AWK to match strings beginning with a number - regex

I want to print all the lines of a file where the first element of each line begins with a number using awk. Below are the details on the data contained in the file and command used:
filename contents:
12.44.4444goad ABCDEF/END
LMNOP/START joker
98.0 kites
command used:
awk '{ $1 ~ /^\d[a-zA-Z0-9]*/ }' filename
After running the above command, no results are displayed on the prompt.
Please let me know if there is any correction that needs to be made to the above command.

To print the lines starting with a digit, you can try the following:
awk '/^[[:digit:]]+/' file
as pointed out by #HenkLangeveld your syntax is incorrect. Also the regex \d is not available in awk.

If you only need to match at least one digit at the start of the line, all you need is ^ to match the start of a line and [0-9] to match a digit.
You can use curly brackets with an if statement:
awk '{if($1 ~ /^[0-9]/) print $0}' filename
But that would just be longhand for this:
awk '$1 ~ /^[0-9]/' filename

From your attempted solution, it looks like you want:
awk 'NF>1 && $1 ~ /^[0-9.]*$/' filename
You need to explicitly match the . if you want to include the decimal point, and you need the $ anchor to make the * meaningful. This will miss lines in which the first column looks like 5e39 or -2.3. You can try to catch those cases with:
awk 'NF>1 && $1 ~ /^-?[0-9.]*(e[0-9*])?$/' filename
but at this point I would tell you to use perl and stop trying to be more robust with awk.
Perhaps (this will print blank lines...not sure which behavior you want):
perl -lane 'use POSIX qw(strtod); my ($num, $end) = strtod($F[0]);
print unless $end;' filename
This uses strtod to parse the number and tells you the number of characters at the end of the string that are not part of it.

Drop the braces and the \d, like this:
awk ' $1 ~ /^[0-9]/ ' filename
Awk programs come in chunks. A chunk is a pattern block pair, where the block
defaults to { print }. (An empty pattern defaults to true.)
The /\d/ is a perl-ism and might work in some versions awk - not in those that I tried*. You need either the traditional /^[0-9]/ or the POSIX /^[[:digit:]]/ notation.
*
gnu and ast

Related

Replace newline in quoted strings in huge files

I have a few huge files with values seperated by a pipe (|) sign.
The strings our quoted but sometimes there is a newline in between the quoted string.
I need to read these files with external table from oracle but on the newlines he will give me errors. So I need to replace them with a space.
I do some other perl commands on these files for other errors, so I would like to have a solution in a one line perl command.
I 've found some other similar questions on stackoverflow, but they don't quite do the same and I can't find a solution for my problem with the solution mentioned there.
The statement I tried but that isn't working:
perl -pi -e 's/"(^|)*\n(^|)*"/ /g' test.txt
Sample text:
4454|"test string"|20-05-1999|"test 2nd string"
4455|"test newline
in string"||"test another 2nd string"
4456|"another string"|19-03-2021|"here also a newline
"
4457|.....
Should become:
4454|"test string"|20-05-1999|"test 2nd string"
4455|"test newline in string"||"test another 2nd string"
4456|"another string"|19-03-2021|"here also a newline "
4457|.....
Sounds like you want a CSV parser like Text::CSV_XS (Install through your OS's package manager or favorite CPAN client):
$ perl -MText::CSV_XS -e '
my $csv = Text::CSV_XS->new({sep => "|", binary => 1});
while (my $row = $csv->getline(*ARGV)) {
$csv->say(*STDOUT, [ map { tr/\n/ /r } #$row ])
}' test.txt
4454|"test string"|20-05-1999|"test 2nd string"
4455|"test newline in string"||"test another 2nd string"
4456|"another string"|19-03-2021|"here also a newline "
This one-liner reads each record using | as the field separator instead of the normal comma, and for each field, replaces newlines with spaces, and then prints out the transformed record.
In your specific case, you can also consider a workaround using GNU sed or awk.
An awk command will look like
awk 'NR==1 {print;next;} /^[0-9]{4,}\|/{print "\n" $0;next;}1' ORS="" file > newfile
The ORS (output record separator) is set to an empty string, which means that \n is only added before lines starting with four or more digits followed with a | char (matched with a ^[0-9]{4,}\| POSIX ERE pattern).
A GNU sed command will look like
sed -i ':a;$!{N;/\n[0-9]\{4,\}|/!{s/\n/ /;ba}};P;D' file
This reads two consecutive lines into the pattern space, and once the second line doesn't start with four digits followed with a | char (see the [0-9]\{4\}| POSIX BRE regex pattern), the or more line break between the two is replaced with a space. The search and replace repeats until no match or the end of file.
With perl, if the file is huge but it can still fit into memory, you can use a short
perl -0777 -pi -e 's/\R++(?!\d{4,}\|)/ /g' <<< "$s"
With -0777, you slurp the file and the \R++(?!\d{4,}\|) pattern matches any one or more line breaks (\R++) not followed with four or more digits followed with a | char. The ++ possessive quantifier is required to make (?!...) negative lookahead to disallow backtracking into line break matching pattern.
With your shown samples, this could be simply done in awk program. Written and tested in GNU awk, should work in any awk. This should work fast even on huge files(better than slurping whole file into memory, having mentioned that OP may use it on huge files).
awk 'gsub(/"/,"&")%2!=0{if(val==""){val=$0} else{print val $0;val=""};next} 1' Input_file
Explanation: Adding detailed explanation for above.
awk ' ##Starting awk program from here.
gsub(/"/,"&")%2!=0{ ##Checking condition if number of " are EVEN or not, because if they are NOT even then it means they are NOT closed properly.
if(val==""){ val=$0 } ##Checking condition if val is NULL then set val to current line.
else {print val $0;val=""} ##Else(if val NOT NULL) then print val current line and nullify val here.
next ##next will skip further statements from here.
}
1 ##In case number of " are EVEN in any line it will skip above condition(gusb one) and simply print the line.
' Input_file ##Mentioning Input_file name here.

