Shell Programming and Scripting

Hi guys,

I was wondering if there was an easy solution, using tcsh, to print out lines that appear twice with a given pattern in a file?

So if I am looking for lines with "test" in a given file that contains:

blah
test blah1
blah //don't print this out as it doesn't have "test" pattern
test blah2
test blah2 //this line appears twice so print this out
test blah4

If I am searching with lines that have "test" in them, I want the script to output:
test blah2

So far I am using grep to find all lines that have the pattern and outputted to a temp file. Then I am using a grep -c on the temp file to see how many of these matched pattern lines exist. If more than 2, it will print the line out. I am close to having this work but I feel like this would take a lot of time and, instead, there is a much simpler and elegant solution.

I feel like I can achieve a solution using awk. I was looking through your forums and found this awk command that counts the occurrences of a pattern.

Code:

 
awk '/,5,/ { count++ } END { print count }' test.txt

Can I extend this further such that it prints out the line with more than two matches? Appreciate any help!

Your example doesn't seem to match your requirement. I'll match the example:
if test appears in the line, look for duplicates (2 only, not 3)

Code:

awk ' /test/ {arr[$0]++} 
       END{ for(i in arr) {if(arr[i]==2){print i}}}' infile > outfile

Thanks jim, you're amazing. That's what I was looking for. I'm guessing I should just do

Code:

 
awk ' /test/ {arr[$0]++} 
       END{ for(i in arr) {if(arr[i]>1){print i}}}' infile > outfile

if I want more than just 1 match right (so there can be three, four, etc matches)?

I've been coding away at this for a few hours now and yet you gave me that line in a few minutes. Anyway, thanks!

Your guess is correct, > 1 finds all repeats.