I have one file Length.txt having multiples names (40) line by line.
I want to write a small shell script where it will count the character count of each line of the file and if the count is less than 9 replace those lines with adding extra 8 spaces and 1 at the end of each line.
For example, if the name is
XXXXXX
replace as
XXXXXX 1
I tried with the below coding. It is working for me, however whenever it's replacing the line it is displaying all the lines at a time.
So suppose I have 40 lines in Length.txt and out of that 4 lines having less than 9 character count then my output has 160 lines.
Can anyone help me to display only 40 line output with the 4 changed lines?
#!/usr/bin/sh
#set -x
while read line;
do
count=`echo $line|wc -m`
if [ $count -lt 9 ]
then
Number=`sed -n "/$line/=" Length.txt`;
sed -e ""$Number"s/$line/$line 1/" Length4.txt
fi
done < Length.txt
A single sed command can do that:
sed -E 's/^.{,8}$/& 1/' file
To modify the contents of the file add -i:
sed -iE 's/^.{,8}$/& 1/' file
Partial output:
94605320 1
105018263
2475218231
7728563 1
1
* Fixed to include only add 8 spaces not 9, and include empty lines. If you don't want to process empty lines, use {1,8}.
$ cat foo.input
I am longer than 9 characters
I am also longer than 9 characters
I am not
Another long line
short
$ while read line; do printf "$line"; (( ${#line} < 9 )) && printf " 1"; echo; done < foo.input
I am longer than 9 characters
I am also longer than 9 characters
I am not 1
Another long line
short 1
Let me show you what is wrong with your script. The only thing missing from your script is that you need to use sed -i to edit file and re-save it after making the replacement.
I'm assuming Length4.txt is just a copy of Length.txt?
I added sed -i to your script and it should work now:
cp Length.txt Length4.txt
while read line;
do
count=`echo $line|wc -m`
if [ $count -lt 9 ]
then
Number=`sed -n "/$line/=" Length.txt`
sed -ie ""$Number"s/$line/$line 1/" Length4.txt
fi
done < Length.txt
However, you don't need sed or wc. You can simplify your script as follows:
while IFS= read -r line
do
count=${#line}
if (( count < 9 ))
then
echo "$line 1"
else
echo "$line"
fi
done < Length.txt > Length4.txt
$ awk -v FS= 'NF<9{$0=sprintf("%s%*s1",$0,8,"")} 1' file
XXXXXX 1
Note how simple it would be to check for a number other than 9 characters and to print some sequence of blanks other than 8.
Related
I am attempting to return the line number of lines that have a break. An input example:
2938
383
3938
3
383
33333
But my script is not working and I can't see why. My script:
input="./input.txt"
declare -i count=0
while IFS= read -r line;
do
((count++))
if [ "$line" == $'\n\n' ]; then
echo "$count"
fi
done < "$input"
So I would expect, 3, 6 as output.
I just receive a blank response in the terminal when I execute. So there isn't a syntax error, something else is wrong with the approach I am taking. Bit stumped and grateful for any pointers..
Also "just use awk" doesn't help me. I need this structure for additional conditions (this is just a preliminary test) and I don't know awk syntax.
The issue is that "$line" == $'\n\n' won't match a newline as it won't be there after consuming an empty line from the input, instead you can match an empty line with regex pattern ^$:
if [[ "$line" =~ ^$ ]]; then
Now it should work.
It's also match easier with awk command:
$ awk '$0 == ""{ print NR }' test.txt
3
6
As Roman suggested, line read by read terminates with a delimiter, and that delimiter would not show up in the line the way you're testing for.
If the pattern you are searching for looks like an empty line (which I infer is how a "double newline" always manifests), then you can just test for that:
while read -r; do
((count++))
if [[ -z "$REPLY" ]]; then
echo "$count"
fi
done < "$input"
Note that IFS is for field-splitting data on lines, and since we're only interested in empty lines, IFS is moot.
Or if the file is small enough to fit in memory and you want something faster:
mapfile -t -O1 foo < i
declare -p foo
for n in "${!foo[#]}"; do
if [[ -z "${foo[$n]}" ]]; then
echo "$n"
fi
done
Reading the file all at once (mapfile) then stepping through an array may be easier on resources than stepping through a file line by line.
You can also just use GNU awk:
gawk -v RS= -F '\n' '{ print (i += NF); i += length(RT) - 1 }' input.txt
By using FS = ".+", it ensures only truly zero-length (i.e. $0 == "") line numbers get printed, while skipping rows consisting entirely of [[:space:]]'s
echo '2938
383
3938
3
383
33333' |
{m,g,n}awk -F'.+' '!NF && $!NF = NR'
3
6
This sed one-liner should do the job at once:
sed -n '/^$/=' input.txt
Simply writes the current line number (the = command) if the line read is empty (the /^$/ matches the empty line).
This question already has answers here:
Count occurrences of a char in a string using Bash
(9 answers)
Closed 3 years ago.