How to use 'sed' to add dynamic prefix to each number in integer list?

How can I use sed to add a dynamic prefix to each number in an integer list?
For example:
I have a string "A-1,2,3,4,5", I want to transform it to string "A-1,A-2,A-3,A-4,A-5" - which means I want to add prefix of first integer i.e. "A-" to each number of the list.
If I have string like "B-1,20,300" then I want to transform it to string "B-1,B-20,B-300".
I am not able to use RegEx Capturing Groups because for global match they do not retain their value in subsequent matches.
When it comes to looping constructs in sed, I like to use newlines as markers for the places I have yet to process. This makes matching much simpler, and I know they're not in the input because my input is a text line.
For example:
$ echo A-1,2,3,4,5 | sed 's/,/\n/g;:a s/^\([^0-9]*\)\([^\n]*\)\n/\1\2,\1/; ta'
A-1,A-2,A-3,A-4,A-5
This works as follows:
s/,/\n/g # replace all commas with newlines (insert markers)
:a # label for looping
s/^\([^0-9]*\)\([^\n]*\)\n/\1\2,\1/ # replace the next marker with a comma followed
# by the prefix
ta # loop unless there's nothing more to do.
The approach is similar to #potong's, but I find the regex much more readable -- \([^0-9]*\) captures the prefix, \([^\n]*\) captures everything up to the next marker (i.e. everything that's already been processed), and then it's just a matter of reassembling it in the substitution.
Don't use sed, just use the other standard UNIX text manipulation tool, awk:
$ echo 'A-1,2,3,4,5' | awk '{p=substr($0,1,2); gsub(/,/,"&"p)}1'
A-1,A-2,A-3,A-4,A-5
$ echo 'B-1,20,300' | awk '{p=substr($0,1,2); gsub(/,/,"&"p)}1'
B-1,B-20,B-300
This might work for you (GNU sed):
sed -E ':a;s/^((([^-]+-)[^,]+,)+)([0-9])/\1\3\4/;ta' file
Uses pattern matching and a loop to replace a number following a comma by the first column prefix and that number.
Assuming this is for shell scripting, you can do so with 2 seds:
set string = "A1,2,3,4,5"
set prefix = `echo $string | sed 's/^\([A-Z]\).*/\1/'`
echo $string | sed 's/,\([0-9]\)/,'$prefix'-\1/g'
Output is
A1,A-2,A-3,A-4,A-5
With
set string = "B-1,20,300"
Output is
B-1,B-20,B-300
Could you please try following(if ok with awk).
awk '
BEGIN{
FS=OFS=","
}
{
for(i=1;i<=NF;i++){
if($i !~ /^A/&&$i !~ /\"A/){
$i="A-"$i
}
}
}
1' Input_file
if your data in 'd' file, tried on gnu sed:
sed -E 'h;s/^(\w-).+/\1/;x;G;:s s/,([0-9]+)(.*\n(.+))/,\3\1\2/;ts; s/\n.+//' d

Using awk or sed to merge / print lines matching a pattern (oneliner?)

