define "$key" use it in a script, but also store it - bash

How can I create a specific line in another file using bash please? Like
echo "Please input the days you want to keep "
$key= ?
touch .beebrc; keep="$key"
where the file ".beebrc" has a line 'keep= x' and "$key" is created in the main script.
But how do I define "$key" please? And write it into ".beebrc" as a new line at position/line 8? The full function is -
function trim {
echo;
read -t "$temi" -n1 -p ""$bldgrn" Do you want to delete some of your download history? [y/n/q/r] $(tput sgr0)" ynqr ;
case "$ynqr" in
[Yy]) echo
read -t "$temi" -n3 -p ""$bldgrn" Please input the days you want to keep $(tput sgr0)" key ## ask
if test -e .beebrc && grep -q "^keep=" .beebrc 2>/dev/null ; then
sed -i "s/^keep=.*/keep=$key/" .beebrc
else
echo "keep=$key" >> .beebrc
#fi
cd /home/$USER/.get_iplayer
eval "$player" --trim-history "$key"; cd; ques;
#echo;;
[Nn]) ques;;
[Qq]) endex;;
[Rr]) exec "$beeb";;
* ) echo ""$bldgrn" Thank you $(tput sgr0)";;
esac
fi
};
Does this help in defining it all? (Sorry, should've put it in at first)

Perhaps:
read -p "Please input the days you want to keep: " key ## Ask.
echo "keep=\"$key\"" > .beebrc ## Store.

Use read to capture user input into a variable, and then write it to your file.
For example:
echo "Please input the days you want to keep "
read key
echo $key > .beebrc

#!/bin/bash
read -p "Please input the days you want to keep: " key
if test -e .beebrc && grep -q "^keep=" .beebrc 2>/dev/null ; then
sed -i "s/^keep=.*/keep=$key/" .beebrc
else
echo "keep=$key" >> .beebrc
fi
This script:
Prompts for input and stores the value in $key
Tests if .beebrc exists and that a line beginning "keep=" exists in it. If so, replace the keep= line with keep=$key
Otherwise append a new line/create the file with keep=$key.
This will need validation added because user input should not be trusted. (this answer might help)

Related

(Ubuntu bash script) Setting rights from a config txt

I am a beginner and trying to write a script that takes a config file (example below) and sets the rights for the users, if that user or group doesn´t exist, they get added.
For every line in the file, I am cutting out the user or the group and check if they exist.
Right now I only check for users.
#!/bin/bash
function SetRights()
{
if [[ $# -eq 1 && -f $1 ]]
then
for line in $1
do
var1=$(cut -d: -f2 $line)
var2=$(cat /etc/passwd | grep $var1 | wc -l)
if [[ $var2 -eq 0 ]]
then
sudo useradd $var1
else
setfacl -m $line
fi
done
else
echo Enter the correct path of the configuration file.
fi
}
SetRights $1
The config file looks like this:
u:TestUser:- /home/temp
g:TestGroup:rw /home/temp/testFolder
u:TestUser2:r /home/temp/1234.txt
The output:
grep: TestGroup: No such file or directory
grep: TestUser: No such file or directory
"The useradd help menu"
If you could give me a hint what I should look for in my research, I would be very grateful.
Is it possible to reset var1 and var2? Using unset didn´t work for me and I couldn´t find variables could only be set once.
It's not clear how you are looping over the contents of the file -- if $1 contains the file name, you should not be seeing the errors you report.
But anyway, here is a refactored version which hopefully avoids your problems.
# Avoid Bash-only syntax for function definition
SetRights() {
# Indent function body
# Properly quote "$1"
if [[ $# -eq 1 && -f "$1" ]]
then
# Read lines in file
while read -r acl file
do
# Parse out user
user=${acl#*:}
user=${user%:*}
# Avoid useless use of cat
# Anchor regex correctly
if ! grep -q "^$user:" /etc/passwd
then
# Quote user
sudo useradd "$user"
else
setfacl -m "$acl" "$file"
fi
done <"$1"
else
# Error message to stderr
echo Enter the correct path of the configuration file. >&2
# Signal failure to the caller
return 1
fi
}
# Properly quote argument
SetRights "$1"

