Bash Shell Scripting - detect the Enter key - bash

I need to compare my input with Enter/Return key...
read -n1 key
if [ $key == "\n" ]
echo "###"
fi
But this is not working.. What is wrong with this code

Several issues with the posted code. Inline comments detail what to fix:
#!/bin/bash
# ^^ Bash, not sh, must be used for read options
read -s -n 1 key # -s: do not echo input character. -n 1: read only 1 character (separate with space)
# double brackets to test, single equals sign, empty string for just 'enter' in this case...
# if [[ ... ]] is followed by semicolon and 'then' keyword
if [[ $key = "" ]]; then
echo 'You pressed enter!'
else
echo "You pressed '$key'"
fi

Also it is good idea to define empty $IFS (internal field separator) before making comparisons, because otherwise you can end up with " " and "\n" being equal.
So the code should look like this:
# for distinguishing " ", "\t" from "\n"
IFS=
read -n 1 key
if [ "$key" = "" ]; then
echo "This was really Enter, not space, tab or something else"
fi

I'm adding below code just for reference if someone will want to use such solution containing countdown loop.
IFS=''
echo -e "Press [ENTER] to start Configuration..."
for (( i=10; i>0; i--)); do
printf "\rStarting in $i seconds..."
read -s -N 1 -t 1 key
if [ "$key" = $'\e' ]; then
echo -e "\n [ESC] Pressed"
break
elif [ "$key" == $'\x0a' ] ;then
echo -e "\n [Enter] Pressed"
break
fi
done

read reads a line from standard input, up to but not including the new line at the end of the line. -n specifies the maximum number of characters, forcing read to return early if you reach that number of characters. It will still end earlier however, when the Return key is pressed. In this case, its returning an empty string - everything up to but not including the Return key.
You need to compare against the empty string to tell if the user immediately pressed Return.
read -n1 KEY
if [[ "$KEY" == "" ]]
then
echo "###";
fi

None of these conditions worked for me and so I've came up with this one:
${key} = $'\0A'
Tested on CentOS with Bash 4.2.46.

Related

Asking for user input in a loop until match found in array of command line arguments