I have a file that contains the following text:
subject:asdfghj
subject:qwertym
subject:bigger1
subject:sage911
subject:mothers
object:cfvvmkme
object:rjo4j2f2
object:e4r234dd
object:uft5ed8f
object:rf33dfd1
I am hoping to achieve the following result using awk or sed (as a oneliner would be a bonus! [Perl oneliner would be acceptable as well]):
subject:asdfghj,object:cfvvmkme
subject:qwertym,object:rjo4j2f2
subject:bigger1,object:e4r234dd
subject:sage911,object:uft5ed8f
subject:mothers,object:rf33dfd1
I'd like to have each line that matches 'subject' and 'object' combined in the order that each one is listed, separated with a comma. May I see an example of this done with awk, sed, or perl? (Preferably as a oneliner if possible?)
I have tried some uses of awk to perform this, I am still learning I should add:
awk '{if ($0 ~ /subject/) pat1=$1; if ($0 ~ /object/) pat2=$2} {print $0,pat2}'
But does not do what I thought it would! So I know I have the syntax wrong. If I were to see an example that would greatly help so that I can learn.
not perl or awk but easier.
$ pr -2ts, file
subject:asdfghj,object:cfvvmkme
subject:qwertym,object:rjo4j2f2
subject:bigger1,object:e4r234dd
subject:sage911,object:uft5ed8f
subject:mothers,object:rf33dfd1
Explanation
-2 2 columns
t ignore print header (filename, date, page number, etc)
s, use comma as the column separator
I'd do it something like this in perl:
#!/usr/bin/perl
use strict;
use warnings;
my #subjects;
while ( <DATA> ) {
m/^subject:(\w+)/ and push #subjects, $1;
m/^object:(\w+)/ and print "subject:",shift #subjects,",object:", $1,"\n";
}
__DATA__
subject:asdfghj
subject:qwertym
subject:bigger1
subject:sage911
subject:mothers
object:cfvvmkme
object:rjo4j2f2
object:e4r234dd
object:uft5ed8f
object:rf33dfd1
Reduced down to one liner, this would be:
perl -ne '/^(subject:\w+)/ and push #s, $1; /^object/ and print shift #s,$_' file
grep, paste and process substitution
$ paste -d , <(grep 'subject' infile) <(grep 'object' infile)
subject:asdfghj,object:cfvvmkme
subject:qwertym,object:rjo4j2f2
subject:bigger1,object:e4r234dd
subject:sage911,object:uft5ed8f
subject:mothers,object:rf33dfd1
This treats the output of grep 'subject' infile and grep 'object' infile like files due to process substitution (<( )), then pastes the results together with paste, using a comma as the delimiter (indicated by -d ,).
sed
The idea is to read and store all subject lines in the hold space, then for each object line fetch the hold space, get the proper subject and put the remaining subject lines back into hold space.
First the unreadable oneliner:
$ sed -rn '/^subject/H;/^object/{G;s/\n+/,/;s/^(.*),([^\n]*)(\n|$)/\2,\1\n/;P;s/^[^\n]*\n//;h}' infile
subject:asdfghj,object:cfvvmkme
subject:qwertym,object:rjo4j2f2
subject:bigger1,object:e4r234dd
subject:sage911,object:uft5ed8f
subject:mothers,object:rf33dfd1
-r is for extended regex (no escaping of parentheses, + and |) and -n does not print by default.
Expanded, more readable and explained:
/^subject/H # Append subject lines to hold space
/^object/ { # For each object line
G # Append hold space to pattern space
s/\n+/,/ # Replace first group of newlines with a comma
# Swap object (before comma) and subject (after comma)
s/^(.*),([^\n]*)(\n|$)/\2,\1\n/
P # Print up to first newline
s/^[^\n]*\n// # Remove first line (can't use D because there is another command)
h # Copy pattern space to hold space
}
Remarks:
When the hold space is fetched for the first time, it starts with a newline (H adds one), so the newline-to-comma substitution replaces one or more newlines, hence the \n+: two newlines for the first time, one for the rest.
To anchor the end of the subject part in the swap, we use (\n|$): either a newline or the end of the pattern space – this is to get the swap also on the last line, where we don't have a newline at the end of the pattern space.
This works with GNU sed. For BSD sed as found in MacOS, there are some changes required:
The -r option has to be replaced by -E.
There has to be an extra semicolon before the closing brace: h;}
To insert a newline in the replacement string (swap command), we have to replace \n by either '$'\n'' or '"$(printf '\n')"'.
Since you specifically asked for a "oneliner" I assume brevity is far more important to you than clarity so:
$ awk -F: -v OFS=, 'NR>1&&$1!=p{f=1}{p=$1}f{print a[++c],$0;next}{a[NR]=$0}' file
subject:asdfghj,object:cfvvmkme
subject:qwertym,object:rjo4j2f2
subject:bigger1,object:e4r234dd
subject:sage911,object:uft5ed8f
subject:mothers,object:rf33dfd1

