This question is based on this thread.
The code
function man()
{
man "$1" > /tmp/manual; less /tmp/manual
}
Problem: if I use even one option, the command does not know where is the wanted-manual
For instance,
man -k find
gives me an error, since the reference is wrong. The command reads -k as the manual.
My attempt to solve the problem in pseudo-code
if no parameters
run the initial code
if one parameter
run: man "$2"
...
In other words, we need to add an option-check to the beginning such that
Pseudo-code
man $optional-option(s) "$n" > /tmp/manual; less /tmp/manual
where $n
n=1 if zero options
n=2 if 1 option
n=3 if 2 options
....
How can you make such an "option-check" that you can alter the value of $n?
Developed Problem: to make two if loops for the situations from n=1 to n=2
How about passing all the arguments
function man()
{
man $# > /tmp/manual; less /tmp/manual
}
What is the bug in less which you mention in the title?
First, you can pass all of your function's arguments to man by using $* or $#. You can read man sh for the precise details on the difference between the two; short story is to almost always use "$#" with double quotes.
Second, the temporary file is unnecessary. You could make this a little cleaner by piping the output of man directly to less:
function man() {
man "$#" | less
}
By the way, if you're just trying to use a different pager (man uses more and you want the fancier less) there's a commonly recognized PAGER environment variable that you can set to override the default pager. You could add this to your ~/.bashrc for instance to tell all programs to use less when displaying multiple screens of output:
export PAGER=less
To answer your precise question, you can check the number of arguments with $#:
if [ $# -eq 0 ]; then
: # No arguments
elif [ $# -eq 1 ]; then
: # One argument
# etc.
You might also find the shift command helpful. It renames $2 to $1, $3 to $2, and so on. It is often used in a loop to process command-line arguments one by one:
while [ $# -gt 1 ]; do
echo "Next argument is: $1"
shift
done
echo "Last argument is: $1"
Related
-- Edit : Resolved. See answer.
Background:
I'm writing a shell that will perform some extra actions required on our system when someone resizes a database.
The shell is written in ksh (requirement), the OS is Solaris 5.10 .
The problem is with one of the checks, which verifies there's enough free space on the underlying OS.
Problem:
The check reads the df -k line for root, which is what I check in this step, and prints it to a file. I then "read" the contents into variables which I use in calculations.
Unfortunately, when I try to run an arithmetic operation on one of the variables, I get an error indicating it is null. And a debug output line I've placed after that line verifies that it is null... It lost it's value...
I've tried every method of doing this I could find online, they work when I run it manually, but not inside the shell file.
(* The file does have #!/usr/bin/ksh)
Code:
df -k | grep "rpool/ROOT" > dftest.out
RPOOL_NAME=""; declare -i TOTAL_SIZE=0; USED_SPACE=0; AVAILABLE_SPACE=0; AVAILABLE_PERCENT=0; RSIGN=""
read RPOOL_NAME TOTAL_SIZE USED_SPACE AVAILABLE_SPACE AVAILABLE_PERCENT RSIGN < dftest.out
\rm dftest.out
echo $RPOOL_NAME $TOTAL_SIZE $USED_SPACE $AVAILABLE_SPACE $AVAILABLE_PERCENT $RSIGN
((TOTAL_SIZE=$TOTAL_SIZE/1024))
This is the result:
DBResize.sh[11]: TOTAL_SIZE=/1024: syntax error
I'm pulling hairs at this point, any help would be appreciated.
The code you posted cannot produce the output you posted. Most obviously, the error is signalled at line 11 but you posted fewer than 11 lines of code. The previous lines may matter. Always post complete code when you ask for help.
More concretely, the declare command doesn't exist in ksh, it's a bash thing. You can achieve the same result with typeset (declare is a bash equivalent to typeset, but not all options are the same). Either you're executing this script with bash, or there's another error message about declare, or you've defined some additional commands including declare which may change the behavior of this code.
None of this should have an impact on the particular problem that you're posting about, however. The variables created by read remain assigned until the end of the subshell, i.e. until the code hits a ), the end of a pipe (left-hand side of the pipe only in ksh), etc.
About the use of declare or typeset, note that you're only declaring TOTAL_SIZE as an integer. For the other variables, you're just assigning a value which happens to consist exclusively of digits. It doesn't matter for the code you posted, but it's probably not what you meant.
One thing that may be happening is that grep matches nothing, and therefore read reads an empty line. You should check for errors. Use set -e in scripts to exit at the first error. (There are cases where set -e doesn't catch errors, but it's a good start.)
Another thing that may be happening is that df is splitting its output onto multiple lines because the first column containing the filesystem name is too large. To prevent this splitting, pass the option -P.
Using a temporary file is fragile: the code may be executed in a read-only directory, another process may want to access the same file at the same time... Here a temporary file is useless. Just pipe directly into read. In ksh (unlike most other sh variants including bash), the right-hand side of a pipe runs in the main shell, so assignments to variables in the right-hand side of a pipe remain available in the following commands.
It doesn't matter in this particular script, but you can use a variable without $ in an arithmetic expression. Using $ substitutes a string which can have confusing results, e.g. a='1+2'; $((a*3)) expands to 7. Not using $ uses the numerical value (in ksh, a='1+2'; $((a*3)) expands to 9; in some sh implementations you get an error because a's value is not numeric).
