[SOLVED] grep - check second pattern only if first does not produce a match

Issue

I am writing a script that accepts as an argument a string. I want to run a particular command, check the output of that command for that input string, first returning lines that match on <input string>$ and only if that does not return any lines, then return all lines that contain <input string> anywhere in the line. I am currently using grep -E but am open to awk or sed.

Consider this output written to file:

> cat command.out
A
A1
B
B1
B2
C1
C2
C3
XYZ
XYZ1
XYZ2

If my input string is 'B' then I want to return

B

not

B
B1
B2

If my input string is 'C' then I want to return

C1
C2
C3

If my input string is 'Z' then I want to return

XYZ

If my input string is 'Y' then I want to return

XYZ
XYZ1
XYZ2

Using a | (or) in the pattern doesn't do what I am after as it would return all lines with B.

What I have works but seems inefficient and I suspect there is a better way.

> command_output="$(cat command.out)"

> matches="$( (print "$command_output"|grep -E 'B$')||(print "$command_output"|grep -E 'B') )"
> print "$matches"
B

> matches="$( (print "$command_output"|grep -E 'C$')||(print "$command_output"|grep -E 'C') )"
> print "$matches"
C1
C2
C3

I have to persist the command output and fire off potentially two greps. I was hoping for a piped one-liner

matches="$(<run command>|grep <first pattern, if no match, second pattern>)"

but maybe that is not possible.

Solution

matches=$(
    <run command> |
    awk '
        $0~r"$" && exact=1;
        !exact && $0~r { inexact[n++] = $0 }
        END {
            if(!exact)
                for(i=0;i<n;i++)
                    print inexact[i]
        }
    ' r='regex'
)

• $ is concatenated to the value of r to form a regex anchored to end of line. If $0 matches this:
  • set flag exact
  • result is non-zero / true, so print line
• if exact has not been set and $0 matches r anywhere:
  • append the line to array inexact
• at end, if exact is unset (ie. no exact match was found), print any stored lines

Note that the value passed in is used as a regex. This corresponds to the grep usage in the question. To match an exact string, rather than a regex, remember to escape any regex metacharacters.

---

An alternative approach using exact string comparison rather than regex (and accumulating to a string rather than an array):

matches=$(
    <run command> |
    awk -v s='input string' '
        BEGIN { len=length(s) }
        idx=index($0,s) {
            if ( idx+len > length ) {
                print
                exact=1
            } else approx = approx $0 ORS
        }
        END {
            ORS=""
            if (!exact) print approx
        }
    '
)

We know the length of the string and the input line. When the string appears in the line, the position + length of the string will be longer than the line length only if it appears at the end:

s=SSS
                  idx+len     length
1234567SSS123 -->   8+3    <  13
1234567SSS1   -->   8+3    == 11
1234567SSS    -->   8+3    >  10

Answered By - jhnc

Answer Checked By - Candace Johnson (WPSolving Volunteer)