I have this
while read -r line
do
echo -e "$line\r"
done <<< $keys
Where $keys, if printed as it is displays 30k rows.
When doing echo as above I get only one line as output.
I need to filter every line and then output it.
Why is that happening (overwriting)
How can I prevent it to happen?
Use More Quotes
while read -r line
do
echo -e "$line\r"
done <<< "$keys"
observe the quotes in <<< "$keys"
Look:
$ printf -v keys '%s\n' "key one" "key two"
$ echo $keys
key one key two
$ # Oh dear
$ # Now with quotes:
$ echo "$keys"
key one
key two
$ # Yeah :)
$ # Same with a here string:
$ while read line; do echo "$line"; done <<< $keys
key one key two
$ # Oh dear :(
$ # Now with quotes:
$ while read line; do echo "Read: $line"; done <<< "$keys"
Read: line one
Read: line two
$ # Done \o/
In this post, you have learned:
to use more quotes,
to use more quotes,
to use more quotes,
to use more quotes,
to use more quotes,
and to use more quotes.
Now remember,
Each time you forget quotes, God kills a kitten
This is happening because of the \r in your echo command.
It will be solved replacing it to:
echo "$line"
It is also important to note that to call the variable you'd better use "$keys" to keep its format.
All together:
while read -r line
do
echo -e "$line"
done <<< "$keys"
Related
I am new to bash and I struggling with a program. I want to write a program that first asks for user input and afterwards prints the words with an \n(blank line) between them. The last echo contains the amount of characters that is written. Also the output can only contain the words and no digits. E.g:
Input: hallo1 user2 Pete4
Ouput: hallo
user
Pete
13 Characters
This is my code for the time beeing.
echo Typ one or multiple words:
read varname
arr=( "${arr[#]}" "$varname" )
for i in "${arr[#]}"; do
echo "$i"
done
echo ${arr[#]}
# printf '%s\n' "${arr[#]}"
Try this. Works for me. I added in the for the sentence to remove the digits.
And after the for, I first remove the spaces between the names and then I count the total of characters using the # in ${#aux}. I added the parameter -n in the first echo too, just to break the line with the second one.
echo Type one or multiple words:
read varname
arr=( "${arr[#]}" "$varname" )
for i in "${arr[#]//[[:digit:]]/}"; do
echo -n "$i"
done
aux=$(echo "${i}" | sed "s/ //g")
echo " " ${#aux} " Characters"
An approach in plain bash without using an array:
#!/bin/bash
echo 'Type one or multiple words on a line:'
read -r
words_without_digits=${REPLY//[0-9]}
line_without_blanks=${words_without_digits//[[:blank:]]}
printf '%s\n' $words_without_digits
echo "${#line_without_blanks} Characters"
echo Type one or multiple words:
read varname
arr=( "${arr[#]}" "$varname" )
for i in "${arr[#]//[[:digit:]]/}"; do
printf '%s\n' $i
done
aux=$(echo "${i}" | sed "s/ //g")
echo ${#aux} " Characters"
I have read the config file which has the below variable:
export BASE_DIR="\usr\usr1"
In the same script I read a file line by line and I wanted to substitute the ${BASE_DIR} with \usr\usr1.
In the script:
while read line; do
echo $line
done <file.txt
${BASE_DIR}\path1 should be printed as \usr\usr1\path1
Tried eval echo and $(( )).
Can use sed, This command will search and replace a value. The dollar sign is the separator.
sed -ie 's$\${BASE_DIR}$\\usr\\usr1$1' hello.txt
You need to set the variable when you read the line that contains the assignment. Then you can replace it later.
#!/bin/bash
while read line; do
if [[ $line =~ ^BASE_DIR= ]]
then basedir=${line#BASE_DIR=}
fi
line=${line/'${BASE_DIR}'/$basedir}
printf "%s\n" "$line"
done < file.txt > newfile.txt
I want to input multiple strings.
For example:
abc
xyz
pqr
and I want output like this (including quotes) in a file:
"abc","xyz","pqr"
I tried the following code, but it doesn't give the expected output.
NextEmail=","
until [ "a$NextEmail" = "a" ];do
echo "Enter next E-mail: "
read NextEmail
Emails="\"$Emails\",\"$NextEmail\""
done
echo -e $Emails
This seems to work:
#!/bin/bash
# via https://stackoverflow.com/questions/1527049/join-elements-of-an-array
function join_by { local IFS="$1"; shift; echo "$*"; }
emails=()
while read line
do
if [[ -z $line ]]; then break; fi
emails+=("$line")
done
join_by ',' "${emails[#]}"
$ bash vvuv.sh
my-email
another-email
third-email
my-email,another-email,third-email
$
With sed and paste:
sed 's/.*/"&"/' infile | paste -sd,
The sed command puts "" around each line; paste does serial pasting (-s) and uses , as the delimiter (-d,).
