CODEWITHSUNDEEP
bash
capgrey7
capgrey7

Reputation: 13

Need to convert single text column to a single row and then split the row based on the pattern

I am very new to the bash programming and need to convert a single text column to a single row and then separate the characters in the row based on the pattern.

I have text document with the column, which has one letter with six digits in each line:

a111111
b222222
c333333
d444444
e555555

I need to transform the column above into the following row:

'a111111','b222222','c333333','d444444','e555555'

Could someone please advise how this can be achieved?

Upvotes: 1

Views: 42

Answers (2)

anubhava
anubhava

Reputation: 785068

You can use awk with printf:

awk -v ORS=, 'NR>1{printf "%s", ORS} {printf "\x27%s\x27", $0}' file

\x27 prints a single quote.

For the 2nd record onwards it will prints ORS (which is set to comma) at start and then the quoted line will be printed.

Output:

'a111111','b222222','c333333','d444444','e555555'

Upvotes: 2

glenn jackman
glenn jackman

Reputation: 246774

Another approach: 

sed -r 's/^|$/\x27/g' file | paste -sd,

sed adds the single quotes at the beginning and end of each line, and paste joins the line together with commas

Or, print a comma for each line, and when you're done back up 1 character and overwrite the last comma with a space:

awk '{printf "'\''%s'\'',", $0} END {printf "\b \n"}' file

Upvotes: 1

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