Fedora Installation :: Bash Shell Could Not Find Kfontview Command
Apr 5, 2009Bash shell could not find "kfontview" command, although kdebase-4.2.1-2.fc10.i386 already installed?
View 2 RepliesBash shell could not find "kfontview" command, although kdebase-4.2.1-2.fc10.i386 already installed?
View 2 Replieshow to install and use tcshell instead of bash shell...
I run this command but it still dosent work
# yum install tcsh
after installing I use this command to setting it as a default login
$ which csh
but it not work yet...
I want to change my default shell to tcsh. I used
Code:
usermod -s /bin/tcsh username
command as given at url
But if I open a new shell, it is still a bash shell.
How do I make my default shell as tcsh?
I'm used to being able to put a script in ~/bin and having it overrule the system version of a command. However for "time" or "kill" since the bash shell implements a version of the command (i.e. /bin/ version is not used) doing this is not enough. How can I get the shell to run my own version instead of the version in the shell.I understand the implications of doing this. I know what I'm doing, I can always /bin/whatever if I want to get the old version (or just chmod -x ~/bin/whatever).
View 6 Replies View RelatedI am unable to use clear or cls command on bash shell. I have recently installed Cygwin and am using that for practicing unix commands.
I see that I can use Ctrl + L to clear the screen. I created an alias in my .bashrc to do the same as
alias cls='^L'
This is how i defined other aliases e.g.
And they work. Hence I assume cls will work too but this is what I get when I try to give cls on command prompt. Am i missing something? Is there a way to do this?
Then someone suggested, You cannot alias keystrokes to commands or vice versa. You could just alias cls to an echo command: echo -en "x0c"
And I added the following to .bashrc,
Sourced the .bashrc file. No errors but cls still does not clear the screen. Infact when I typed the echo -en "x0c on command prompt as well, nothing happened. What does this command do?
Cis 140 student.how to use the test command to evaluate whether the shell variable I create contains a referance to the bash shell? and use the echo command to determine the result.
View 1 Replies View RelatedJust trying to execute cd command in a .cshrc file (bash shell in SUSE Linux) it says."No such file or directory".Do you see any reasons for this
View 3 Replies View RelatedI want to use ssh to execute a command and to wait endlessly to log everything (in file) that comes as a stream of the connected server. But unfortunately, in the manual its written "If command is specified, it is executed on the remote host instead of a login shell"
So what happens is that when I specify my command:
ssh user@server "my_command"
It executed the command and the flow of execution returns to bash shell. So basically my session ends right after the command is executed. This happens only in case I specify command in the command line. If I login into ssh manually and then type "my_command", then the session doesn't end. I want the ssh not to exit, because after "my_command" executes, I want to capture everything in the session.
I have a command that outputs n lines of text, and I want to place each line into an array element, but I can't seem to get the syntax correct
So my command is this:
cat $configfile | sed -n '/cluster:'$clustername'/,/cluster/ p' | awk /host/
Which produces many lines depending on the value of $clustername. I'd like to get each line as elements of an array.
I wonder if there is anyway to make a user-defined bash shell function global, meaning the function can be use in any bash shell scripts, interactively or not. This is what I attempted:
Code:
$ tail -n 3 /etc/bashrc
echotm () {
echo "[`date`] $@"
}
[code]....
I am trying to fix a perl script, and I really suck at perl. But I think this problem will be easy for people who know it.
The problem is, I have an old setup script someone wrote many years ago. It fails if the standard shell is dash and not bash. The only way I've gotten it to work is to point /bin/sh to bash. I looked thru the script and it uses "system" many places, and I think that's the problem.
I searched for it and found this link:url
My plan is to include this function:
Code:
sub system_bash {
my @args = ( "bash", "-c", shift );
system(@args);
}
Then I could simply change all calls to system into system_bash and it should work?
The parameter to the system calls is usually some variable. What if the parameter is a list already? Do I need to test for it somehow, and if it's a list, prepend "bash" and "-c" to the list? How do I do that?
In the script there are lots of places like this:
my $error = system($cmd);
if ($error) {
die/warn "some error message";
}
Shouldn't there be a return in the system_bash function?