If input is from standard input (and not a file), you can just remove the input filename (infile) from the command; to store in a file, add a redirection at the end (> outfile).
If you can withstand a trailing comma, then printf can convert an array, with no loop required...
$ readarray -t a < <(printf 'abc\nxyx\npqr\n' )
$ declare -p a
declare -a a=([0]="abc" [1]="xyx" [2]="pqr")
$ printf '"%s",' "${a[#]}"; echo
"abc","xyx","pqr",
(To be fair, there's a loop running inside bash, to step through the array, but it's written in C, not bash. :) )
If you wanted, you could replace the final line with:
$ printf -v s '"%s",' "${a[#]}"
$ s="${s%,}"
$ echo "$s"
"abc","xyx","pqr"
This uses printf -v to store the imploded text into a variable, $s, which you can then strip the trailing comma off using Parameter Expansion.
It seems everybody else wants to remove any additional whitespace, however I have the opposite problem.
I have a file, call it some_file.txt that looks like
a b c d
and some more
and I'm reading it line-by-line with sed,
num_lines=$(cat some_file.txt | wc -l)
for i in $(seq 1 $num_lines); do
echo $(sed "${i}q;d" $file)
string=$(sed "${i}q;d" $file)
echo $string
done
I would expect the number of whitespace characters to stay the same, however the output I get is
a b c d
a b c d
and some more
and some more
So it seems that the problem is with sed removing the extra whitespace between chars, anyway to fix this?
Have a look at this example:
$ echo Hello World
Hello World
$ echo "Hello World"
Hello World
sed is not your problem, your problem is that bash removes the whitespaces when passing the output of sed into echo.
You just need to surround whatever echo is supposed to print with double quotation marks. So instead of
echo $(sed "${i}q;d" $file)
echo $string
You write
echo "$(sed "${i}q;d" $file)"
echo "$string"
The new script should look like this:
#!/usr/bin/env bash
file=some_file.txt
num_lines=$(cat some_file.txt | wc -l)
for i in $(seq 1 $num_lines); do
echo "$(sed "${i}q;d" $file)"
string=$(sed "${i}q;d" $file)
echo "$string"
done
prints the correct output:
a b c d
a b c d
and some more
and some more
However, if you just want to go through your file line by line, I strongly recommend something like this:
while IFS= read -r line; do
echo "$line"
done < some_file.txt
Question from the comments: What to do if you only want 33 lines starting from line x. One possible solution is this:
#!/usr/bin/env bash
declare -i s=$1
declare -i e=${s}+32
sed -n "${s},${e}p" $file | while IFS= read -r line; do
echo "$line"
done
(Note that I would probably include some validation of $1 in there as well.)
I declare s and e as integer variables, then even bash can do some simple arithmetic on them and calculate the actual last line to print.
I have a script variable which is multi-line.
How do i traverse this variable to read it line by line and process each line the way I want?
Consider the following multi-line variable
x=$(echo -e "a\nb\nc d e")
and a simple process for each line: just echo it with a prefix=LINE: and with single quotes around the line. Either of the following codes will satisfy that requirement:
while read line; do echo "LINE: '${line}'"; done <<< "$x"
or
while read line; do echo "LINE: '${line}'"; done < <(echo "$x")
Neither creates a subshell (so you can, e.g., set variables in the loop and access them outside of it), and both output
LINE: 'a'
LINE: 'b'
LINE: 'c d e'
But suppose instead you have
x=$(echo -e "a \n b\nc d e")
# note--------^--^
and that leading and trailing whitespace matter for your application (e.g., parsing Git porcelain). Both the above codes will give exactly the same output for the latter variable/data as for the former, which is not what you want. To preserve leading and trailing whitespace, replace while read line with while IFS= read -r line . I.e., either of the following codes
while IFS= read -r line; do echo "LINE: '${line}'"; done <<< "$x"
or
while IFS= read -r line; do echo "LINE: '${line}'"; done < <(echo "$x")
will produce
LINE: 'a '
LINE: ' b'
LINE: 'c d e'
See Greg Wooledge's excellent Bash FAQ for details.
Although I typically use "while read" for processing multi-line variables, I recently had an instance where it removed the leading space from each line in a file. Using this instead fixed my issue:
printf '%s\n' "$var" | while IFS= read -r line
do
echo "$line"
done
Code taken from this Unix Stack Exchange answer.
Edit: updated to fix last line issue as suggested by Nicolae Iotu