke
    I've run into an unusual quirk in a program I'm writing, and I was trying to figure out if anyone knew the cause. Note that fixing the issue is easy enough. I just can't figure out why it is happening in the first place. I have a WinForms program written in VB.NET that is displaying a subset of data. It contains a few labels that show numeric values (the .Text property of the labels are being assigned directly from the Decimal values). These numbers are being returned by a DLL I wrote in C#. The DLL calls a webservice which initially returns the values in question. It returns one as a string, the other as a decimal (I don't have any control over the webservice, I just consume it). The DLL assigns these to properties on an object (both of which are decimals) then returns that object back to the WinForm program that called the DLL. Obviously, there's a lot of other data being consumed from the webservice, but no other operations are happening which could modify these properties. So, the short version is: WinForm requests a new Foo from the DLL. DLL creates object Foo. DLL calls webservice, which returns SomeOtherFoo. //Both Foo.Bar1 and Foo.Bar2 are decimals Foo.Bar1 = decimal.Parse(SomeOtherFoo.Bar1); //SomeOtherFoo.Bar1 is a string equal to "2.9000" Foo.Bar2 = SomeOtherFoo.Bar2; //SomeOtherFoo.Bar2 is a decimal equal to 2.9D DLL returns Foo to WinForm. WinForm.lblMockLabelName1.Text = Foo.Bar1 //Inspecting Foo.Bar1 indicates my value is 2.9D WinForm.lblMockLabelName2.Text = Foo.Bar2 //Inspecting Foo.Bar2 also indicates I'm 2.9D So, what's the quirk? WinForm.lblMockLabelName1.Text displays as "2.9000", whereas WinForm.lblMockLabelname2.Text displays as "2.9". Now, everything I know about C# and VB indicates that the format of the string which was initially parsed into the decimal should have no bearing on the outcome of a later decimal.ToString() operation called on the same decimal. I would expect that decimal.Parse(someDecimalString).ToString() would return the string without any trailing zeroes. Everything I find online seems to corroborate this (there are countless Stack Overflow questions asking exactly the opposite...how to keep the formatting from the initial parsing). At the moment, I've just removed the trailing zeroes from the initial string that gets parsed, which has hidden the quirk. However, I'd love to know why it happens in the first place.

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  • Apache is sending php files to my browser instead of parsing

    - by justen doherty
    I have to set up PHP on an existing web host. I have made a virtual host entry, but for some reason Apache is sending the PHP to the browser instead of parsing.. from googling around it looks like it's a problem with the mimetypes, but I'm not an Apache expert by any means, so if anyone could help it would be appreciated... I have the following in my httpd.conf: AddHandler php5-script php DirectoryIndex index.html index.phtml index.php index.phps AddType application/x-httpd-php .phtml AddType application/x-httpd-php .php AddType application/x-httpd-php-source .phps The PHP module is loaded into Apache: /usr/sbin/apachectl -M Loaded Modules: core_module (static) mpm_prefork_module (static) http_module (static) so_module (static) auth_basic_module (shared) auth_digest_module (shared) authn_file_module (shared) authn_alias_module (shared) authn_anon_module (shared) authn_dbm_module (shared) authn_default_module (shared) authz_host_module (shared) authz_user_module (shared) authz_owner_module (shared) authz_groupfile_module (shared) authz_dbm_module (shared) authz_default_module (shared) ldap_module (shared) authnz_ldap_module (shared) include_module (shared) log_config_module (shared) logio_module (shared) env_module (shared) ext_filter_module (shared) mime_magic_module (shared) expires_module (shared) deflate_module (shared) headers_module (shared) usertrack_module (shared) setenvif_module (shared) mime_module (shared) dav_module (shared) status_module (shared) autoindex_module (shared) info_module (shared) dav_fs_module (shared) vhost_alias_module (shared) negotiation_module (shared) dir_module (shared) actions_module (shared) speling_module (shared) userdir_module (shared) alias_module (shared) rewrite_module (shared) proxy_module (shared) proxy_balancer_module (shared) proxy_ftp_module (shared) proxy_http_module (shared) proxy_connect_module (shared) cache_module (shared) suexec_module (shared) disk_cache_module (shared) file_cache_module (shared) mem_cache_module (shared) cgi_module (shared) version_module (shared) fcgid_module (shared) perl_module (shared) php5_module (shared) proxy_ajp_module (shared) ssl_module (shared) And this is my virtual host entry: <VirtualHost 10.16.140.113:80> ServerName viridor-cms.co.uk ServerAlias www.viridor-cms.co.uk UseCanonicalName Off DocumentRoot /var/www/vhosts/viridor-cms.co.uk/httpdocs CustomLog /var/www/vhosts/viridor-cms.co.uk/cms-access_log common ErrorLog /var/www/vhosts/viridor-cms.co.uk/cms-error_log DirectoryIndex index.php index.html <IfModule sapi_apache2.c> php_admin_flag engine on php_admin_flag safe_mode on </IfModule> <IfModule mod_php5.c> php_admin_flag engine on php_admin_flag safe_mode on </IfModule> AddType application/x-httpd-php .php AddType application/x-httpd-php-source .phps </VirtualHost> Please help, my head is so sore from banging it against the table and the wall!