how to use sed, awk, or gawk to print only what is matched?

I see lots of examples and man pages on how to do things like search-and-replace using sed, awk, or gawk.
But in my case, I have a regular expression that I want to run against a text file to extract a specific value. I don't want to do search-and-replace. This is being called from bash. Let's use an example:
Example regular expression:
.*abc([0-9]+)xyz.*
Example input file:
a
b
c
abc12345xyz
a
b
c
As simple as this sounds, I cannot figure out how to call sed/awk/gawk correctly. What I was hoping to do, is from within my bash script have:
myvalue=$( sed <...something...> input.txt )
Things I've tried include:
sed -e 's/.*([0-9]).*/\\1/g' example.txt # extracts the entire input file
sed -n 's/.*([0-9]).*/\\1/g' example.txt # extracts nothing
My sed (Mac OS X) didn't work with +. I tried * instead and I added p tag for printing match:
sed -n 's/^.*abc\([0-9]*\)xyz.*$/\1/p' example.txt
For matching at least one numeric character without +, I would use:
sed -n 's/^.*abc\([0-9][0-9]*\)xyz.*$/\1/p' example.txt
You can use sed to do this
sed -rn 's/.*abc([0-9]+)xyz.*/\1/gp'
-n don't print the resulting line
-r this makes it so you don't have the escape the capture group parens().
\1 the capture group match
/g global match
/p print the result
I wrote a tool for myself that makes this easier
rip 'abc(\d+)xyz' '$1'
I use perl to make this easier for myself. e.g.
perl -ne 'print $1 if /.*abc([0-9]+)xyz.*/'
This runs Perl, the -n option instructs Perl to read in one line at a time from STDIN and execute the code. The -e option specifies the instruction to run.
The instruction runs a regexp on the line read, and if it matches prints out the contents of the first set of bracks ($1).
You can do this will multiple file names on the end also. e.g.
perl -ne 'print $1 if /.*abc([0-9]+)xyz.*/' example1.txt example2.txt
If your version of grep supports it you could use the -o option to print only the portion of any line that matches your regexp.
If not then here's the best sed I could come up with:
sed -e '/[0-9]/!d' -e 's/^[^0-9]*//' -e 's/[^0-9]*$//'
... which deletes/skips with no digits and, for the remaining lines, removes all leading and trailing non-digit characters. (I'm only guessing that your intention is to extract the number from each line that contains one).
The problem with something like:
sed -e 's/.*\([0-9]*\).*/&/'
.... or
sed -e 's/.*\([0-9]*\).*/\1/'
... is that sed only supports "greedy" match ... so the first .* will match the rest of the line. Unless we can use a negated character class to achieve a non-greedy match ... or a version of sed with Perl-compatible or other extensions to its regexes, we can't extract a precise pattern match from with the pattern space (a line).
You can use awk with match() to access the captured group:
$ awk 'match($0, /abc([0-9]+)xyz/, matches) {print matches[1]}' file
12345
This tries to match the pattern abc[0-9]+xyz. If it does so, it stores its slices in the array matches, whose first item is the block [0-9]+. Since match() returns the character position, or index, of where that substring begins (1, if it starts at the beginning of string), it triggers the print action.
With grep you can use a look-behind and look-ahead:
$ grep -oP '(?<=abc)[0-9]+(?=xyz)' file
12345
$ grep -oP 'abc\K[0-9]+(?=xyz)' file
12345
This checks the pattern [0-9]+ when it occurs within abc and xyz and just prints the digits.
perl is the cleanest syntax, but if you don't have perl (not always there, I understand), then the only way to use gawk and components of a regex is to use the gensub feature.
gawk '/abc[0-9]+xyz/ { print gensub(/.*([0-9]+).*/,"\\1","g"); }' < file
output of the sample input file will be
12345
Note: gensub replaces the entire regex (between the //), so you need to put the .* before and after the ([0-9]+) to get rid of text before and after the number in the substitution.
If you want to select lines then strip out the bits you don't want:
egrep 'abc[0-9]+xyz' inputFile | sed -e 's/^.*abc//' -e 's/xyz.*$//'
It basically selects the lines you want with egrep and then uses sed to strip off the bits before and after the number.
You can see this in action here:
pax> echo 'a
b
c
abc12345xyz
a
b
c' | egrep 'abc[0-9]+xyz' | sed -e 's/^.*abc//' -e 's/xyz.*$//'
12345
pax>
Update: obviously if you actual situation is more complex, the REs will need to me modified. For example if you always had a single number buried within zero or more non-numerics at the start and end:
egrep '[^0-9]*[0-9]+[^0-9]*$' inputFile | sed -e 's/^[^0-9]*//' -e 's/[^0-9]*$//'