I have a data in file.txt like below
N4*1
NM1*IL*2
PER*IC*XM*
how can i read line by line and get the character * count ?
Pseudocode
#!bash/sh
cd `dirname $0`
filelinecount=echo '$(wc -l file.txt)'
if [ $filelinecount -gt 0 ] ; then
for (int i=0, i++); do
fileline=$i (STORES LINE IN VARIABLE)
charactercount= cat '$fileline | wc [*] $fileline'
(GET CHARACTER [*] COUNT AND STORED IN VARIABLE)
echo $charactercount
done
else
echo "file.txt don't contain any lines"
fi
Expected output:
'For' Loop should read line by line from file and store each line in variable "fileline" then count the characters [*] and store in variable "charactercount" then print the variable $charactercount. This loop has to repeat for for all the files in the file. How can i achieve this in 'for' loop ?
1
2
3
This is not a duplicate question as this question clearly asked count of characters using "for" loop.
"Count occurrences of a char in a string using Bash" post don't have answer to this post
awk '{print gsub(/\*/,$0)}' file
To achieve the same in a loop:
#! /bin/bash
while read line
do
grep -o '*' <<<"$line" | grep -c .
done < file
This should print * count per line.
Update as per the comment:
#! /bin/bash
while read line
do
echo "$line" | awk -F"[*]" '{print NF-1}'
done < file
[ ! -s file ] && echo "no lines in file"
[ ! -s file ] has nothing to do with loop. -s flag checks if file has contents. If it has t returns true. But in your case you want opposite behaviour so we used !. So when file is empty, it returns true and && causes the next command to execute i.e. echo "no lines in file”.
tr -d -c '*\n' file.txt | awk '{print length}'
Remove everything except stars and newlines from the file. Then print the line lengths.
In A.csv, there are
1
2
3
4
How should I read this file and create variables $B and $C so that:
echo $B
echo $C
returns:
1 2 3 4
1,2,3,4
So far I am trying:
cat A.csv | while read A;
do
echo $A
done
It only returns
1
2
3
4
Assuming bash 4.x, the following is efficient, robust, and native:
# Read each line of A.csv into a separate element of the array lines
readarray -t lines <A.csv
# Generate a string B with a comma after each item in the array
printf -v B '%s,' "${lines[#]}"
# Prune the last comma from that string
B=${B%,}
# Generate a string C with a space after each item in the array
printf -v B '%s ' "${lines[#]}"
As #Cyrus said
B=$(cat A.csv)
echo $B
Will output:
1 2 3 4
Because bash will not carry the newlines if the variable is not wrapped in quotes. This is dangerous if A.csv contains any characters which might be affected by bash glob expansion, but should be fine if you are just reading simple strings.
If you are reading simple strings with no spaces in any of the elements, you can also get your desired result for $C by using:
echo $B | tr ' ' ','
This will output:
1,2,3,4
If lines in A.csv may contain bash special characters or spaces then we return to the loop.
For why I've formatted the file reading loop as I have, refer to: Looping through the content of a file in Bash?
B=''
C=''
while read -u 7 curr_line; do
if [ "$B$C" == "" ]; then
B="$curr_line"
C="$curr_line"
else
B="$B $curr_line"
C="$C,$curr_line"
fi
done 7<A.csv
echo "$B"
echo "$C"
Will construct the two variables as you desire using a loop through the file contents and should prevent against unwanted globbing and splitting.
B=$(cat A.csv)
echo $B
Output:
1 2 3 4
With quotes:
echo "$B"
Output:
1
2
3
4
I would read the file into a bash array:
mapfile -t array < A.csv
Then, with various join characters
b="${array[*]}" # space is the default
echo "$b"
c=$( IFS=","; echo "${array[*]}" )
echo "$c"
Or, you can use paste to join all the lines with a specified separator:
b=$( paste -d" " -s < A.csv )
c=$( paste -d"," -s < A.csv )
Try this :
cat A.csv | while read A;
do
printf "$A"
done
Regards!
Try This(Simpler One):
b=$(tr '\n' ' ' < file)
c=$(tr '\n' ',' < file)
You don't have to read File for that. Make sure you ran dos2unix file command. If you are running in windows(to remove \r).
Note: It will modify the file. So, make sure you copied from original file.
I'd like to rearrange a file in two columns after the nth line.
For example, say I have a file like this here:
This is a bunch
of text
that I'd like to print
as two
columns starting
at line number 7
and separated by four spaces.
Here are some
more lines so I can
demonstrate
what I'm talking about.
And I'd like to print it out like this:
This is a bunch and separated by four spaces.
of text Here are some
that I'd like to print more lines so I can
as two demonstrate
columns starting what I'm talking about.
at line number 7
How could I do that with a bash command or function?
