Find the pattern (YYYY-MM-DD) and replace it with the same value concatenating with apostrophes - bash

Question

I have this kind of data:
1,1990-01-01,2,A,2015-02-09
1,NULL,2,A,2015-02-09
1,1990-01-01,2,A,NULL
And looking for solution which will replace each date in the file with the old value but adding apostrophes. Basically expected result from the example will be:
1,'1990-01-01',2,A,'2015-02-09'
1,NULL,2,A,'2015-02-09'
1,'1990-01-01',2,A,NULL
I have found the way how to find the pattern which match my date, but still can't get with what I can then replace it.
sed 's/[0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]/????/' a.txt > b.txt

Answer 1

Catch the date in a group by surrounding the pattern with parentheses (). Then you can use this catched group with \1 (second group would be \2 etc.).
sed "s/\([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]\)/'\1'/g"
Note the g at the end, which ensures that all matches are replaced (if there are more than one in one line).
If you add -r switch to sed, the awkward backslashes before () can be omitted:
sed -r "s/([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9])/'\1'/g"
This can be further simplified using quantifiers:
sed -r "s/([0-9]{4}-[0-9]{2}-[0-9]{2})/'\1'/g"
Or even:
sed -r "s/([0-9]{4}-([0-9]{2}){2})/'\1'/g"
As mentioned in the comments: Also, in this particular case, you may use & instead of \1, which matches the whole looked-up expression, and omit the ():
sed -r "s/[0-9]{4}(-[0-9]{2}){2}/'&'/g"

Answer 2

You need to use a capture group, as well as replace all matching occurrences with the g flag.
sed 's/\([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]\)/'"'"'\1'"'"'/g' a.txt > b.txt
The replacement text is a bit confusing because a single-quoted string in shell cannot contain a single quote, so you have to close the single-quoted string, then use a double-quoted single-quote. Using $'...'-style quoting in bash simplies it a bit, at the cost of needing to escape the backslashes.
sed $'s/\\([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]\\)/\'\1\'/g' a.txt > b.txt
Or, you can simply double-quote the script, since there's nothing currently in it that is subject to expansion:
sed "s/\([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]\)/'\1'/g" a.txt > b.txt
There is also the special & replacement text, which expands to whatever the regular expressions matches, so you can avoid an explicit capture group:
sed "s/[0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]/'&'/g" a.txt > b.txt

Answer 3

With GNU sed:
sed -E 's/([0-9]{2,4}-?){3}/'\''&'\''/g' file
Depending on your file content, the dates may also be described as 1 or 2 followed by a combination of nine dashes or digits:
sed -E 's/[12][-0-9]{9}/'\''&'\''/g" file

Answer 4

Here is one in awk:
$ awk -v q="'" '
BEGIN { FS=OFS="," } # set selimiters
{
for(i=1;i<=NF;i++) # loop all fields
if($i~/[0-9]{4}-[0-9]{2}-[0-9]{2}/) # if field has a date looking string
$i=q $i q # quote it
}1' file
Output:
1,'1990-01-01',2,A,'2015-02-09'
1,NULL,2,A,'2015-02-09'
1,'1990-01-01',2,A,NULL

Answer 5

Could you please try following.(REGEX mentioned inside match could be written as [0-9]{4}-[0-9]{2}-[0-9]{2} too but since my awk is of old version so couldn't test it, you could try it once)
awk -v s1="'" '
{
while(match($0,/[0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]/)){
val=val substr($0,1,RSTART-1) s1 substr($0,RSTART,RLENGTH) s1
$0=substr($0,RSTART+RLENGTH)
}
print val
val=""
}' Input_file
Output will be as follows.
1,'1990-01-01',2,A,'2015-02-09'
1,NULL,2,A,'2015-02-09'
1,'1990-01-01'

Answer 6

With Perl, it is simple
perl -pe ' s/(\d{4}-\d\d-\d\d)/\x27$1\x27/g '
with inputs - \x27 is used for single quotes
$ cat liubo.txt
1,1990-01-01,2,A,2015-02-09
1,NULL,2,A,2015-02-09
1,1990-01-01,2,A,NULL
$ perl -pe ' s/(\d{4}-\d\d-\d\d)/\x27$1\x27/g ' liubo.txt
1,'1990-01-01',2,A,'2015-02-09'
1,NULL,2,A,'2015-02-09'
1,'1990-01-01',2,A,NULL
$
If you want to use single quotes, then escape $ and wrap the command in double quotes
$ perl -pe " s/(\d{4}-\d\d-\d\d)/\'\$1\'/g " liubo.txt
1,'1990-01-01',2,A,'2015-02-09'
1,NULL,2,A,'2015-02-09'
1,'1990-01-01',2,A,NULL
$