Variable scope in Bash [duplicate]

Please explain to me why the very last echo statement is blank? I expect that XCODE is incremented in the while loop to a value of 1:
#!/bin/bash
OUTPUT="name1 ip ip status" # normally output of another command with multi line output
if [ -z "$OUTPUT" ]
then
echo "Status WARN: No messages from SMcli"
exit $STATE_WARNING
else
echo "$OUTPUT"|while read NAME IP1 IP2 STATUS
do
if [ "$STATUS" != "Optimal" ]
then
echo "CRIT: $NAME - $STATUS"
echo $((++XCODE))
else
echo "OK: $NAME - $STATUS"
fi
done
fi
echo $XCODE
I've tried using the following statement instead of the ++XCODE method
XCODE=`expr $XCODE + 1`
and it too won't print outside of the while statement. I think I'm missing something about variable scope here, but the ol' man page isn't showing it to me.
Because you're piping into the while loop, a sub-shell is created to run the while loop.
Now this child process has its own copy of the environment and can't pass any
variables back to its parent (as in any unix process).
Therefore you'll need to restructure so that you're not piping into the loop.
Alternatively you could run in a function, for example, and echo the value you
want returned from the sub-process.
http://tldp.org/LDP/abs/html/subshells.html#SUBSHELL
The problem is that processes put together with a pipe are executed in subshells (and therefore have their own environment). Whatever happens within the while does not affect anything outside of the pipe.
Your specific example can be solved by rewriting the pipe to
while ... do ... done <<< "$OUTPUT"
or perhaps
while ... do ... done < <(echo "$OUTPUT")
This should work as well (because echo and while are in same subshell):
#!/bin/bash
cat /tmp/randomFile | (while read line
do
LINE="$LINE $line"
done && echo $LINE )
One more option:
#!/bin/bash
cat /some/file | while read line
do
var="abc"
echo $var | xsel -i -p # redirect stdin to the X primary selection
done
var=$(xsel -o -p) # redirect back to stdout
echo $var
EDIT:
Here, xsel is a requirement (install it).
Alternatively, you can use xclip:
xclip -i -selection clipboard
instead of
xsel -i -p
I got around this when I was making my own little du:
ls -l | sed '/total/d ; s/ */\t/g' | cut -f 5 |
( SUM=0; while read SIZE; do SUM=$(($SUM+$SIZE)); done; echo "$(($SUM/1024/1024/1024))GB" )
The point is that I make a subshell with ( ) containing my SUM variable and the while, but I pipe into the whole ( ) instead of into the while itself, which avoids the gotcha.
#!/bin/bash
OUTPUT="name1 ip ip status"
+export XCODE=0;
if [ -z "$OUTPUT" ]
----
echo "CRIT: $NAME - $STATUS"
- echo $((++XCODE))
+ export XCODE=$(( $XCODE + 1 ))
else
echo $XCODE
see if those changes help
Another option is to output the results into a file from the subshell and then read it in the parent shell. something like
#!/bin/bash
EXPORTFILE=/tmp/exportfile${RANDOM}
cat /tmp/randomFile | while read line
do
LINE="$LINE $line"
echo $LINE > $EXPORTFILE
done
LINE=$(cat $EXPORTFILE)