Actually, pr can do almost exactly this:
pr --output-tabs=' 1' -2 -t tmp1
↓
This is a bunch and separated by four spaces.
of text Here are some
that I'd like to print more lines so I can
as two demonstrate
columns starting what I'm talking about.
at line number 7
-2 for two columns; -t to omit page headers; and without the --output-tabs=' 1', it'll insert a tab for every 8 spaces it added. You can also set the page width and length (if your actual files are much longer than 100 lines); check out man pr for some options.
If you're fixed upon “four spaces more than the longest line on the left,” then perhaps you might have to use something a bit more complex;
The following works with your test input, but is getting to the point where the correct answer would be, “just use Perl, already;”
#!/bin/sh
infile=${1:-tmp1}
longest=$(longest=0;
head -n $(( $( wc -l $infile | cut -d ' ' -f 1 ) / 2 )) $infile | \
while read line
do
current="$( echo $line | wc -c | cut -d ' ' -f 1 )"
if [ $current -gt $longest ]
then
echo $current
longest=$current
fi
done | tail -n 1 )
pr -t -2 -w$(( $longest * 2 + 6 )) --output-tabs=' 1' $infile
↓
This is a bunch and separated by four spa
of text Here are some
that I'd like to print more lines so I can
as two demonstrate
columns starting what I'm talking about.
at line number 7
… re-reading your question, I wonder if you meant that you were going to literally specify the nth line to the program, in which case, neither of the above will work unless that line happens to be halfway down.
Thank you chatraed and BRPocock (and your colleague). Your answers helped me think up this solution, which answers my need.
function make_cols
{
file=$1 # input file
line=$2 # line to break at
pad=$(($3-1)) # spaces between cols - 1
len=$( wc -l < $file )
max=$(( $( wc -L < <(head -$(( line - 1 )) $file ) ) + $pad ))
SAVEIFS=$IFS;IFS=$(echo -en "\n\b")
paste -d" " <( for l in $( cat <(head -$(( line - 1 )) $file ) )
do
printf "%-""$max""s\n" $l
done ) \
<(tail -$(( len - line + 1 )) $file )
IFS=$SAVEIFS
}
make_cols tmp1 7 4
Could be optimized in many ways, but does its job as requested.
Input data (configurable):
file
num of rows borrowed from file for the first column
num of spaces between columns
format.sh:
#!/bin/bash
file=$1
if [[ ! -f $file ]]; then
echo "File not found!"
exit 1
fi
spaces_col1_col2=4
rows_col1=6
rows_col2=$(($(cat $file | wc -l) - $rows_col1))
IFS=$'\n'
ar1=($(head -$rows_col1 $file))
ar2=($(tail -$rows_col2 $file))
maxlen_col1=0
for i in "${ar1[#]}"; do
if [[ $maxlen_col1 -lt ${#i} ]]; then
maxlen_col1=${#i}
fi
done
maxlen_col1=$(($maxlen_col1+$spaces_col1_col2))
if [[ $rows_col1 -lt $rows_col2 ]]; then
rows=$rows_col2
else
rows=$rows_col1
fi
ar=()
for i in $(seq 0 $(($rows-1))); do
line=$(printf "%-${maxlen_col1}s\n" ${ar1[$i]})
line="$line${ar2[$i]}"
ar+=("$line")
done
printf '%s\n' "${ar[#]}"
Output:
$ > bash format.sh myfile
This is a bunch and separated by four spaces.
of text Here are some
that I'd like to print more lines so I can
as two demonstrate
columns starting what I'm talking about.
at line number 7
$ >
i am trying to read a text file, say file.txt and it contains multiple lines.
say the output of file.txt is
$ cat file.txt
this is line 1
this is line 2
this is line 3
I want to store the entire output as a variable say, $text.
When the variable $text is echoed, the expected output is:
this is line 1 this is line 2 this is line 3
my code is as follows
while read line
do
test="${LINE}"
done < file.txt
echo $test
the output i get is always only the last line. Is there a way to concatenate the multiple lines in file.txt as one long string?
You can translate the \n(newline) to (space):
$ text=$(tr '\n' ' ' <file.txt)
$ echo $text
this is line 1 this is line 2 this is line 3
If lines ends with \r\n, you can do this:
$ text=$(tr -d '\r' <file.txt | tr '\n' ' ')
Another one:
line=$(< file.txt)
line=${line//$'\n'/ }
test=$(cat file.txt | xargs)
echo $test
You have to append the content of the next line to your variable:
while read line
do
test="${test} ${LINE}"
done < file.txt
echo $test
Resp. even simpler you could simply read the full file at once into the variable:
test=$(cat file.txt)
resp.
test=$(tr "\n" " " < file.txt)
If you would want to keep the newlines it would be as simple as:
test=<file.txt
I believe it's the simplest method:
text=$(echo $(cat FILE))
But it doesn't preserve multiple spaces/tabs between words.
Use arrays
#!/bin/bash
while read line
do
a=( "${a[#]}" "$line" )
done < file.txt
echo -n "${a[#]}"
output:
this is line 1 this is line 2 this is line 3
See e.g. tldp section on arrays