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  • Delphi - How do I split a string into an array of strings based on a delimiter?

    - by Ryan
    Hello all, I'm trying to find a Delphi function that will split an input string into an array of strings based on a delimiter. I've found a lot on Google, but all seem to have their own issues and I haven't been able to get any of them to work. I just need to split a string like: "word:doc,txt,docx" into an array based on ':'. The result would be ['word', 'doc,txt,docx']. Does anyone have a function that they know works? Thank you

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  • Haskell: Left-biased/short-circuiting function

    - by user2967411
    Two classes ago, our professor presented to us a Parser module. Here is the code: module Parser (Parser,parser,runParser,satisfy,char,string,many,many1,(+++)) where import Data.Char import Control.Monad import Control.Monad.State type Parser = StateT String [] runParser :: Parser a -> String -> [(a,String)] runParser = runStateT parser :: (String -> [(a,String)]) -> Parser a parser = StateT satisfy :: (Char -> Bool) -> Parser Char satisfy f = parser $ \s -> case s of [] -> [] a:as -> [(a,as) | f a] char :: Char -> Parser Char char = satisfy . (==) alpha,digit :: Parser Char alpha = satisfy isAlpha digit = satisfy isDigit string :: String -> Parser String string = mapM char infixr 5 +++ (+++) :: Parser a -> Parser a -> Parser a (+++) = mplus many, many1 :: Parser a -> Parser [a] many p = return [] +++ many1 p many1 p = liftM2 (:) p (many p) Today he gave us an assignment to introduce "a left-biased, or short-circuiting version of (+++)", called (<++). His hint was for us to consider the original implementation of (+++). When he first introduced +++ to us, this was the code he wrote, which I am going to call the original implementation: infixr 5 +++ (+++) :: Parser a -> Parser a -> Parser a p +++ q = Parser $ \s -> runParser p s ++ runParser q s I have been having tons of trouble since we were introduced to parsing and so it continues. I have tried/am considering two approaches. 1) Use the "original" implementation, as in p +++ q = Parser $ \s - runParser p s ++ runParser q s 2) Use the final implementation, as in (+++) = mplus Here are my questions: 1) The module will not compile if I use the original implementation. The error: Not in scope: data constructor 'Parser'. It compiles fine using (+++) = mplus. What is wrong with using the original implementation that is avoided by using the final implementation? 2) How do I check if the first Parser returns anything? Is something like (not (isNothing (Parser $ \s - runParser p s) on the right track? It seems like it should be easy but I have no idea. 3) Once I figure out how to check if the first Parser returns anything, if I am to base my code on the final implementation, would it be as easy as this?: -- if p returns something then p <++ q = mplus (Parser $ \s -> runParser p s) mzero -- else (<++) = mplus Best, Jeff

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  • Can I use regular expressions with String.Replace in C#?

