Deleting a line from a specific variable - bash

Using sed I'm able to delete a line of code inside the command prompt. When i go to use it inside a bash script and use a variable it doesn't delete it.
function remove_user {
echo "Input user you would like to delete"
read rui
if grep -q $rui database.txt ;
then
echo "Are you sure yu want to delete this user?"
echo "(Y/N)"
read input3
if [ $input3 == "Y" ] || [ $input3 == "y" ] ;
then
sed -i '/"$rui"/d' ./databse.txt
echo "User Deleted"
echo "returning to the main menu..."
echo "$rui"
#sleep 1 ; clear
elif [ $input3 == "N" ] || [$input3 == "n" ] ;
then
echo "returning to main menu..."
sleep 1 ; clear
else
echo "input invalid"
echo "returning to main menu..."
sleep 1 ; clear
fi
else
echo " user not found"
echo "returning to main menu..."
sleep 1 ; clear
fi
My database looks like this
Larry:Larry#hotmail.com:Larry Bob:ATC:4.0
Not sure what the issue could be because the code works just not with a variable

You need to make the variable visible to the shell by breaking back out of the single-quote delimited sed script briefly. You do that by adding a terminating single quote before the variable, and a starting one again after it:
sed -i '/'"$rui"'/d' ./databse.txt
The reason you don't just use double quotes around the whole script is that sets you up for surprises later when the shell interprets you're whole script instead of just the variable you want it to expand. e.g.:
$ echo "money is nice" | sed 's/money/$$/'
$$ is nice
$ foo="money"; echo "money is nice" | sed 's/'"$foo"'/$$/'
$$ is nice
$ foo="money"; echo "money is nice" | sed "s/$foo/$$/"
6756 is nice
That last one happens because the double quotes expose your whole script to the shell before sed sees it and the shell interprets $$ as the current PID.

Related

How to execute a file that is located in $PATH

I am trying to execute a hallo_word.sh that is stored at ~/bin from this script that is stored at my ~/Desktop. I have made both scripts executable. But all the time I get the problem message. Any ideas?
#!/bin/sh
clear
dir="$PATH"
read -p "which file you want to execute" fl
echo ""
for fl in $dir
do
if [ -x "$fl" ]
then
echo "executing=====>"
./$fl
else
echo "Problem"
fi
done
This line has two problems:
for fl in $dir
$PATH is colon separated, but for expects whitespace separated values. You can change that by setting the IFS variable. This changes the FIELD SEPARATOR used by tools like for and awk.
$fl contains the name of the file you want to execute, but you overwrite its value with the contents of $dir.
Fixed:
#!/bin/sh
clear
read -p "which file you want to execute" file
echo
IFS=:
for dir in $PATH ; do
if [ -x "$dir/$file" ]
then
echo "executing $dir/$file"
exec "$dir/$file"
fi
done
echo "Problem"
You could also be lazy and let a subshell handle it.
PATH=(whatever) bash command -v my_command
if [ $? -ne 0 ]; then
# Problem, could not be found.
else
# No problem
fi
There is no need to over-complicate things.
command(1) is a builtin command that allows you to check if a command exists.
The PATH value contains all the directories in which executable files can be run without explicit qualification. So you can just call the command directly.
#!/bin/sh
clear
# r for raw input, e to use readline, add a space for clarity
read -rep "Which file you want to execute? " fl || exit 1
echo ""
"$fl" || { echo "Problem" ; exit 1 ; }
I quote the name as it could have spaces.
To test if the command exists before execution use type -p
#!/bin/sh
clear
# r for raw input, e to use readline, add a space for clarity
read -rep "Which file you want to execute? " fl || exit 1
echo ""
type -p "$fq" >/dev/null || exit 1
"$fl" || { echo "Problem" ; exit 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)

