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 have a file that contains 10 lines with this sort of content:
aaaa,bbb,132,a.g.n.
I wanna walk throw every line, char by char and put the data before the " , " is met in an output file.
if [ $# -eq 2 ] && [ -f $1 ]
then
echo "Read nr of fields to be saved or nr of commas."
read n
nrLines=$(wc -l < $1)
while $nrLines!="1" read -r line || [[ -n "$line" ]]; do
do
for (( i=1; i<=$n; ++i ))
do
while [ read -r -n1 temp ]
do
if [ temp != "," ]
then
echo $temp > $(result$i)
else
fi
done
paste -d"\n" $2 $(result$i)
done
nrLines=$($nrLines-1)
done
else
echo "File not found!"
fi
}
In parameter $2 I have an empty file in which I will store the data from file $1 after I extract it without the " , " and add a couple of comments.
Example:
My input_file contains:
a.b.c.d,aabb,comp,dddd
My output_file is empty.
I call my script: ./script.sh input_file output_file
After execution the output_file contains:
First line info: a.b.c.d
Second line info: aabb
Third line info: comp
(yes, without the 4th line info)
You can do what you want very simply with parameter-expansion and substring-removal using bash alone. For example, take an example file:
$ cat dat/10lines.txt
aaaa,bbb,132,a.g.n.
aaaa,bbb,133,a.g.n.
aaaa,bbb,134,a.g.n.
aaaa,bbb,135,a.g.n.
aaaa,bbb,136,a.g.n.
aaaa,bbb,137,a.g.n.
aaaa,bbb,138,a.g.n.
aaaa,bbb,139,a.g.n.
aaaa,bbb,140,a.g.n.
aaaa,bbb,141,a.g.n.
A simple one-liner using native bash string handling could simply be the following and give the following results:
$ while read -r line; do echo ${line%,*}; done <dat/10lines.txt
aaaa,bbb,132
aaaa,bbb,133
aaaa,bbb,134
aaaa,bbb,135
aaaa,bbb,136
aaaa,bbb,137
aaaa,bbb,138
aaaa,bbb,139
aaaa,bbb,140
aaaa,bbb,141
Paremeter expansion w/substring removal works as follows:
var=aaaa,bbb,132,a.g.n.
Beginning at the left and removing up to, and including, the first ',' is:
${var#*,} # bbb,132,a.g.n.
Beginning at the left and removing up to, and including, the last ',' is:
${var##*,} # a.g.n.
Beginning at the right and removing up to, and including, the first ',' is:
${var%,*} # aaaa,bbb,132
Beginning at the left and removing up to, and including, the last ',' is:
${var%%,*} # aaaa
Note: the text to remove above is represented with a wildcard '*', but wildcard use is not required. It can be any allowable text. For example, to only remove ,a.g.n where the preceding number is 136, you can do the following:
${var%,136*},136 # aaaa,bbb,136 (all others unchanged)
To print 2016 th line from a file named file.txt u have to run a command like this-
sed -n '2016p' < file.txt
More-
sed -n '2p' < file.txt
will print 2nd line
sed -n '2011p' < file.txt
2011th line
sed -n '10,33p' < file.txt
line 10 up to line 33
sed -n '1p;3p' < file.txt
1st and 3th line
and so on...
For more detail, please have a look in this tutorial and this answer.
In native bash the following should do what you want, assuming you replace the contents of your script.sh with the below:
#!/bin/bash
IN_FILE=${1}
OUT_FILE=${2}
IFS=\,
while read line; do
set -- ${line}
for ((i=1; i<=${#}; i++)); do
((${i}==4)) && continue
((n+=1))
printf '%s\n' "Line ${n} info: ${!i}"
done
done < ${IN_FILE} > ${OUT_FILE}
This will not print the 4th field of each line within the input file, on a new line in the output file (I assume this is your requirement as per your comment?).
