Unix Record Extract by Column Value

[jraymond]

I have a file of thousands of records. In column 87 it can contain 001, 002, 003, or 004. I need a quick easy way to create an output with all the records greater than 002. What unix command can do this? Or, do I need to write a quick program.

[rski]

You do not write what is the column separator, assuming it is a ':' yo can do something like that

awk 'BEGIN{$IFS=":"}$87=!/00[1,2]/{print}' <filename>
 

I guess saying grater than 002 you mean rows which do not contain 001and 002

[jraymond]

There are no column separators just a string of alphanumeric data. Basically I have a datafile being transmitted to us and I neet to strip off certain records.

[rski]

@jraymond
If there are no column separator, how do you recognize that 001 is in column 87?
Can you put here some data from that datafile and show where do you think 87th column is.

[ennoil]

You can use the cut (-b) command and specify what position in the string that you want to look at...From there you can use an if statement:

for i in `cat <file>`
do
  section=`echo ${i} | cut -b 87-89`
  if [[ ${section} == "003" -o ${section} == "004"
  then
    echo ${section}
  fi
done

Just for a start...