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Shell Programming and Scripting

pattern checking

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# 1  
Old 06-23-2008
pattern checking
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  
Old 06-23-2008
Code:
grep -E '^\"[[:digit:]]{10}\"$'

example:
Code:
# 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  
Old 06-23-2008
The following sed script will print out the line number and contents of any line that does not meet the pattern requirements.

Code:
#!/usr/bin/ksh

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

# 4  
Old 06-28-2008
hi,

thanks a lot... it works perfectly..

@fpmurphy

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

Krips.Smilie
# 5  
Old 06-29-2008
!{=;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.
Code:
10
FPM1234567

This output is then piped to a second invocation of sed which concatinates these 2 lines into a single line i.e.
Code:
10 FPM1234567

# 6  
Old 06-29-2008
thanks a lot ...Smilie
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