[wspace#wspace sandbox]$ awk -F"," 'BEGIN{OFS="\n"}{for(i=1; i<=NF-1; i++){print "line Info: "$i}}' data.txt
line Info: a.b.c.d
line Info: aabb
line Info: comp
This little snippet can ignore the last field.
updated:
#!/usr/bin/env bash
if [ ! -f "$1" -o $# -ne 2 ];then
echo "Usage: $(basename $0) input_file out_file"
exit 127
fi
input_file=$1
output_file=$2
: > $output_file
if [ "$(wc -l < $1)" -ne 0 ];then
while true
do
read -r -n1 char
if [ "$char" == "" ];then
break
elif [ $char != "," ];then
temp=$temp$char
else
echo "line info: $temp" >> $output_file
temp=""
fi
done < $input_file
else
echo "file $1 is empty"
fi
Maybe this is what you want
Did you try
sed "s|,|\n|g" $1 | head -n -1 > $2
I assume that only the last word would not have a comma on its right.
Try this (tested with you sample line) :
#!/bin/bash
# script.sh
echo "Number of fields to save ?"
read nf
while IFS=$',' read -r -a arr; do
newarr=${arr[#]:0:${nf}}
done < "$1"
for i in ${newarr[#]};do
printf "%s\n" $i
done > "$2"
Execute script with :
$ ./script.sh inputfile outputfile
Number of fields ?
3
$ cat outputfile
a.b.c.d
aabb
comp
All words separated with commas are stored into an array $arr
A tmp array $newarr removes last $n element ($n get the read command).
It loops over new array and prints result in $2, the outputfile.
I'm learning Bash and am looking at the read command. I thought the difference between the -N and the -n option was that -N would overwrite the IFS while -n wouldn't. In the following example I expected var6 to take the value of "ijfz", because I thought the space would act as field separator.
But it seems to have value "ijfz e". The space wasn't used as field separator
printf "%s\n" "ijfz eszev enacht" | {
read -n 6 var6
printf "%s\n" "$var6"
}
I wanted to see what was $IFS, but the following printf command doesn't learn me too much:
printf ":%s:\n" "$IFS"
gives the following output
:
:
What am I not understanding...?
By delimiter it means using option -d. See this difference:
printf "%s\n" "ijfz eszev enacht" | { read -d ' ' -N 6 var6; printf "[%s]\n" "$var6"; }
[ijfz e]
printf "%s\n" "ijfz eszev enacht" | { read -d ' ' -n 6 var6; printf "[%s]\n" "$var6"; }
[ijfz]
In first case when using -N it ignores -d ' ' and reads exactly 6 characters.
In second case when using -n it respects -d ' ' and reads until a space is read.
I wish to take names of two files as command line arguments in bash shell script and then for each word (words are comma separated and the file has more than one line) in the first file I need to count its occurrence in the second file.
I wrote a shell script like this
if [ $# -ne 2 ]
then
echo "invalid number of arguments"
else
i=1
a=$1
b=$2
fp=*$b
while[ fgetc ( fp ) -ne EOF ]
do
d=$( cut -d',' -f$i $a )
echo "$d"
grep -c -o $d $b
i=$(( $i + 1 ))
done
fi
for example file1 has words abc,def,ghi,jkl (in first line )
mno,pqr (in second line)
and file2 has words abc,abc,def
Now the output should be like abc 2
def 1
ghi 0
To read a file word by word separated by comma use this snippet:
while read -r p; do
IFS=, && for w in $p; do
printf "%s: " "$w"
tr , '\n' < file2 | grep -Fc "$w"
done
done < file1
Another approach:
words=( `tr ',' ' ' < file1`) #split the file1 into words...
for word in "${words[#]}"; do #iterate in the words
printf "%s : " "$word"
awk 'END{print FNR-1}' RS="$word" file2
# split file2 with 'word' as record separator.
# print number of lines == number of occurrences of the word..
done
This is probably pretty basic, I want to read in a occurrence file.
Then the program should find all occurrences of "CallTilEdb" in the file Hendelse.logg:
CallTilEdb 8
CallCustomer 9
CallTilEdb 4
CustomerChk 10
CustomerChk 15
CallTilEdb 16
and sum up then right column. For this case it would be 8 + 4 + 16, so the output I would want would be 28.
