I'm trying to compare 2 strings. One comes from a file through the grep command and the other one never changes because I'm always comparing it with the ones I create while reading file texts. If they are equal, the program should print the data associated with the content that the new string contains. I've tried with all the syntax that bash allows (cause I'm new at bash) but it is just not working like I expected. It looks like the second if doesn't work, because I tried earlier without that and only print the strings (echo $text) and it worked, but not the proper way as the exercise I'm doing asks for. I have to show in the console only the pid and state of the processes that are running.
cd
cd /proc
run="State: S (sleeping)"
for i in $( ls -d */);
do
cd $i
if [ -f /proc/$i/status ];
then text="`grep -w "S" status`"
if [ "$text" == "$run" ]
then grep -w "Pid" status
grep -w "State" status
fi
cd /proc
else cd /proc
fi
done
;;
Your run variable contains State:<space>S (sleeping), but /proc/<pid>/status files contain State:<tab>S (sleeping) (at least, on my system). So you should replace that space character with tab character.
Related
I have a script which write some warnings to separate file (it's name is passed as an argument). I want to make this script fail if there is a warning.
So, I need to pass some file name which must raise an error if someone try to write there.
I want to use some more or less idiomatic name (to include it to man page of my script).
So, let say my script is
# myScript.sh
echo "Hello" > $1
If I call it with
./myScript.sh /dev/stdin
it is not fail because /dev/stdin is not read-only device (surprisingly!)
After
./myScript.sh /
it is failed, as I want it (because / is a directory, you can't write there). But is is not idiomatic.
Is there some pretty way to do it?
if [ -w "$1" ]
then
echo "$Hello" > "$1" # Mind the double-quotes
fi
is what you're looking for. Below would even be better in case you've only
one argument.
if [ -w "$*" ]
then
echo "$Hello" > "$*" # Mind the double-quotes
fi
$* is used to accommodate nonstandard file names. "$*" combines multiple arguments into a single word. Check [ this ].
In a book I'm reading the below line
ls "$1" 2>/dev/null | grep "$1" 2>/dev/null 1>&2
when written in a script - by the book it says "The command is executed to check whether the file passed as the command line argument exists. The standard error is redirected to /dev/null (the unix black hole), and standard output is redirected to standard error by using 1>&2. Thus, the command does not produce any output or error message; its only puprose is to set the command returns status value $?."
But running the code:
if [ $? -eq 0 ]
would I not know it otherwise, I have tried without the cmd at beginning and with it as well with having no impact on the results. I'm sure the author would have written for some purpose. I cannot just figure what?
This looks like a very bad book, giving code that noone sane would ever write to poorly illustrate concepts that are generally used in completely different ways in shell scripts.
The line:
ls "$1" 2>/dev/null | grep "$1" 2>/dev/null 1>&2
is as described -- it has no visible effect other than setting the return code. Is your question about what this does in detail to get a return code or something else?
The line:
if [ $? -eq 0 ]
is an incomplete fragment that checks the return code of the previous command. It's incomplete as there is no then or fi, without which the shell will reject it as a syntax error and not do anything (if you type the above at a prompt, you'll get the secondary prompt, telling you the shell is waiting for more input to get a complete command). So without more code there's no apparent effect. Something more complete like:
if [ $? -eq 0 ]; then echo YES; else echo NO; fi
would output YES or NO based on that return code.
A more sensible way of doing the 6 lines starting with the ls would be:
if [ ! -e "$1" ]; then
echo "$1: not found"
exit 1
fi
As to what the ls line actually does, it runs ls (list files) with the name in $1 as an argument, then uses grep to search that listing for the same filename.
So if the file does not exist, ls gives an error and outputs nothing, so the grep fails (setting $? to 1). If the filename exists and is not a directory, the grep will succeed (setting $? to 0). Finally, if the filename exists and is a directory, it will search the contents of that directory, looking for any file or subdirectory with the same name as a substring -- which is probably just a bug. In addition, if $1 is a string beginning with -, it will do something fairly useless and unpredictable.
Overall, a prime example of a shell script that should never be written -- any student that turned in such a monstrosity should get an immediate F.
Im wondering why awk print different output when run in background
My script:
#!/bin/bash
echo "Name of shell is $SHELL"
relase=`uname -r`
echo "Release is: $relase"
if [ $SHELL != "/bin/bash" ] || [ $relase != "3.13.0-32-generic" ] ; then
echo "Warning, different configuration"
fi
if [ $# -eq 0 ] ; then
echo "Insert name of shell"
read sname
else
sname=$1
fi
awk -v sname="$sname" 'BEGIN {FS=":"} {if ($7 == sname) print $1 }' </etc/passwd &
When i run awk without ampersand, output is:
petr#PetrLinux-VirtualBox:~/Documents$ ./script1 /bin/bash
Name of shell is /bin/bash
Release is: 3.13.0-32-generic
root
petr
but when i run awk with ampersand - in background, output is folowing:
petr#PetrLinux-VirtualBox:~/Documents$ ./script1 /bin/bash
Name of shell is /bin/bash
Release is: 3.13.0-32-generic
petr#PetrLinux-VirtualBox:~/Documents$ root
petr
First record (root) is not printed on single line. Please tell me why ańd if there is way how to print on single line while running on background. Thanks.
What you see is a mix of two outputs. The first output is of your shell, printing the command prompt (petr#PetrLinux-VirtualBox:~/Documents$). The second output is root from your script.
