Change ouput for loop in bash

Question

I've these data :

2020-01-01-00-00
2020-01-01-06-00
2020-01-01-12-00
2020-01-01-18-00

I would like to display these data like this :

[ 2020-01-01-00-00, 2020-01-01-06-00, 2020-01-01-12-00, 2020-01-01-18-00 ]

I try this :

for i in $(cat Test2.txt)
do
tr -d "\n" <<< $i," "
done

The output is :

2020-01-01-00-00, 2020-01-01-06-00, 2020-01-01-12-00, 2020-01-01-18-00,

Then I try :

for i in $(cat Test2.txt)
do
echo " [ `tr -d "\n" <<< "'$i'"," "` ]"
done

But the output is :

[ '2020-01-01-00-00',  ]
[ '2020-01-01-06-00',  ]
[ '2020-01-01-12-00',  ]
[ '2020-01-01-18-00',  ]

Could you show me how to do that ?

What is this data? How do you get this? and who is the receiver for this formatted data? knowing these information would help us provide better and more meaningful solutions — Inian
– Inian, Commented Jan 27, 2020 at 13:21

tripleee · Accepted Answer · 2020-01-27 12:39:49Z

2

Don't read lines with for.

A common arrangement is to use a separator prefix which changes after the first iteration.

prefix='['
while read -r line; do
    printf '%s %s' "$prefix" "$line"
    prefix=','
done <Test2.txt
printf ' ]\n'

I'll second the suggestion to use a JSON-specific tool if your task is to generate valid JSON, though. This is pesky and somewhat brittle.

answered Jan 27, 2020 at 12:39

tripleee

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Comments

oguz ismail · Accepted Answer · 2020-01-27 12:34:06Z

1

Your desired output looks like JSON, if so you can use jq for this. E.g:

jq -Rn '[inputs]' Test2.txt

answered Jan 27, 2020 at 12:34

oguz ismail

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Ivan · Accepted Answer · 2020-01-27 12:55:16Z

0

Using printf

data="
    2020-01-01-00-00
    2020-01-01-06-00
    2020-01-01-12-00
    2020-01-01-18-00
"

printf -v data %s,\  $data
printf "[ ${data%, } ]"

answered Jan 27, 2020 at 12:55

Ivan

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Comments

glenn jackman · Accepted Answer · 2020-01-27 13:41:41Z

0

a tricky bit of perl:

perl -00 -pe 'chomp; s/\n/, /g; BEGIN {print "[ "} END {print " ]\n"}' Test2.txt

answered Jan 27, 2020 at 13:41

glenn jackman

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Comments

Paul Hodges · Accepted Answer · 2020-01-27 14:14:05Z

0

In sed,

$: sed -n '$!H; ${H; x; s/\n/, /g; s/$/ ]\n/; s/^,/[/; p;}' infile

In bash,

$: dat="$(printf "%s, " $(<infile))";  printf "[ ${dat%, } ]\n";

in 'awk',

$: awk 'BEGIN{ printf "[ "; sep=""; } { printf "%s%s", sep, $0; sep=", "; } END{ print " ]"; }' infile

answered Jan 27, 2020 at 14:14

Paul Hodges

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