Sometimes I need to find a specific serial in a box with many items, so I wrote a simple Bash script that allows me to use a barcode scanner to scan hundreds of barcodes until a match is found, at which point the screen flashes (so I can see it from the corner of my eyes while looking at the box).
The script works great, but it only checks against one specific serial number provided by the user. Here's the code:
#!/bin/bash
INPUT=''
SCAN=''
SN=''
I='0'
clear
printf "Enter serial\n"
read INPUT
SN=`printf "${INPUT}" | tr '[:lower:]' '[:upper:]'`
# Keep comparing scans to needed serial until a match is found
while [[ "${SCAN}" != *"${SN}"* ]];
do
clear
printf "Looking for [ "${SN}" ]\n"
printf "Please scan barcode\n"
read INPUT
SCAN=`printf "${INPUT}" | tr '[:lower:]' '[:upper:]'`
done
# Flash screen when match is found
while [[ "${I}" -lt 3 ]];
do
printf '\e[?5h' && sleep 0.3
printf '\e[?5l' && sleep 0.3
I=$[${I}+1]
done
printf "FOUND\n"
Today I spent hours trying to implement a way to pass multiple possible serial numbers as command line arguments, but I can't seem to get it working. I would like to be able to pass a small, manageable number of possible serials, like this:
$ ./script.sh sn1 sn2 sn3 sn4 sn5
And for the script continue asking for input until I come across the item I am looking for.
I've studied the handling of shell arguments, but I can't seem to "massage" the above while loop to get it to check if the scanned serial exists in the array (created from the command line arguments passed):
#!/bin/bash
snList=( "$#" )
INPUT=''
SCAN=''
SN=''
I='0'
clear
#displaying "things" so I can see what each variable contains (debugging)
printf "$#\n"
printf "$0\n"
printf "$*\n"
printf "$0\n"
printf "$1\n"
printf "$2\n"
printf "$3\n"
printf "snList: $snList\n"
printf "snList[#]: ${snList[#]}\n"
printf "snList[*]: ${snList[*]}\n"
# Keep comparing scans to needed serial until a match is found
while [[ ! " ${snList[*]} " =~ "${SCAN}" ]];
do
clear
printf "Looking for [ "$*" ]\n"
printf "Please scan barcode\n"
read INPUT
SCAN=`printf "${INPUT}" | tr '[:lower:]' '[:upper:]'`
done
I've tried using ${snList[#]} in the loop as well, same result, it behaves like a match was found immediately, without even asking for a scan (indicating that the content of the while loop is not being executed).
Any help will be immensely appreciated, I think I am close, but I can't figure out what I am doing wrong.
Thanks in advance!
Something like this maybe?
#!/usr/bin/env bash
to_compare_input=("$#")
exglob_pattern_input=$(IFS='|'; printf '%s' "#(${to_compare_input[*]})")
until [[ $user_input == $exglob_pattern_input ]]; do
read -r user_input
done
Run the script with the the following arguments.
bash -x ./myscript foo bar baz more
Output
+ to_compare_input=("$#")
++ IFS='|'
++ printf %s '#(foo|bar|baz|more)'
+ exglob_pattern_input='#(foo|bar|baz|more)'
+ [[ '' == #(foo|bar|baz|more) ]]
+ read -r user_input
papa
+ [[ papa == #(foo|bar|baz|more) ]]
+ read -r user_input
mama
+ [[ mama == #(foo|bar|baz|more) ]]
+ read -r user_input
baz
+ [[ baz == #(foo|bar|baz|more) ]]
The first user input is empty since the builtinread has not been executed to ask for the user's input. As shown at the debug message.
+ [[ '' == #(foo|bar|baz|more) ]]
The second (assuming the user has entered papa) is papa
The third (assuming the user has entered mama) is mama
The last is baz which breaks out of off the until loop, because it belongs to the $extglob_pattern_input, which is an extglob feature.
A regex is also an alternative using the =~ operator.
#!/usr/bin/env bash
to_compare_input=("$#")
regex_pattern_input=$(IFS='|'; printf '%s' "^(${to_compare_input[*]})$")
until [[ $user_input =~ $regex_pattern_input ]]; do
read -r user_input
done
Run the script same as before.
Using two loops which was suggested in the comments section.
#!/usr/bin/env bash
to_compare_input=("$#")
inarray() {
local n=$1 h
shift
for h; do
[[ $n == "$h" ]] && return
done
return 1
}
until inarray "$user_input" "${to_compare_input[#]}"; do
read -r user_input
done
As for the tr if your version of bash supports the ^^ and ,, for uppercase and lowercase parameter expansion. use ${user_input^^}
until [[ ${user_input^^} == $exglob_pattern_input ]]; do
until [[ ${user_input^^} =~ $regex_pattern_input ]]; do
until inarray "${user_input^^}" "${to_compare_input[#]}"; do
Assuming no spaces in the bar code texts. You can do something like this
while read -r INPUT
do
#Append spaces to prevent substring matching
if [[ $(echo " $# " | grep -i " ${INPUT} " | wc -l) -eq 1 ]]
then
break
fi
done

Bash: Masking user input for a password, with * support for backspace and special characters