how do i assigne out put of command as variable

Hi i am very new to scripting. please My apologizes if i am pointing in wrong way.
I am trying to develop a script which take the backup of given path. Below is my script.
The problem I facing is that I am trying to assign a variable "s" to an command "mkdir" and it do not work. Please help me,how i can correct this syntax?
#!/bin/bash
# to back up the given folder "
i="`date | awk '{ print $1$2$4}'`"
echo " please enter the full path of folder you want to back up"
read foldern
echo " $foldern is of `du -sh $foldern`. Do you want to back up this folder"
echo "yes / no"
read ans
if [ $ans = yes ]
then
echo " enter the back up folder name"
read bpn
s=$(mkdir $bpn$i) # here I am trying to assign a variable "s" for out put of mkdir but dosent work Please help me #
echo $s
cp -R "$foldern" "$s"
else
echo "no back up is taken"
fi
mkdir does not create an output or print the directory it has created. Manually create the string instead:
s=$bpn$i
mkdir -- "$s"
echo "$s"
-- is an optional option-argument separator so files beginning with - is not misinterpreted as a bad option to mkdir.
Adding -p can also be helpful if you don't want mkdir to show an error if the directory already exists.
mkdir -p -- "$s"
Suggestion:
#!/bin/bash
# To back up the given folder.
date=$(date | awk '{ print $1$2$4}') ## Consider date=$(date '+%F-%T')
read -p "Please enter the full path of folder you want to back up: " folder_name
size=$(du -sh "$folder_name")
read -p "$folder_name is of $size. Do you want to back up this folder (Yes/No)? "
if [[ $ans == [yY][eE][sS] ]]; then
read -p "Enter the back up folder name: " backup_name
backup_name+=$date
echo "$backup_name"
mkdir -p "$backup_name" && cp -R "$folder_name" "$backup_name"
else
echo "No back up is taken."
fi

Finding and adding a word to a particular line using shell script with exact format of a file

My problem is to add a username to a file, I really stuck to proceed, please help.
Problem: I am having a file called usrgrp.dat. The format of this file is like:
ADMIN:srikanth,admin
DEV:dev1
TEST:test1
I am trying to write a shell script which should give me the output like:
Enter group name: DEV
Enter the username: dev2
My expected output is:
User added to Group DEV
If I see the contents of usrgrp.dat, it should now look like:
DEV:dev1,dev2
TEST:test1
And it should give me error saying user already present if I am trying to add already existing user in that group. I am trying this out with the following script:
#!/bin/sh
dispgrp()
{
groupf="/home/srikanth/scm/auths/group.dat"
for gname in `cat $groupf | cut -f1 -d:`
do
echo $gname
done
echo "Enter the group name:"
read grname
for gname in `cat $groupf | cut -f1 -d:`
do
if [ "$grname" = "$gname" ]
then
echo "Enter the username to be added"
read uname
for grname in `cat $groupf`
do
$gname="$gname:$uname"
exit 1
done
fi
done
}
echo "Group display"
dispgrp
I am stuck and need your valuable help.
#!/bin/sh
dispgrp()
{
groupf="/home/srikanth/scm/auths/group.dat"
tmpfile="/path/to/tmpfile"
# you may want to pipe this to more or less if the list may be long
cat "$groupf" | cut -f1 -d:
echo "Enter the group name:"
read grname
if grep "$grname" "$groupf" >/dev/null 2>&1
then
echo "Enter the username to be added"
read uname
if ! grep "^$grname:.*\<$uname\>" "$groupf" >/dev/null 2>&1
then
sed "/^$grname:/s/\$/,$uname/" "$groupf" > "$tmpfile" && mv "$tmpfile" "$groupf"
else
echo "User $uname already exists in group $grname"
return 1
fi
else
echo "Group not found"
return 1
fi
}
echo "Group display"
dispgrp
You don't need to use loops when the loops are done for you (e.g. cat, sed and grep).
Don't use for to iterate over the output of cat.
Don't use exit to return from a function. Use return.
A non-zero exit or return code signifies an error or failure. Use 0 for normal, successful return. This is the implicit action if you don't specify one.
Learn to use sed and grep.
Since your shebang says #!/bin/sh, the changes I made above are based on the Bourne shell and assume POSIX utilities (not GNU versions).
Something like (assume your shell is bash):
adduser() {
local grp="$1"
local user="$2"
local gfile="$3"
if ! grep -q "^$grp:" "$gfile"; then
echo "no such group: $grp"
return 1
fi
if grep -q "^$grp:.*\\<$user\\>" "$gfile"; then
echo "User $user already in group $grp"
else
sed -i "/^$grp:/s/\$/,$user/" "$gfile"
echo "User $user added to group $grp"
fi
}
read -p "Enter the group name: " grp
read -p "Enter the username to be added: " user
adduser "$grp" "$user" /home/srikanth/scm/auths/group.dat

