Make awk print only if matches are greater than 3 and not print if less than 3

Question

How can I make awk print only if it matches 3 or more and NOT print if it's less than 3.

I'm not strong in awk programming. But here is what I tried.

    awk '/DEF/ {count++;print} { if ( count >  2 ) print count  } ' file
    DEF

DEF is the result from the command above. It is not suppose to print because there is only one match in the file, not 3 or more.

The pattern DEF is in the file once, but awk still prints it even though I wrote in the if statement to print only if it matches 3 or more. Any ideas what I am doing wrong?

Answer 1

This will print number of line with DEF found in the file more than 2 times (3 or more):

awk '/DEF/ {++c} END {if (c>2) print c}' file

What if there are two DEF on one line, count as one line or two DEF ?

Print the lines found when there are more than two DEF lines found.

awk '/DEF/ {++c;a[NR]=$0} END {if (c>2) {for (i in a) print a[i]}}' file

This will print a line with number of DEF lines found as well as the line.

awk '/DEF/ {++c;a[NR]=$0} END {if (c>2) {print "DEF found "c" times";for (i in a) print a[i]}}' file

This will print number of hits and line number its found in as well:

awk '/DEF/ {++c;a[NR]=$0} END {if (c>2) {print "DEF found "c" times\n-----";for (i in a) {print ++t". line "i": "a[i]}}}' file
DEF found 3 times
-----
1. line 4: DEF first time
2. line 8: red DEF second time
3. line 16: DEF more