As your shell script runs in the background, you now have two processes writing to your terminal window: the bash (printing the prompt), and your script, printing the awk-output. This then just mixes up.
The only way to prevent that is to redirect the output of the script to a file or other device, instead of your console. For example:
$ ./script1 /bin/bash &> output.txt &
The output is the same. It just appears to be different because two processes write on the same channel (your terminal) and mix their output. One process is the awk script and the other is your shell which prints a new prompt.
There is no way to determine the precise point in which the output will switch from one process to the other. It can be different on different systems (with the same software), it can also depend on the load of the computer and lots of other things.
The only decent solution is to redirect the output into a different stream, e. g. a file using > outfile.
I wrote a zsh function to help me do some grepping at my job.
function rgrep (){
if [ -n "$1" ] && [ -n "$2" ]
then
exec grep -rnw $1 -r $2
elif [ -n "$1" ]
then
exec grep -rnw $1 -r "./"
else
echo "please enter one or two args"
fi
}
Works great, however, grep finishes executing I don't get thrown back into the shell. it just hangs at [process complete] any ideas?
I have the function in my .zshrc
In addition to getting rid of the unnecessary exec, you can remove the if statement as well.
function rgrep (){
grep -rwn "${1:?please enter one or two args}" -r "${2:-./}"
}
If $1 is not set (or null valued), an error will be raised and the given message displayed. If $2 is not set, a default value of ./ will be used in its place.
Do not use exec as it replace the existing shell.
exec [-cl] [-a name] [command [arguments]]
If command is supplied, it replaces the shell without creating a new process. If the -l option is supplied, the shell places a dash at the beginning of the zeroth argument passed to command. This is what the login program does. The -c option causes command to be executed with an empty environment. If -a is supplied, the shell passes name as the zeroth argument to command. If no command is specified, redirections may be used to affect the current shell environment. If there are no redirection errors, the return status is zero; otherwise the return status is non-zero.
Try this instead:
rgrep ()
{
if [ -n "$1" ] && [ -n "$2" ]
then
grep -rnw "$1" -r "$2"
elif [ -n "$1" ]
then
grep -rnw "$1" -r "./"
else
echo "please enter one or two args"
fi
}
As a completely different approach, I like to build command shortcuts like this as minimal shell scripts, rather than functions (or aliases):
% echo 'grep -rwn "$#"' >rgrep
% chmod +x rgrep
% ./rgrep
Usage: grep [OPTION]... PATTERN [FILE]...
Try `grep --help' for more information.
%
(This relies on a traditional behavior of Unix: executable text files without #! lines are considered shell scripts and are executed by /bin/sh. If that doesn't work on your system, or you need to run specifically under zsh, use an appropriate #! line.)
One of the main benefits of this approach is that shell scripts in a directory in your PATH are full citizens of the environment, not local to the current shell like functions and aliases. This means they can be used in situations where only executable files are viable commands, such as xargs, sudo, or remote invocation via ssh.
This doesn't provide the ability to give default arguments (or not easily, anyway), but IMAO the benefits outweigh the drawbacks. (And in the specific case of defaulting grep to search PWD recursively, the real solution is to install ack.)
I'm completing the following for one of my assignments using Korn shell.
For each argument in the argument list (which becomes the current pathname):
Check whether the current pathname is a directory, and if so:
Initialize a variable maxsubdir with the null (empty) string, and
a maxentries variable to 0;
For each entry in the directory check if that entry represents a
directory and if so, find the numbers of entries in that
subdirectory with a pipe consisting of ls -l and wc, and save the
result in a variable named curentries.
Compare curentries with maxentries, and if curentries is greater,
update maxsubdir and maxentries. (--10 points)
When the for cycle for a directory is completed, display (with
echo) the directory name, maxsubdir and maxentries (with appropriate
explanatory text.)
If the pathname in a) is not a directory, display the pathname
and an explanatory text saying that the pathname does not represent
a directory.
Go to the next command line argument (pathname) and repeat 1-7
The execution of the script ends when all pathnames are processed (the while is completed )
This is the code I have for it so far (EDITED):
#!/bin/ksh
directoy=$1
while [ $# -ne 0 ]; do
if [ -d $1 ]; then
maxsubdir=
maxentries=0
for x in $1; do
echo "Checking if $1 represents a directory..\n"
curentries="ls -l | wc"
if [ $curentries > $maxentries ]; then
maxentries=$curentries
maxsubdir=$curentries
fi;
done
echo "The directory structure of $1 is … \n"
echo "Maximum sub directories: \n"
echo "$maxsubdir\n"
echo "Maximum directory entries: \n"
echo "$maxentries"
fi
done
Where do I need to insert the "shift" command since I Unix can only handle a limited number of arguments?
Is my syntax appropriate? Or do I have syntax errors on sort lines?
Script seems to run but does not produce output to screen? Perhaps it's endless?
Have a look here and see if this helps out. Explanations are in the code.
#!/bin/ksh
directory=$1
# check whether the entered path is a directory
if [ -d $1 ];then # yes, it's a directory
maxsubdir=null
maxentries=0
echo "$1 is a directory"
# you are only counting lines, add -l to wc
# also you have to not count the first line. it's returns the size
curentries=`ls -l $1 | wc -l`
echo ${curentries}
fi
You don't.
You do have some errors.
Or perhaps, it never reaches that code?
Your assignment says specifically to use a for loop, and you've implemented a while loop.
I'll get you started:
for directory in $*; do
cd "$directory"
curentries=$(ls -1 | wc -l)
for entry in $(ls -1); do
...
done
done