Shell Script not reading input

I have written a script that backs up and restores files. I have a problem in that when the user enters '2' for a restore the program says that this is an invalid input, all other options work fine. I feel it is something small that I have missed but I cant fix it
Update and Restore Script
#!/bin/bash
ROOT="/Users/Rory/Documents"
ROOT_EXCLUDE="--exclude=/dev --exclude=/proc --exclude=/sys --exclude=/temp --exclude=/run --exlucde=/mnt --exlcude=/media --exlude=/backup2.tgz"
DESTIN="/Users/Rory/test/"
BACKUP="backup2.tgz"
CREATE="/dev /proc /sys /temp /run /mnt /media "
if [ "$USER" != "root" ]; then
echo "You are not the root user"
echo "To use backup please use: sudo backup"
exit
fi
clear
echo "************************************************"
echo "********* Backup Menu **************************"
echo "************************************************"
OPTIONS="BACKUP RESTORE DESTINATION EXIT"
LIST="1)BACKUP 2)RESTORE 3)DESTINATION 4)EXIT"
select opt in $OPTIONS; do
if [ "$opt" = "EXIT" ]; then
echo "GOODBYE!"
sleep 3
clear
exit
elif [ "$opt" = "BACKUP" ]; then
echo "BACKING UP FILES..."
sleep 2
tar cvpfz $DESTIN/backup.`date +%d%m%y_%k:%M`.tgz $ROOT $ROOT_EXCLUDE_DIRS
echo "BACKUP COMPLETE"
sleep 2
exit
elif [ "$opt" = "RESTORE" ]; then
echo "RESTOTING FILES..."
sleep 2
tar xvpfz $BACKUP_FILE -C /
sleep2
echo "RESTORE COMPLETE..."
if [[ -e "/proc" ]]; then
echo "$CREATE_DIRS allready exists! "
else
mkdir $CREATE_DIRS
echo "$CREATE_DIRS are created! "
fi
exit
elif [ "$opt" = "DESTINATION" ]; then
echo "CURRENT DESTINATION: $DEST_DIR/backup.`date +%d/%m/%y_%k:%M`.tgz "
echo "TO CHANGE ENTER THE NEW DESTINATION..."
echo "TO LEAVE IT AS IS JUST PRESS ENTER..."
read NEW_DESTIN
#IF GREATER THEN 0 ASSIGN NEW DESTINATION
if [ ${#NEW_DESTIN} -gt 0 ]; then
DESTIN = "$NEW_DESTIN"
fi
clear
echo $BANNER1
echo $BANNER2
echo $BANNER3
echo $LIST
else
clear
echo "BAD INPUT!"
echo "ENTER 1 , 2, 3 or 4.."
echo $LIST
fi
done
Except where you missed the ending quote where you set ROOT_EXCLUDE (line #4), it looks okay to me. I take it the missing quote is a transcription error or your program wouldn't really work at all.
I've tried out the program and it seems to work.
A debugging trick is to put set -xv to turn on debugging in your script and set +xv to turn it off. The -x means to print out the line before executing, and the -v means to print out the line once the shell interpolates the line.
I'm sure that you'll immediately see the issue once you have set -xv in your program.
As part of this, you can set PS4 to the line prompt to print when the debugging information is printed. I like setting PS4 like this:
export PS4="[\$LINENO]> "
This way, the line prompt prints out the line it's executing which is nice.
In your case, I would put set -xv right before you set OPTIONS and then at the very end of the program. This way, you can see the if comparisons and maybe spot your issue.
export PS4="[\$LINENO]> "
set -xv
OPTIONS="BACKUP RESTORE DESTINATION EXIT"
LIST="1)BACKUP 2)RESTORE 3)DESTINATION 4)EXIT"
select opt in $OPTIONS; do
if [ "$opt" = "EXIT" ]; then
echo "GOODBYE!"
set +xv
By the way, it's better to use double square brackets like [[ ... ]] for testing rather than the single square brackets like [ ... ]. This has to do with the way the shell interpolates the values in the test.
The [ ... ] is an alias to the built in test command. The shell interpolates the line as is and the entire line is executed.
The [[ ... ]] are a compound statement where the shell will interpolate variables, but not the entire line. The line is kept as whole:
foo="one thing"
bar="another thing"
This will work:
if [ "$foo" = "$bar" ]
then
echo "Foo and bar are the same"
fi
This won't:
if [ $foo = $bar ]
then
echo "Foo and bar are the same"
fi
The shell interpolates the line as is:
if [ one thing = another thing ]
And this is the same as:
if test one thing = another thing
The test command looks at the first item to see if it's a standard test, or assumes three items and the second item is a comparison. In this case, neither is true.
However, this will work:
if [[ $foo = $bar ]] # Quotes aren't needed
then
echo "Foo and bar are the same"
fi
With the [[ ... ]] being a compound command, the $foo and $bar are replaced with their values, but their positions are kept. Thus, the = is recognized as a comparison operator.
Using [[ ... ]] instead of [ ... ] has solved a lot of hard to find shell scripting bugs I have.