I have this snippet below based on the idea to mask the input for a password by #SiegeX and #mklement0 from this question.
It's great and my only desired addition was to delete for length of entered chars only, so we're not wiping out the entire line.
I don't understand this very well, so have run into bugs.
With below, entering "12345" and backspacing, the numbers don't "backspace"; no error.
Entering "123FourFive" and backspacing, produces error: line 9: [[: 123FourFiv: value too great for base (error token is "123FourFiv")
Entering "OneTwo345" and backspacing, seems to work fine.
Entering symbols one might expect in a password and then backspacing produces error: line 9: [[: OneTwo./?: syntax error: invalid arithmetic operator (error token is "./?")
Also pressing arrow keys during input creates wild screen behaviour after backspacing...
How to improve this so we're masking user input, and only deleting what has been entered?
Or are we "reinventing the wheel"? Is there something else out there that will do what I'm trying to do already (which is mask user input to obtain a password in a bash script to put it in a variable)?
User environment is Linux Mint 19.3 with Cinnamon.
#!/bin/bash
printf "\n\tPlease enter password: "
# mask the input for the password by #SiegeX and #mklement0 (https://stackoverflow.com/questions/4316730)
while IFS= read -r -s -n1 char; do
[[ -z "${char}" ]] && { printf '\n'; break; } # ENTER pressed; output \n and break.
if [[ "${char}" == $'\x7f' ]]; then # backspace was pressed
# #nooblag, only delete for length of entered chars?
if [[ "${password}" -lt "${#password}" ]]; then printf '\b \b'; fi # erase one '*' to the left.
[[ -n $password ]] && password=${password%?} # remove last char from output variable
else
# add typed char to output variable
password+="${char}"
# print '*' in its stead
printf '*'
fi
done
printf "\tPassword: ${password}\n\n"
Update: askpass as suggested here nearly does what I'm after, but if the user tries to abort/kill it with Ctrl+C it messes up the terminal...
This might be the solution! Taken from here.
#!/bin/bash
#
# Read and echo a password, echoing responsive 'stars' for input characters
# Also handles: backspaces, deleted and ^U (kill-line) control-chars
#
unset PWORD
PWORD=
echo -n 'password: ' 1>&2
while true; do
IFS= read -r -N1 -s char
# Note a NULL will return a empty string
# Convert users key press to hexadecimal character code
code=$(printf '%02x' "'$char") # EOL (empty char) -> 00
case "$code" in
''|0a|0d) break ;; # Exit EOF, Linefeed or Return
08|7f) # backspace or delete
if [ -n "$PWORD" ]; then
PWORD="$( echo "$PWORD" | sed 's/.$//' )"
echo -n $'\b \b' 1>&2
fi
;;
15) # ^U or kill line
echo -n "$PWORD" | sed 's/./\cH \cH/g' >&2
PWORD=''
;;
[01]?) ;; # Ignore ALL other control characters
*) PWORD="$PWORD$char"
echo -n '*' 1>&2
;;
esac
done
echo
echo $PWORD

Bash problems with string comparison

I have a problem with writing bash script. The problem is in comparison of strings. When I launch it, there's no errors. However in result, it is always changing the variable client.
So if for an example we have two lines in file
apple A
orange D
and if I give the who=A I expect to see in result apple, or if at D - orange
But no matter of what I choose A or D it is always giving me the result - orange
No matter of the strings, it always change the variable client, like ignoring the comparison. Please help.
while read line
do
IFS=" "
set -- $line
echo $2" "$who":"$1
if [[ "$2"="$who" ]]
then
echo "change"
client=$1
fi
done < $file
echo $client
So now I changed the code as in one of the comment below, but now the caparison always false therefore the variable client is always empty
while read -r line
do
#IFS=" "
#set -- $line
#echo $2" "$who":"$1
#if [[ "$2" = "$who" ]]
a="${line% *}"
l="${line#* }"
if [[ "$l" == "$who" ]]
then
echo "hi"
client="$a"
fi
done < $file
If you have data in a file with each line like apple D and you want to read the file and separate then items, the parameter expansion/substring extraction is the correct way to process the line. For example (note $who is taken from your problem statement):
while read -r line
do
fruit="${line% *}" # remove from end to space
letter="${line#* }" # remove from start to space
if [[ "$letter" == "$who" ]]
then
echo "change"
client="$fruit"
fi
done < $file
Short Example
Here is a quick example of splitting the words with parameter expansion/substring extraction:
#!/bin/bash
while read -r line
do
fruit="${line% *}"
letter="${line#* }"
echo "fruit: $fruit letter: $letter"
done
exit 0
input
$ cat dat/apple.txt
Apple A
Orange D
output
$ bash apple.sh <dat/apple.txt
fruit: Apple letter: A
fruit: Orange letter: D
Change if [[ "$2"="$who" ]] to
if [[ "$2" = "$who" ]]
spaces around =
Example (for clarification):
who=A
while read line
do
IFS=" "
set -- $line
echo $2" "$who":"$1
if [[ "$2" = "$who" ]]
then
echo "change"
client=$1
fi
done < file #this is the file I used for testing
echo $client
Output:
A A:apple
change
D A:orange
apple
For who=D:
A D:apple
D D:orange
change
orange
You do need spaces around that = operator.
However, I think you're facing yet another issue as you're trying to change the value of the client variable from inside the while loop (which executes in a subshell). I don't think that will work; see this quesion for details.