The OP's case doesn't specify that there can be multiple matches on a single line, but for the Google traffic, I'll add an example for that too.
Since the OP's need is to extract a group from a pattern, using grep -o will require 2 passes. But, I still find this the most intuitive way to get the job done.
$ cat > example.txt <<TXT
a
b
c
abc12345xyz
a
abc23451xyz asdf abc34512xyz
c
TXT
$ cat example.txt | grep -oE 'abc([0-9]+)xyz'
abc12345xyz
abc23451xyz
abc34512xyz
$ cat example.txt | grep -oE 'abc([0-9]+)xyz' | grep -oE '[0-9]+'
12345
23451
34512
Since processor time is basically free but human readability is priceless, I tend to refactor my code based on the question, "a year from now, what am I going to think this does?" In fact, for code that I intend to share publicly or with my team, I'll even open man grep to figure out what the long options are and substitute those. Like so: grep --only-matching --extended-regexp
why even need match group
gawk/mawk/mawk2 'BEGIN{ FS="(^.*abc|xyz.*$)" } ($2 ~ /^[0-9]+$/) {print $2}'
Let FS collect away both ends of the line.
If $2, the leftover not swallowed by FS, doesn't contain non-numeric characters, that's your answer to print out.
If you're extra cautious, confirm length of $1 and $3 both being zero.
** edited answer after realizing zero length $2 will trip up my previous solution
there's a standard piece of code from awk channel called "FindAllMatches" but it's still very manual, literally, just long loops of while(), match(), substr(), more substr(), then rinse and repeat.
If you're looking for ideas on how to obtain just the matched pieces, but upon a complex regex that matches multiple times each line, or none at all, try this :
mawk/mawk2/gawk 'BEGIN { srand(); for(x = 0; x < 128; x++ ) {
alnumstr = sprintf("%s%c", alnumstr , x)
};
gsub(/[^[:alnum:]_=]+|[AEIOUaeiou]+/, "", alnumstr)
# resulting str should be 44-chars long :
# all digits, non-vowels, equal sign =, and underscore _
x = 10; do { nonceFS = nonceFS substr(alnumstr, 1 + int(44*rand()), 1)
} while ( --x ); # you can pick any level of precision you need.
# 10 chars randomly among the set is approx. 54-bits
#
# i prefer this set over all ASCII being these
# just about never require escaping
# feel free to skip the _ or = or r/t/b/v/f/0 if you're concerned.
#
# now you've made a random nonce that can be
# inserted right in the middle of just about ANYTHING
# -- ASCII, Unicode, binary data -- (1) which will always fully
# print out, (2) has extremely low chance of actually
# appearing inside any real word data, and (3) even lower chance
# it accidentally alters the meaning of the underlying data.
# (so intentionally leaving them in there and
# passing it along unix pipes remains quite harmless)
#
# this is essentially the lazy man's approach to making nonces
# that kinda-sorta have some resemblance to base64
# encoded, without having to write such a module (unless u have
# one for awk handy)
regex1 = (..); # build whatever regex you want here
FS = OFS = nonceFS;
} $0 ~ regex1 {
gsub(regex1, nonceFS "&" nonceFS); $0 = $0;
# now you've essentially replicated what gawk patsplit( ) does,
# or gawk's split(..., seps) tracking 2 arrays one for the data
# in between, and one for the seps.
#
# via this method, that can all be done upon the entire $0,
# without any of the hassle (and slow downs) of
# reading from associatively-hashed arrays,
#
# simply print out all your even numbered columns
# those will be the parts of "just the match"
if you also run another OFS = ""; $1 = $1; , now instead of needing 4-argument split() or patsplit(), both of which being gawk specific to see what the regex seps were, now the entire $0's fields are in data1-sep1-data2-sep2-.... pattern, ..... all while $0 will look EXACTLY the same as when you first read in the line. a straight up print will be byte-for-byte identical to immediately printing upon reading.
Once i tested it to the extreme using a regex that represents valid UTF8 characters on this. Took maybe 30 seconds or so for mawk2 to process a 167MB text file with plenty of CJK unicode all over, all read in at once into $0, and crank this split logic, resulting in NF of around 175,000,000, and each field being 1-single character of either ASCII or multi-byte UTF8 Unicode.
you can do it with the shell
while read -r line
do
case "$line" in
*abc*[0-9]*xyz* )
t="${line##abc}"
echo "num is ${t%%xyz}";;
esac
done <"file"
For awk. I would use the following script:
/.*abc([0-9]+)xyz.*/ {
print $0;
next;
}
{
/* default, do nothing */
}
gawk '/.*abc([0-9]+)xyz.*/' file