Check if file exists [BASH]

How do I check if file exists in bash?
When I try to do it like this:
FILE1="${#:$OPTIND:1}"
if [ ! -e "$FILE1" ]
then
echo "requested file doesn't exist" >&2
exit 1
elif
<more code follows>
I always get following output:
requested file doesn't exist
The program is used like this:
script.sh [-g] [-p] [-r FUNCTION_ID|-d FUNCTION_ID] FILE
Any ideas please?
I will be glad for any help.
P.S. I wish I could show the entire file without the risk of being fired from school for having a duplicate. If there is a private method of communication I will happily oblige.
My mistake. Fas forcing a binary file into a wrong place. Thanks for everyone's help.
Little trick to debugging problems like this. Add these lines to the top of your script:
export PS4="\$LINENO: "
set -xv
The set -xv will print out each line before it is executed, and then the line once the shell interpolates variables, etc. The $PS4 is the prompt used by set -xv. This will print the line number of the shell script as it executes. You'll be able to follow what is going on and where you may have problems.
Here's an example of a test script:
#! /bin/bash
export PS4="\$LINENO: "
set -xv
FILE1="${#:$OPTIND:1}" # Line 6
if [ ! -e "$FILE1" ] # Line 7
then
echo "requested file doesn't exist" >&2
exit 1
else
echo "Found File $FILE1" # Line 12
fi
And here's what I get when I run it:
$ ./test.sh .profile
FILE1="${#:$OPTIND:1}"
6: FILE1=.profile
if [ ! -e "$FILE1" ]
then
echo "requested file doesn't exist" >&2
exit 1
else
echo "Found File $FILE1"
fi
7: [ ! -e .profile ]
12: echo 'Found File .profile'
Found File .profile
Here, I can see that I set $FILE1 to .profile, and that my script understood that ${#:$OPTIND:1}. The best thing about this is that it works on all shells down to the original Bourne shell. That means if you aren't running Bash as you think you might be, you'll see where your script is failing, and maybe fix the issue.
I suspect you might not be running your script in Bash. Did you put #! /bin/bash on the top?
script.sh [-g] [-p] [-r FUNCTION_ID|-d FUNCTION_ID] FILE
You may want to use getopts to parse your parameters:
#! /bin/bash
USAGE=" Usage:
script.sh [-g] [-p] [-r FUNCTION_ID|-d FUNCTION_ID] FILE
"
while getopts gpr:d: option
do
case $option in
g) g_opt=1;;
p) p_opt=1;;
r) rfunction_id="$OPTARG";;
d) dfunction_id="$OPTARG";;
[?])
echo "Invalid Usage" 1>&2
echo "$USAGE" 1>&2
exit 2
;;
esac
done
if [[ -n $rfunction_id && -n $dfunction_id ]]
then
echo "Invalid Usage: You can't specify both -r and -d" 1>&2
echo "$USAGE" >2&
exit 2
fi
shift $(($OPTIND - 1))
[[ -n $g_opt ]] && echo "-g was set"
[[ -n $p_opt ]] && echo "-p was set"
[[ -n $rfunction_id ]] && echo "-r was set to $rfunction_id"
[[ -n $dfunction_id ]] && echo "-d was set to $dfunction_id"
[[ -n $1 ]] && echo "File is $1"
To (recap) and add to #DavidW.'s excellent answer:
Check the shebang line (first line) of your script to ensure that it's executed by bash: is it #!/bin/bash or #!/usr/bin/env bash?
Inspect your script file for hidden control characters (such as \r) that can result in unexpected behavior; run cat -v scriptFile | fgrep ^ - it should produce NO output; if the file does contain \r chars., they would show as ^M.
To remove the \r instances (more accurately, to convert Windows-style \r\n newline sequences to Unix \n-only sequences), you can use dos2unix file to convert in place; if you don't have this utility, you can use sed 's/'$'\r''$//' file > outfile (CAVEAT: use a DIFFERENT output file, otherwise you'll destroy your input file); to remove all \r instances (even if not followed by \n), use tr -d '\r' < file > outfile (CAVEAT: use a DIFFERENT output file, otherwise you'll destroy your input file).
In addition to #DavidW.'s great debugging technique, you can add the following to visually inspect all arguments passed to your script:
i=0; for a; do echo "\$$((i+=1))=[$a]"; done
(The purpose of enclosing the value in [...] (for example), is to see the exact boundaries of the values.)
This will yield something like:
$1=[-g]
$2=[input.txt]
...
Note, though, that nothing at all is printed if no arguments were passed.
Try to print FILE1 to see if it has the value you want, if it is not the problem, here is a simple script (site below):
#!/bin/bash
file="${#:$OPTIND:1}"
if [ -f "$file" ]
then
echo "$file found."
else
echo "$file not found."
fi
http://www.cyberciti.biz/faq/unix-linux-test-existence-of-file-in-bash/
Instead of plucking an item out of "$#" in a tricky way, why don't you shift off the args you've processed with getopts:
while getopts ...
done
shift $(( OPTIND - 1 ))
FILE1=$1

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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