shell script if else loop not working

Given the following programme which reads in user input twice
function search_grep
{
if [ "$2" == "" ];then
for x in "${title[#]}"
do
value=$(echo $x | grep "$1")
if [ "$value" != "" ];then
echo "$value"
fi
done
elif [ "$1" == "" ];then
hello="123"
echo "$hello"
fi
}
echo -n "Enter title : "
read book_title
echo -n "Enter author : "
read author
title=(CatchMe HappyDay)
search_grep $book_title $author
it works as expected when i enter followed by HappyDay HOWEVER
When i enter foo followed by , I would expect console output to be
123
instead I am getting
Can someone explain to me , the programme is not executing the second elif loop though second input is
In both of your cases cases, the following:
search_grep $book_title $author
expands to a call with a single argument. Hence, the "then" clause is activated. The reason is that an unquoted argument consisting of whitespace expands to nothing and disappears. That is the way of bash.
If you want to pass search_grep two arguments, then you need to quote the variables:
search_grep "$book_title" "$author"
As shown here, you might try using = instead of ==
Or for an empty string comparison try -z

Bash read backspace button behavior problem

When using read in bash, pressing backspace does not delete the last character entered, but appears to append a backspace to the input buffer. Is there any way I can change it so that delete removes the last key typed from the input? If so how?
Here's a short example prog I'm using it with if it's of any help:
#!/bin/bash
colour(){ #$1=text to colourise $2=colour id
printf "%s%s%s" $(tput setaf $2) "$1" $(tput sgr0)
}
game_over() { #$1=message $2=score
printf "\n%s\n%s\n" "$(colour "Game Over!" 1)" "$1"
printf "Your score: %s\n" "$(colour $2 3)"
exit 0
}
score=0
clear
while true; do
word=$(shuf -n1 /usr/share/dict/words) #random word from dictionary
word=${word,,} #to lower case
len=${#word}
let "timeout=(3+$len)/2"
printf "%s (time %s): " "$(colour $word 2)" "$(colour $timeout 3)"
read -t $timeout -n $len input #read input here
if [ $? -ne 0 ]; then
game_over "You did not answer in time" $score
elif [ "$input" != "$word" ]; then
game_over "You did not type the word correctly" $score;
fi
printf "\n"
let "score+=$timeout"
done
The option -n nchars turns the terminal into raw mode, so your best chance is to rely on readline (-e) [docs]:
$ read -n10 -e VAR
BTW, nice idea, although I would leave the end of the word to the user (it's a knee-jerk reaction to press return).
I know the post is old, still this can be useful for someone. If you need specific response to a single keypress on backspace, something like this can do it (without -e):
backspace=$(cat << eof
0000000 005177
0000002
eof
)
read -sn1 hit
[[ $(echo "$hit" | od) = "$backspace" ]] && echo -e "\nDo what you want\n"

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