#!/usr/bin/ksh
set -e
typeset -i TOTAL_SIZE=0 USED_SPACE=0 AVAILABLE_SPACE=0 AVAILABLE_PERCENT=0
df -Pk | grep "rpool/ROOT" | read RPOOL_NAME TOTAL_SIZE USED_SPACE AVAILABLE_SPACE AVAILABLE_PERCENT RSIGN
echo $RPOOL_NAME $TOTAL_SIZE $USED_SPACE $AVAILABLE_SPACE $AVAILABLE_PERCENT $RSIGN
((TOTAL_SIZE=TOTAL_SIZE/1024))
Strange...when I get rid of your "declare" line, your original code seems to work perfectly well (at least with ksh on Linux)
The code :
#!/bin/ksh
df -k | grep "/home" > dftest.out
read RPOOL_NAME TOTAL_SIZE USED_SPACE AVAILABLE_SPACE AVAILABLE_PERCENT RSIGN < dftest.out
\rm dftest.out
echo $RPOOL_NAME $TOTAL_SIZE $USED_SPACE $AVAILABLE_SPACE $AVAILABLE_PERCENT $RSIGN
((TOTAL_SIZE=$TOTAL_SIZE/1024))
print $TOTAL_SIZE
The result :
32962416 5732492 25552588 19% /home
5598
Which are the value a simple df -k is returning. The variables seem to last.
For those interested, I have figured out that it is not possible to use "read" the way I was using it.
The variable values assigned by "read" simply "do not last".
To remedy this, I have applied the less than ideal solution of using the standard "while read" format, and inside the loop, echo selected variables into a variable file.
Once said file was created, I just "loaded" it.
(pseudo code:)
LOOP START
echo "VAR_A="$VAR_A"; VAR_B="$VAR_B";" > somefile.out
LOOP END
. somefile.out
I wrote configuration of fish shell like this:
# One or more argument(s) will be given
function run
set -l src $argv[1]
set -l var
switch "$src"
case *
set var "$src"
end
echo $var
end
I expected the first argument is printed in any case if one or more argument is given. However, $var becomes $argv[1] if file with the same name as $argv[1] exists, otherwise it becomes empty string.
Could someone tell me why?
The * in case * is interpreted as a glob. Quote it if you do not want that.
Faho already answered your question but I wanted to point out that your approach is more complicated than necessary. If you just want to print the first argument if one or more were provided do this:
set -q argv[1]
and echo $argv[1]
The first statement checks if argv has at least one value. The second echoes it if the previous statement returned success (i.e., $status set to zero).
Using fish shell, I'm writing very simple script that checks the command execution
#!/usr/bin/fish
command
if $status
echo "Oops error"
else
echo "Worked OK"
#...
end
And get the error message:
fish: Variables may not be used as commands. Instead, define a function like “function status; 0 $argv; end”. See the help section for the function command by typing “help function”.
The message looks pretty straight forward but no "defining function like..." nor "help function" helps solving the problem.
There is also a 'test' command, that sounds promising. But docs say it is to be used to check files...
How this simple thing should be done with fish shell?
Heh... And why all documentation is SO misleading?..
P.S. Please, don't write about 'and' command.
Fish's test command currently works exactly like POSIX test (i.e. the one you'll find in bash or similar shells). It has a couple of operations, including "-gt", "-eq", "-lt" to check if a number is bigger, equal or less than another number, respectively.
So if you want to use test, you'll do if test $status -eq 0 (a 0 traditionally denotes success). Otherwise, you can check the return value of a command by putting it in the if clause directly like if command (which will be true if the command returns 0) - that's what fish is trying to do here, which is why it complains about a variable being used in place of a command.
I've been searching all over the place and since I'm taking my first steps in PERL this might be one of he dumbest questions but here it goes.
So I'm creating a script to manage my windows and later bind it to keyboard shortcuts, so I I'm trying to run a command and passing some variables:
my $command = `wmctrl -r :ACTIVE: -e 0,0,0,$monitors->{1}->{'width'}/2,$monitors->{1}->{'height'}`;
But I get an error saying I'm not passing the right parameters to the command, but if I do this, everything works great:
my $test = $monitors->{1}->{'width'}/2;
my $command = `wmctrl -r :ACTIVE: -e 0,0,0,$test,$monitors->{1}->{'height'}`;
So do I really have to do this? assign it first to a variable and then pass it, or there's a more elegant way of doing it?
The backticks operator (or the qx{}) accepts A string which is (possibly) interpolated. So accepts string and not expression like $var/2.
Thats mean than the $variables ($var->{1}->{some} too) are expanded but not the arithmetic expressions.
Therefore your 2 step variant works, but not the first.
If you want evaluate an expression inside the string you can use the next:
my $ans=42;
print "The #{[ $ans/2 ]} is only the half of answer\n";
prints
The 21 is only the half of answer
but it is not very readable, so better and elegant is what you're already doing - calculate the command argument in andvace, and to the qx{} or backticks only pass the calculated $variables.
I'm re-learning scripting and found an example the book doesn't explain very well.
e.g.:
if [ ! $# -eq 0 ]
This is testing to determine if '$#' does not equal zero, yes?
But what then is the value of '$#'?
Are there others?
'##', '#?' ?
Thank you
$# is the number of the arguments you passed to your script.
For example, your have script called a.sh,
#!/bin/bash
echo $#
And you run it like
/bin/bash a.sh 1 2 3
you will get 3.
There are others like $#.