Bash variable scope

Please explain to me why the very last echo statement is blank? I expect that XCODE is incremented in the while loop to a value of 1:
#!/bin/bash
OUTPUT="name1 ip ip status" # normally output of another command with multi line output
if [ -z "$OUTPUT" ]
then
echo "Status WARN: No messages from SMcli"
exit $STATE_WARNING
else
echo "$OUTPUT"|while read NAME IP1 IP2 STATUS
do
if [ "$STATUS" != "Optimal" ]
then
echo "CRIT: $NAME - $STATUS"
echo $((++XCODE))
else
echo "OK: $NAME - $STATUS"
fi
done
fi
echo $XCODE
I've tried using the following statement instead of the ++XCODE method
XCODE=`expr $XCODE + 1`
and it too won't print outside of the while statement. I think I'm missing something about variable scope here, but the ol' man page isn't showing it to me.
Because you're piping into the while loop, a sub-shell is created to run the while loop.
Now this child process has its own copy of the environment and can't pass any
variables back to its parent (as in any unix process).
Therefore you'll need to restructure so that you're not piping into the loop.
Alternatively you could run in a function, for example, and echo the value you
want returned from the sub-process.
http://tldp.org/LDP/abs/html/subshells.html#SUBSHELL
The problem is that processes put together with a pipe are executed in subshells (and therefore have their own environment). Whatever happens within the while does not affect anything outside of the pipe.
Your specific example can be solved by rewriting the pipe to
while ... do ... done <<< "$OUTPUT"
or perhaps
while ... do ... done < <(echo "$OUTPUT")
This should work as well (because echo and while are in same subshell):
#!/bin/bash
cat /tmp/randomFile | (while read line
do
LINE="$LINE $line"
done && echo $LINE )
One more option:
#!/bin/bash
cat /some/file | while read line
do
var="abc"
echo $var | xsel -i -p # redirect stdin to the X primary selection
done
var=$(xsel -o -p) # redirect back to stdout
echo $var
EDIT:
Here, xsel is a requirement (install it).
Alternatively, you can use xclip:
xclip -i -selection clipboard
instead of
xsel -i -p
I got around this when I was making my own little du:
ls -l | sed '/total/d ; s/ */\t/g' | cut -f 5 |
( SUM=0; while read SIZE; do SUM=$(($SUM+$SIZE)); done; echo "$(($SUM/1024/1024/1024))GB" )
The point is that I make a subshell with ( ) containing my SUM variable and the while, but I pipe into the whole ( ) instead of into the while itself, which avoids the gotcha.
#!/bin/bash
OUTPUT="name1 ip ip status"
+export XCODE=0;
if [ -z "$OUTPUT" ]
----
echo "CRIT: $NAME - $STATUS"
- echo $((++XCODE))
+ export XCODE=$(( $XCODE + 1 ))
else
echo $XCODE
see if those changes help
Another option is to output the results into a file from the subshell and then read it in the parent shell. something like
#!/bin/bash
EXPORTFILE=/tmp/exportfile${RANDOM}
cat /tmp/randomFile | while read line
do
LINE="$LINE $line"
echo $LINE > $EXPORTFILE
done
LINE=$(cat $EXPORTFILE)

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