CODEWITHSUNDEEP
regexbashsed
aloha
aloha

Reputation: 4774

Replace hyphen with space using bash

I'm reading a filename that has this data:

2017-03-23
2018-01-23
2017-07-31

I want to read each line and replace the hyphens with spaces. Given the above example, I should get the following:

2017 03 23
2018 01 23
2017 07 31

Here's my code:

#!/bin/sh

filename=extract_dates.dat
line=1
totline=`wc -l < $filename`

while [ $line -le $totline ]
do
    date=`sed -n -e ''"$line"'p' $filename | awk '{print $line}'`
    echo $date
    test=`sed 's/([0-9])-([0-9])/\1\2/g' $date`
    echo $test
    line=`expr $line + 1`
done

I get the following error: 

sed: 1: "s/([0-9])-([0-9])/\1\2/g": \1 not defined in the RE

Upvotes: 6

Views: 8211

Answers (4)

HemanthJabalpuri
HemanthJabalpuri

Reputation: 384

We can use string replacement of bash instead

#!/bin/bash

input="2017-03-23
2018-01-23
2017-07-31"

output="${input//-/ }"
echo "$output"

Outputs

2017 03 23
2018 01 23
2017 07 31

Upvotes: 2

aloha
aloha

Reputation: 4774

tr will replace hyphens with spaces, for example:

test=$(echo $date | tr "-" " ")

Upvotes: 12

Hari Menon
Hari Menon

Reputation: 35405

How about this:

tr '-' ' ' < filename > output

If you want to use sed, you could use that as well. The reason your sed command doesn't work is that you need to escape your paranthesis. Try this:

sed 's/\([0-9]\)-\([0-9]\)/\1\2/g' $date

Upvotes: 0

chepner
chepner

Reputation: 531075

You don't need explicit counters or sed or awk; bash itself can handle this. (Your use of awk, especially, doesn't really do anything.)

filename=extract_dates.dat

while IFS=- read -r year month day; do
    echo "Year: $year"
    echo "Month: $month"
    echo "Day: $day"
done < "$filename"

Upvotes: 2

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