Matching Regex Of Multiple Lines In Awk. && Operator?

I am not sure if the && operator works in regular expressions. What I am trying to do is match a line such that it starts with a number and has the letter 'a' AND the next

Solution 1:

[Update based on clarification.]

One high order bit is that Awk is a line-oriented language, so you won't actually be able to do a normal pattern match to span lines. The usual way to do something like this is to match each line separately, and have a later clause / statement figure out if all the right pieces have been matched.

What I'm doing here is looking for an a in the second field on one line, a b in the second field on another line, and a c in the second field on a third line. In the first two cases, I stash away the contents of the line as well as what line number it occurred on. When the third line is matched and we haven't yet found the whole sequence, I go back and check to see if the other two lines are present and with acceptable line numbers. If all's good, I print out the buffered previous lines and set a flag indicating that everything else should print.

Here's the script:

$2 == "a" { a = $0; aLine = NR; }
$2 == "b" { b = $0; bLine = NR; }
$2 == "c" && !keepPrinting {
    if ((bLine == (NR - 1)) && (aLine == (NR - 2))) {
        print a;
        print b;
        keepPrinting = 1;
    }
}
keepPrinting { print; }

And here's a file I tested it with:

JUNK UP HERE NOT STARTING WITH NUMBER
1a0.1100.0692a0.0620.0883a0.0620.1214b0.0620.1215     c           0.0320.1006     d           0.0320.1007     e           0.0320.1008a0.0990.1219b0.0980.12110    c           0.0970.10011    x           0.0000.200

Here's what I get when I run it:

$ awk -f blort.awk blort.txt
3     a           0.0620.1214     b           0.0620.1215c0.0320.1006     d           0.0320.1007     e           0.0320.1008     a           0.0990.1219     b           0.0980.12110c0.0970.10011    x           0.0000.200

Solution 2:

No it doesn't work. You could try something like this:

/(^[0-9]+.*a[^\n]*)\n([0-9]+.*b[^\n]*)\n([0-9]+.*c[^\n]*)/

And repeat that for as many letters as you need.

The [^\n]* will match as much non-linebreak characters in a row as possible (so up to the linebreak).

Solution 3:

A friend wrote this awk program for me. It is a state machine. And it works.

#!/usr/bin/awk -f

BEGIN {
    # We start out in the "idle" state.state = "idle"
}

/^[0-9]+[[:space:]]+q/ {
    # Everytime we encounter a "# q" we either print it or go to the# "q_found" state.if (state != "printing") {
        state = "q_found"
        line_q = $0
    }
}

/^[0-9]+[[:space:]]+r/ {
    # If we are in the q_found state and "# r" immediate follows,# advance to the r_found state.  Else, return to "idle" and # wait for the "# q" to start us off.if (state == "q_found") {
        state = "r_found"
        line_r = $0
    } elseif (state != "printing") {
        state = "idle"
    }
}

/^[0-9]+[[:space:]]+l/ {
    # If we are in the r_found state and "# l" immediate follows,# advance to the l_found state.  Else, return to "idle" and # wait for the "# q" to start us off.if (state == "r_found") {
        state = "l_found"
        line_l = $0
    } elseif (state != "printing") {
        state = "idle"
    }
}

/^[0-9]+[[:space:]]+i/ {
    # If we are in the l_found state and "# i" immediate follows,# we're ready to start printing.  First, display the lines we# squirrelled away then move to the "printing" state.  Else,# go to "idle" and wait for the "# q" to start us off.if (state == "l_found") {
        state = "printing"print line_q
        print line_r
        print line_l
        line = 0
    } elseif (state != "printing") {
        state = "idle"
    }
}

/^[0-9]+[[:space:]]+/ {
    # If in state "printing", print 50 lines then stop printingif (state == "printing") {
        if (++line < 48) print
    }
}