I'm not sure how to do this, and this is as far as I have gotten with vistid.sh:
#!/bin/bash
declare -t filename=hendelse.logg
declare -t occurance="$1"
declare -i sumTime=0
while read -r line
do
if [ "$occurance" = $(cut -f1 line) ] #line 10
then
sumTime+=$(cut -f2 line)
fi
done < "$filename"
so the execution in terminal would be
vistid.sh CallTilEdb
but the error I get now is:
/home/user/bin/vistid.sh: line 10: [: unary operator expected
You have a nice approach, but maybe you could use awk to do the same thing... quite faster!
$ awk -v par="CallTilEdb" '$1==par {sum+=$2} END {print sum+0}' hendelse.logg
28
It may look a bit weird if you haven't used awk so far, but here is what it does:
-v par="CallTilEdb" provide an argument to awk, so that we can use par as a variable in the script. You could also do -v par="$1" if you want to use a variable provided to the script as parameter.
$1==par {sum+=$2} this means: if the first field is the same as the content of the variable par, then add the second column's value into the counter sum.
END {print sum+0} this means: once you are done from processing the file, print the content of sum. The +0 makes awk print 0 in case sum was not set... that is, if nothing was found.
In case you really want to make it with bash, you can use read with two parameters, so that you don't have to make use of cut to handle the values, together with some arithmetic operations to sum the values:
#!/bin/bash
declare -t filename=hendelse.logg
declare -t occurance="$1"
declare -i sumTime=0
while read -r name value # read both values with -r for safety
do
if [ "$occurance" == "$name" ]; then # string comparison
((sumTime+=$value)) # sum
fi
done < "$filename"
echo "sum: $sumTime"
So that it works like this:
$ ./vistid.sh CallTilEdb
sum: 28
$ ./vistid.sh CustomerChk
sum: 25
first of all you need to change the way you call cut:
$( echo $line | cut -f1 )
in line 10 you miss the evaluation:
if [ "$occurance" = $( echo $line | cut -f1 ) ]
you can then sum by doing:
sumTime=$[ $sumTime + $( echo $line | cut -f2 ) ]
But you can also use a different approach and put the line values in an array, the final script will look like:
#!/bin/bash
declare -t filename=prova
declare -t occurance="$1"
declare -i sumTime=0
while read -a line
do
if [ "$occurance" = ${line[0]} ]
then
sumTime=$[ $sumtime + ${line[1]} ]
fi
done < "$filename"
echo $sumTime
For the reference,
id="CallTilEdb"
file="Hendelse.logg"
sum=$(echo "0 $(sed -n "s/^$id[^0-9]*\([0-9]*\)/\1 +/p" < "$file") p" | dc)
echo SUM: $sum
prints
SUM: 28
the sed extract numbers from a lines containing the given id, such CallTilEdb
and prints them in the format number +
the echo prepares a string such 0 8 + 16 + 4 + p what is calculation in RPN format
the dc do the calculation
another variant:
sum=$(sed -n "s/^$id[^0-9]*\([0-9]*\)/\1/p" < "$file" | paste -sd+ - | bc)
#or
sum=$(grep -oP "^$id\D*\K\d+" < "$file" | paste -sd+ - | bc)
the sed (or the grep) extracts and prints only the numbers
the paste make a string like number + number + number (-d+ is a delimiter)
the bc do the calculation
or perl
sum=$(perl -slanE '$s+=$F[1] if /^$id/}{say $s' -- -id="$id" "$file")
sum=$(ID="CallTilEdb" perl -lanE '$s+=$F[1] if /^$ENV{ID}/}{say $s' "$file")
Awk translation to script:
#!/bin/bash
declare -t filename=hendelse.logg
declare -t occurance="$1"
declare -i sumTime=0
sumtime=$(awk -v entry=$occurance '
$1==entry{time+=$NF+0}
END{print time+0}' $filename)