Trying to create a small script that will read user's input, test if user entered some input and if not display some message or display a text using user's input.
The script is the following but i get an error saying "[: 6: =: argument expected"
shell scripting in Fedora14I want a script"Find in curent folder for files, and it copy first file he find with name gived by user, if name already exist then echo error message and finish"command usage " bash scriptname copyASname"
smthing like Code: #!/bin/bash
for files in /home/user/*
do
[code]....
I'm running Fedora 10 [KDE] on an Acer laptop and am having problems configuring "hot keys" for it. First thing that needs to be mentioned is that the hotkeys used to work when I was running it on Gnome without me having to do anything. I assumed that this will be the case with KDE too. In any case, I think driver installation is supposed to be quite straight forward. There are two packages: acer_acpi and acerhk and different installation instructions for each one of them. The problem I am having is when I run the makefile script for either one of these, I get the following output in the terminal:
Quote:
Makefile: line 5: KERNELSRC?=/lib/modules/2.6.27.9-159.fc10.x86_64/build: No such file or directory
Makefile: line 6: KERNELSRC: command not found
Makefile: line 6: KERNELSRC: command not found
Makefile: line 6: shell: command not found
Makefile: line 6: shell: command not found
Makefile: line 8: KERNELVERSION: command not found
[Code].....
I think build-essential and some other packages are normally needed, but since you can't get them for Fedora (?), I groupinstalled "Development Tools" and some other group.
I'm relatively experienced with UNIX and Linux, but this has me thrown for quite a loop, and it seemed like such a simple question. How would I go about finding the newest file in a file system? I thought something like:
Code:
ls -ltr `find /usr -type f`
would work, but I seem to be exceeding the argument maximum for ls:
ksh: 0403-029 There is not enough memory available now
I thought something involving xargs might work, but I really suck with that command.
I'd like to say I'm very impressed with Fedora 11. I'm a long time Linux user and I've tried many distros. But, I usually keep only the best on my laptop. For a long time that was Ubuntu but, I think Fedora 11 has made some key improvements over Ubuntu and I'm eager to switch. The problem is: I haven't been able to run Fedora as anything other than on the Live CD. Everything works perfectly and it installs but, when I reboot, Grub begins. Instead of booting, however, Grub drops into its minimal shell and gives me a command line.
I've tried installing it a number of ways now and have read much about the problems with Ext4 on Grub and took special care to see that Grub has its own, separate, /boot ext3 partition. Even then, no luck. My hardware should work fine. I've got an HP DV-5 with 4GB RAM, AMD Turion 64-bit dual-core @ 2 Ghz, and an IDE 250GB hard drive. I'm working with the 64-bit Fedora 11 Live disc with KDE as the Gui.
Is there any linux command to find out if a user exists? It should something like this: if user exists it returns 1, if he doesn't it returns 0.
View 3 Replies View RelatedI'm fairly new to writing bash scripts and haven't been able to find a an example of effectively using a the find command in a bash script.
I want to run some git commands on any sub directory that has a .git project in it. Getting to the directory is not my problem, its how to find them
What i want to do is execute
Code:
find -name .git
Then act on each response line that is printed out. E.G Navigate to the directory and run git status.
How do i use the output of the find command in the bash script?
Linux command to find files changed in last n seconds. shell script,that we can run from cli or command.
View 3 Replies View RelatedI'm having problems with Tomboy. I have a few hundred note files and I need to go through all of them and replace all instances of "<link:broken>a</link:broken>" with "a". Is there a bash command I can use to do this?
View 2 Replies View RelatedIs there a non-root shell command that can tell me if a user's account is disabled or not? note that there is a fine distinction between LOCKING and DISABLED:
LOCKING is where you prepend ! or * or !! to the password field of the /etc/passwd file. On Linux systems that shadow the passwords, this marker flag may be placed in /etc/shadow instead of /etc/passwd. Password locking can be done (at a shell prompt) via password -l username (as root) to lock the account of username, and the use of the option -u will unlock it.
