Can awk print lines that do not have a pattern?

awk match pattern in column
awk pattern matching
awk '(print column matching pattern)
awk does not contain
awk multiple patterns
awk 'nr==1
awk print line if column matches
awk 'begin

Can awk print all lines that did not match one of the patterns?

In other words, I want to transform some lines but leave the rest unchanged. So, if a /pattern/ matched I would provide a custom block to print the line. I just need to provide a default matcher (like an else) to print the other lines.


Yes, just use any non-zero number and awk will do its default thing which is to print the line:

awk '7' file

If you want it as an "else", put "next" after whatever lines you select for special processing so this one isn't executed for them too.

awk '/pattern/{special processing; next} 7' file

awk print lines that doesn't have a pattern, So far I can print the lines that have Mozilla, but I am unsure how to do this for multiple words and to not print them instead of printing them. Code:. who | awk '{print $1}' awk prints the first field and discards the rest of the line. We can print as many fields as we like. If we add a comma as a separator, awk prints a space between each field. We type the following to also print the time the person logged in (field four): who | awk '{print $1,$4}' There are a couple of special field


You can negate the pattern to get else like behavior:

awk '
    /pattern/ {
        # custom block to print the line    
    }
    !/pattern/ {
        # else do other things
    }
'

AWK negative regular expression, awk '!/domain/ {print}' <<EOF This is domain test This is do test This is test EOF That leaves the lines to process which are the negation of the pattern which awk does not support PCRE (Perl Compatible Regular Expression), so you I need to write a negative regexp, but I must admin I don't remember how I got to that. Lets say we need to print only strings between two lines that contain patterns ‘BEGIN’ and ‘END’. Print Lines Between Two Patterns with SED. With the sed command, we can specify the starting pattern and the ending pattern, to print the lines between strings with these patterns. The syntax and the example are shown below. Syntax:


You can do:

awk '/pattern/ {do something with line} 1' file

Here the 1 will print all lines, both the changed and the not changed line.


Just to show the solution Askan posted using else if

awk '{
    if (/pattern/)
        print "Line number:",NR,"pattern matched"
    else if (/Second/) 
        print "Line number:",NR,"Second matched"
    else 
        print "Line number:",NR,"Another line matched"
    }' file

AWK Cheat Sheet, awk '/pattern/ {action}' file↵. Execute action for matched pattern 'pattern' on file ' file' awk '$1 !~ /regex/' file↵. Print line if field 1 does NOT match regex in file. Variables II END {print n+0}' file↵. Print total number of lines that contain foo. awk� Match pattern=string1 How to print match lines but only from string2 to string5 with awk? Unfortunately this is not processing in line : awk '/string2/,/string5/' file Let's assume there are few long lines with string2 and string5 in different places.


You can use switch if you are using gawk for example

awk '{switch ($0) {
case /pattern/:
    print "Line number:",NR,"pattern matched"
    break

case /Second/:
    print "Line number:",NR,"Second matched"
    break

default:
    print "Line number:",NR,"Another line matched"

}}' input.txt

input.txt

This line matches the pattern
Second line does not match
Hello
This line also matches the pattern
Another line

Output:

Line number: 1 pattern matched
Line number: 2 Second matched
Line number: 3 Another line matched
Line number: 4 pattern matched
Line number: 5 Another line matched

You can also group the cases by removing the break between them. more info

Awk One-Liners Explained, Part III: Selective Printing and Deleting , The "NR > 1" is true only for lines greater than one, so it does not get executed on When the pattern is missing, Awk executes the statement group for all lines. The following snippet uses awk to do the searching and copes with file names containing : (and even newlines, but for these it produces ambiguous output). Note that awk uses extended regular expressions, like grep -E (with minor variations, but not really more than you get between implementations of grep or of awk). find . -type f -exec awk


echo -n "one\ntwo\nthree\nfour" | 
  awk '{a=1} /one/ {print 1;a=0} /three/ {print 3;a=0} a'

1
two
3
four

AWK one-liner collection, $0 the entire line without the newline at the end; $1 to $9, $10 to , the fields AWK ranges: Print the lines from a file starting at the line matching "start" until the Note: make sure that the stop pattern does not match the start line otherwise� Does it have to be awk? If you want to show lines that do not contain a pattern, “grep -v” can be handy. With “-v” grep works like the opposite of grep. When you say “remove lines that do NOT contain a certain pattern”, are you in effect saying, s


Print line X lines before/after found line, which will output m lines before the match, and n lines after the match. After looking into this question a little bit I think I have to revise my comment about the If you don't mind using awk I could offer the following solution. The script tries to seperate multiple pattern contexts in the file by printing lines with --- in between .


Regular expressions in AWK programming: What, Why, and How, Today we will introduce you to the regular expressions in AWK A regular expression doesn't search for a pattern that begins on one line and ends on expression string to print all lines that contain the ly pattern in them.


USD:19 Awk, Printing An action may have no pattern, in which case the action is executed for is actually a complete awk program which will print all lines which contain any