awk, print lines which start with four digits

问题:

I want to print all lines from a file which begin with four digits. I tried this allredy but it does not work:

cat data.txt | awk --posix '{ if ($1 ~ /^[0-9]{4}/) print $1}'

No output is generated

The next line prints all lins which start with a number:

cat data.txt | awk --posix '{ if ($1 ~ /^[0-9]/) print $1}'

回答1:

With awk:

$ awk '/^[0-9][0-9][0-9][0-9]/ {print $1}' your_file

that is, check for 4 digits starting the line.

Update: 5th character not to be a digit.

$ awk '/^[0-9][0-9][0-9][0-9]([^0-9].*)?$/ {print $1}' your_file

Note it is not necessary to use the { if ($1 ~ /^[0-9]/) sentence, it is done with just /^.../.

回答2:

For printing lines that match a given regexp grep is the first tool to grab:

grep -Eo '^[0-9]{4}[^0-9]*\s' file

回答3:

I cannot see the problem.

awk --posix '{ if ($1 ~ /^[0-9]{4}/) print $1}'<<EOT
1234 qwer
234 asdf
34456 qwe
EOT

gets

1234
34456

As expected...

If You need match exactly four digits, then You could use:

awk --posix '$1~/^[0-9]{4}$/{print $1}'<<EOT
1234 qwer
234 asdf
34456 qwe
EOT

Output:

1234

But actually You do not need to run awk if you are in bash:

while read f rest; do
  [[ $f =~ ^[[:digit:]]{4}$ ]] && echo $f
done <<EOT
1234 qwer
234 asdf
34456 qwe
EOT

Output:

1234