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Regex - Return numbers of exactly 8 digits

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# 1  
Old 08-22-2012
Regex - Return numbers of exactly 8 digits
Hi All

I am new to this forum and also regex.

I am using bash scripting and have a file like this
Code:
"0012","efgh","12345678","adfdf", "36598745"
"87654321","hijk","lmno"

I want the ouput to be
Code:
12345678
36598745
87654321

Criteria like this
- number
- 8 carachters long

Please let me know what regexc a satisy my reqs

Thanks in advance!

B


Moderator's Comments:
Mod Comment Please use code tags next time for your code and data.
Last edited by zaxxon; 08-22-2012 at 03:31 AM.. Reason: code tags
# 2  
Old 08-22-2012
Welcome to this forum...Smilie

try this...

Code:
awk -F '"' '{ for (i=1;i<=NF;i++) { if ( ($i ~ /[0-9]/) && (length($i) == 8))   { print $i } } } ' file2

# 3  
Old 08-22-2012
Hi Pamu

Thanks a mil for the quick response!

I have tried in both ways

Code:
cat file1.txt | awk -F '"' '{ for (i=1;i<=NF;i++) { if ( ($i ~ /[0-9]/) && (length($i) == 8))   { print $i } } } '

and

Code:
awk -F '"' '{ for (i=1;i<=NF;i++) { if ( ($i ~ /[0-9]/) && (length($i) == 8))   { print $i } } } ' file1.txt

(file1.txt being the text example in the 1st post)

...but i get

Code:
awk: can't open { for (i=1;i<=NF;i++) { if ( ($i ~ /[0-9]/) && (length($i) == 8))   { print $i } } }

please advise (and thanks again!)


Moderator's Comments:
Mod Comment Please use code tags next time for your code and data.
Last edited by zaxxon; 08-22-2012 at 03:32 AM.. Reason: code tags
# 4  
Old 08-22-2012
Quote:
Originally Posted by pamu
Welcome to this forum...Smilie

try this...

Code:
awk -F '"' '{ for (i=1;i<=NF;i++) { if ( ($i ~ /[0-9]/) && (length($i) == 8))   { print $i } } } ' file2

Hi pamu,
Your script will print any 8 character quoted field as long as it contains at least one digit. In addition to strings like "12345678" it will also match a string like "1bcdefgh" and "abc4efgh". If someone just wants 8 character quoted strings with each character being a digit, this should work:
Code:
awk -F '"' '{ for(i=1;i<=NF;i++) if($i ~ "^([0-9]){8}$") print $i }' input

# 5  
Old 08-22-2012
Quote:
Originally Posted by pamu
Welcome to this forum...Smilie

try this...

Code:
awk -F '"' '{ for (i=1;i<=NF;i++) { if ( ($i ~ /[0-9]/) && (length($i) == 8))   { print $i } } } ' file2

The output of that code is not limited to 8 digit numbers. It will print out any field of 8 characters with at least one digit.

You can set RS and then just use a single regex:
Code:
awk '/^[0-9]{8}$/' RS=\"

Or, without using awk:
Code:
tr \" \\n | grep '^[0-9]\{8\}$'

Regards,
Alister
# 6  
Old 08-22-2012
Quote:
Originally Posted by buttseire
Hi Pamu

Thanks a mil for the quick response!

I have tried in both ways

cat file1.txt | awk -F '"' '{ for (i=1;i<=NF;i++) { if ( ($i ~ /[0-9]/) && (length($i) == 8)) { print $i } } } '

and

awk -F '"' '{ for (i=1;i<=NF;i++) { if ( ($i ~ /[0-9]/) && (length($i) == 8)) { print $i } } } ' file1.txt

(file1.txt being the text example in the 1st post)

...but i get

awk: can't open { for (i=1;i<=NF;i++) { if ( ($i ~ /[0-9]/) && (length($i) == 8)) { print $i } } }

please advise (and thanks again!)
I wouldn't expect to see that error message unless you had a -f option after the -F option and its option-argument. Are you sure you entered the command exactly as shown above? What shell are you using? What is the output from the command uname -a on the system you're using?
# 7  
Old 08-22-2012
A perl alternative:
Code:
perl -lne 'while (m/(\d{8})/g){print $1}' infile
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