DISABLING an account is done by setting the expiration time of the user account to some point in the past. This can be done with chage -E 0 username, which sets the expiration date to 0 days after the Unix epoch. Setting it to -1 will disable the use of the expiration date.
The effect of locking to to prevent the login process from using a supplied password to hash correctly against the saved hash (by virtue of the fact that the pre-pended marker character(s) are not valid output character(s) for the hash, thus no possible input can ever be used to generate a hash that would match it). The effect of disabling is to prevent any process from using an account because the expiration date of the account has already passed.For my situation, the use of locking is not sufficient because a user might still be able to login, e.g. using ssh authentication tokens, and processes under that user can still spawn other processes. Thus, we have accounts that are enabled or disabled, not just locked. We already know how to disable and enable the account - it requires root access and the use of chage, as shown above.To repeat my question: is there a shell command which can be run without root privileges which can output the status of this account expiration info for a given user? this is intended for use on a Red Hat Enterprise 5.4 system.The output is being returned to a java process which can then parse the output as needed, or make use of the return code.
I wanted to find and replace a string from a perl file. I have written a script in bash which runs the following command.
perl -pi -e "s/$findstring/$replacestring/" testfile
where as $findstring = print F_WC_TMP"$line
";
and $replaceString = $line = join ' ', split ' ', $line; print F_WC_TMP"$line
";
But when I am running the above command, i think it is replacing the $findstring with the above mentioned string and hence it contains a $line, it is looking for the variable $line and not finding the exact string. I am confused about how to search for a string that contains $ in it and replace it with another $string.
What options should I use when I'm using the sort command to sort the top 5 CPU processes (ps -eo user,pid,ppid,%cpu,%mem,fname | sort ??? | head -5) showing max to min usage?
View 2 Replies View Relatedkernel 2.6.21.5, slackware 12.0
GNU bash 3.1.17
Code:
As you can see, /usr/local/bin is in the path. However, bash does not look for nasm in /usr/local/bin.
If I am root, things go well:
Code:
I know I was asked about wanting to change/upgrade my shell during the upgrade from Lenny to Squeeze, but I can't remember what it said, and I can't find any documentation about it online. But did something change to bash? Is this documented somehwere?Most notably I see progress messages on one line, overwriting eachother (for instance during an aptitude safe-upgrade).
View 7 Replies View RelatedIs there some type of functional way to read things in the Python shell interpreter similar to less or more in the bash (and other) command line shells?
Example:
Code:
>>> import subprocess
>>> help(subprocess)
...
[pages of stuff to read]
...
I'm hoping so as I hate scrolling and love how less works with simple keystrokes for page-up/page-down/searching etc.
I was giving the found the following shell script. I was told it was suppose to ensure only that only one script of Test.sh can run..
However, I get it looks like it has a error when i run it... As i get Test.sh: line 9: kill: (20831) - No such process
what is going on in this script can someone explain it to me... I thought it suppose to work like a singleton for my script creating a file .run-test-sdolan. However, i don't see how or where .run-test-sdolan is create?
sdolan@staging:$ vi Test.sh
#!/bin/sh
MYDIR=`dirname $0`
CONFDIR=$HOME/
code....
I use command "find" in my bash script: if the filename exist command find work quiet, and if the filename not exist I see the message "find: /tmp/filename: No such file or directory". My problem is following, i want to have in my script something like this:
find "/tmp/filename" -type f -delete | "if no_any_errors execute command1" , if file_not_found execute command2"
how come I can create a shell script file with two functions, I can execute the file, but when running declare -f, the functions are not on memory, and when invoking the function bash returns invalid. In the other hand, I can copy & paste the two functions at the end of my /etc/bashrc file.... then I can called the function by name.... and the commands within that function run on my session. here is a print of all my bash packets:
[Code]....
Does Fedora has restrictions on shell scripting? I haven't touch bash in seven years, so if things have change on it I'm behind on it, and sorry for my ignorance.
I recently discovered this forum and I really hope you can help me better understand this new parameter OS:)I wanted to ask you, using only debian shell and then the various applications will be installed and used only by the command line, is there any program that can tell me when to un'applicaizone installed an update is available online?
View 1 Replies View Related