    - by Taz
    For example I have code below string txt="I have strings like West, and West; and west, and Western." I would like to replace the word west or West with some other word. But I would like not to replace West in Western. Can I use regular expression in string.replace? I used inputText.Replace("(\\sWest.\\s)",temp); It dos not work.

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  • Provide a URI as a parameter for an ODBC or OleDB provider connection string?

    - by womp
    Is there any way of specifying a URI (rather than a physical path) as the "Data Source" parameter in an OleDB connection string, or the "Dbq" parameter in an ODBC connection string, such that the default ADO.Net providers will work with it? I've tried file://, and it's a no go, so I just wanted to confirm whether or not this is possible. I'm guessing not, but any confirmation or documentation links would be appreciated.

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  • How to define function in a string in Android?

    - by androidNoob
    I have an event handler that I want to be connected to a button through xml. Basically, I want to define it as a string in my xml file, then use the property inspector to select the name of the handler for my button (the "on Click" property). How do I define a call to the function in the string?

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  • How do I pass a string or data object between two view controllers?

    - by Jonathan
    In my last question i asked how to best send a string from one view controller to another, both which were on a navigation stack: http://stackoverflow.com/questions/2898860/pass-string-from-tableviewcontroller-to-viewcontroller-in-navigation-stack However I just realised I can either pass the path to the file in the app's document's folder as the first (the table view) has already accessed the data in the file should I pass viewcontroller the data to the pushed VC?

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  • PHP - How to remove the first number in a string?

    - by Dodi300
    Hello. How can I remove the first number in a string? Say if I had these 48 numbers seperated with a ',' (comma): 8,5,8,10,15,20,27,25,60,31,25,39,25,31,26,28,80,28,27,31,27,29,26,35,8,5,8,10,15,20,27,25,60,31,25,39,25,31,26,28,80,28,27,31,27,29,26,35 How would I remove the "8," from the string? Thanks.

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  • I'm writing a spellchecking program, how do I replace ch in a string?

    - by Ajay Hopkins
    What am I doing wrong/what can I do? import sys import string def remove(file): punctuation = string.punctuation for ch in file: if len(ch) > 1: print('error - ch is larger than 1 --| {0} |--'.format(ch)) if ch in punctuation: ch = ' ' return ch else: return ch ref = (open("ref.txt","r")) test_file = (open("test.txt", "r")) dictionary = ref.read().split() file = test_file.read().lower() file = remove(file) print(file) This is in Python 3.1.2

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  • About to migrate :string but I'm thinking :text might be better. Performance/Purpose?

    - by Sam
    class CreateScrapes < ActiveRecord::Migration def self.up create_table :scrapes do |t| t.text :saved_characters t.text :sanitized_characters t.string :href t.timestamps end end def self.down drop_table :scrapes end end I'm about to rake db:migrate and I'm think about the attribute type if I should be using text or string. Since saved_characters and sanitized_characters will be arrays with thousands of unicode values, its basically comma delimited data, I'm not sure if `:text' is really the right way to go here. What would you do?

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  • Java Regular Expression. Check if String contains ONLY Letters

    - by user69514
    How do I check if a String contains only letters in java? I want to write an if statement that will return false if there is a white space, a number, a symbol or anything else other than a-z A-Z. My string must contain ONLY letters. I thought I could do it this way, but I'm doing it wrong: if( ereg("[a-zA-Z]+", $myString)) return true; else return false;

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  • im writing a spellchecking program, how do i replace ch in a string..eg..

    - by Ajay Hopkins
    what am i doing wrong/what can i do?? import sys import string def remove(file): punctuation = string.punctuation for ch in file: if len(ch) > 1: print('error - ch is larger than 1 --| {0} |--'.format(ch)) if ch in punctuation: ch = ' ' return ch else: return ch ref = (open("ref.txt","r")) test_file = (open("test.txt", "r")) dictionary = ref.read().split() file = test_file.read().lower() file = remove(file) print(file) p.s, this is in Python 3.1.2

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