Use to awk to match pattern, and print the pattern

Shell Programming and Scripting
1

Hi,

I know how to use awk to search some expressions like five consecutive numbers, [0-9][0-9][0-9][0-9][0-9], this is easy.
However, how do I make awk print the pattern that is been matched?

For example:
input: usa,canada99292,japan222,france59664,egypt223
output:99292,59664

2

Should work for you :rolleyes:

awk -F, '{for(i=0;++i<=NF;){gsub("[a-z]","",$i);if(length($i)==5){a=(a)?a FS$i:$i}}{print a;a=""}}' file

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3

Hi Dan,

Thanks a lot for your reply. But the code won't work in instances such as this:
input:

usa,66 canada99292,japan222,44 france59664,egypt223

because within each column, the field separator actually changes to space

Is it possible to have multiple field separator in place simultaneously, in this case: space and comma. If this is possible, then your code would work .

4

If your version of awk supports regex as RS, try this:

awk -v RS="[ ,]" -v ORS="," '/[0-9][0-9][0-9][0-9]/ {gsub(/[a-z]/,"");print}' file
5

With a slightly change of the code of danmero:

awk -F",| " '{for(i=0;++i<=NF;){gsub("[a-z]","",$i);if(length($i)==5){a=(a)?a "," $i:$i}}{print a;a=""}}'
6

if you have other options like grep or sed:

grep -o pattern file

or

sed -n '/pattern/ s/.*\(pattern\).*/\1/p' file
7

Yes but you should check the OP required output.