Problem with regular expression using grep

I've got some textfiles that hold names, phone numbers and region codes. One combination per line.
The syntax is always "Name Region_code number"
With any number of spaces between the 3 variables.
What I want to do is search for specific region codes, like 23 or 493, forexample.
The problem is that these numbers might appear in the longer numbers too, which might enable a return that shouldn't have been returned.
I was thinking of this sort of command:
grep '04' numbers.txt
But if I do that, a line that contains 04 in the number but not as region code will show as a result too... which is not correct.
I'm sure you are about to get buried in clever regular expressions, but I think in this case all you need to do is include one of the spaces on each side of your region code in the grep.
grep ' 04 ' numbers.txt
I'd do:
awk '$2 == "04"' < numbers.txt
and with grep:
grep -e '^[^ ]*[ ]*04[ ]*[^ ]*$' numbers.txt
If you want region codes alone, you should use:
grep "[[:space:]]04[[:space:]]"
this way it will only look for numbers on the middle column, while start or end of strings are considered word breaks.
You can even do:
function search_region_codes {
grep "[[:space:]]${1}[[:space:]]" FILE
}
replacing FILE with the name of your file,
and use
search_region_codes 04
or even
function search_region_codes {
grep "[[:space:]]${1}[[:space:]]" $2
}
and using
search_region_codes NUMBER FILE
Are you searching for an entire region code, or a region code that contains the subpattern?
If you want the whole region code, and there is at least one space on either side, then you can format the grep by adding a single space on either side of the specific region code. There are other ways to indicate word boundaries using regular expressions.
grep ' 04 ' numbers.txt
If there can be spaces in the name or phone number fields, than that solution might not work. Also, if you the pattern can be a sub-part of the region code, then awk is a better tool. This assumes that the 'name' field contains no spaces. The matching operator '==' requires that the pattern exactly match the field. This can be tricky when there is whitespace on either side of the field.
awk '$2 == "04" {print $0}' < numbers.txt
If the file has a delimiter, than can be set in awk using the '-F' argument to awk to set the field separator character. In this example, a comma is used as the field separator. In addition, the matching operator in this example is a '~' allowing the pattern to be any part of the region code (if that is applicable). The "/y" is a way to match work boundaries at the beginning and end of the expression.
awk -F , '$2 ~ /\y04\y/ {print $0}' < numbers.txt
In both examples, the {print $0} is optional, if you want the full line to be printed. However, if you want to do any formatting on the output, that can be done inside that block.
use word boundaries. not sure if this works in grep, but in other regex implementations i'd surround it with whitespace or word boundary patterns
'\s+04\s+' or '\b04\b'
Something like that