pattern checking

Shell Programming and Scripting
1

hi ,

i am having one file which is having the contents like this :

"1111111111"
"2222222222"
"4444232344"
...

i want to check if all the patterns in the file follows this sequence like starting with " ending with " inside that 10 digits...

can anyone help me in checking that..

thanks,

Krips.

2 
grep -E '^\"[[:digit:]]{10}\"$'

example:

# cat aaa | grep -E '^\"[[:digit:]]{10}\"$'
"1111111111"
"2222222222"
"4444232344"
# cat >bbb
"1111111111"
"2222222222"
"4444232d33"
"f666666666"
"6455"
# cat bbb | grep -E '^\"[[:digit:]]{10}\"$'
"1111111111"
"2222222222"
3

The following sed script will print out the line number and contents of any line that does not meet the pattern requirements.

#!/usr/bin/ksh

sed -n '/^"[0-9]\{10\}"$/!{=;p;}' file | \
sed '{
   N;
   s/\n/ /
}'
4

hi,

thanks a lot... it works perfectly..

@fpmurphy

can you please clear my doubt .. in that sed command what is !{=;p;} checking for..

Krips.:slight_smile:

5

!{=;p;} means if a line does not match the regular expression i.e. exactly 10 digits, then print the line number followed by the line itself i.e.

10
FPM1234567

This output is then piped to a second invocation of sed which concatinates these 2 lines into a single line i.e.

10 FPM1234567
6

thanks a lot ...:slight_smile: