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print field before and after matching pattern of single line

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

2

My file contains the data below:

tail fn0 logfile
more tail3 fn0 logfile1
get than tail4 fn0 logfile2

I want to get output from the field before fn0 and the field after fn0.

Expected output:

tail logfile
tail3 logfile1
tail4 logfile2

linux text-processing

share|improve this question

edited 10 hours ago

Jeff Schaller♦

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asked 11 hours ago

rahulrahul

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2

My file contains the data below:

tail fn0 logfile
more tail3 fn0 logfile1
get than tail4 fn0 logfile2

I want to get output from the field before fn0 and the field after fn0.

Expected output:

tail logfile
tail3 logfile1
tail4 logfile2

linux text-processing

share|improve this question

edited 10 hours ago

Jeff Schaller♦

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asked 11 hours ago

rahulrahul

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add a comment | 

My file contains the data below:

tail fn0 logfile
more tail3 fn0 logfile1
get than tail4 fn0 logfile2

I want to get output from the field before fn0 and the field after fn0.

Expected output:

tail logfile
tail3 logfile1
tail4 logfile2

linux text-processing

share|improve this question

edited 10 hours ago

Jeff Schaller♦

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asked 11 hours ago

rahulrahul

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tail fn0 logfile
more tail3 fn0 logfile1
get than tail4 fn0 logfile2

tail logfile
tail3 logfile1
tail4 logfile2

share|improve this question

edited 10 hours ago

Jeff Schaller♦

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asked 11 hours ago

rahulrahul

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share|improve this question

edited 10 hours ago

Jeff Schaller♦

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rahulrahul

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edited 10 hours ago

Jeff Schaller♦

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edited 10 hours ago

Jeff Schaller♦

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asked 11 hours ago

rahulrahul

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4 Answers
4

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4

You can use awk:

awk -v pattern="fn0" 'for (i=0;i<=NF;i++) if ($i==pattern) print $(i-1),$(i+1) ' file

or if you want to use a regex pattern:

awk -v pattern="^fn0$" 'for (i=0;i<=NF;i++) if ($i~pattern) print $(i-1),$(i+1) ' file

Output:

tail logfile
tail3 logfile1
tail4 logfile2

share|improve this answer

edited 9 hours ago

answered 9 hours ago

pLumopLumo

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3

With perl

$ perl -lane '($i) = grep $F[$_] eq "fn0" 0..$#F;
 print "$F[$i-1] $F[$i+1]"' ip.txt
tail logfile
tail3 logfile1
tail4 logfile2

• 
  -l will remove newline character from input line, and add it back while printing


• 
  -a split input line based on whitespaces, @F array will have the data


• 
  ($i) = grep $F[$_] eq "fn0" 0..$#F get index of the element whose exact content is fn0
  


• 
  print "$F[$i-1] $F[$i+1]" print the required fields

With sed that supports ERE

$ sed -E 's/^(.* )?([^ ]+) fn0 ([^ ]+).*/2 3/' ip.txt
tail logfile
tail3 logfile1
tail4 logfile2

• 
  ^(.* )? optional fields at start of line


• 
  ([^ ]+) fn0 ([^ ]+) capture fields before and after fn0 (assuming single space as field separator)


• 
  .* rest of the line


• 
  2 3 required fields in output

share|improve this answer

answered 8 hours ago

SundeepSundeep

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0

a solution with grep and sed:

egrep -oh '[a-zA-Z0-9]+ fn0 [a-zA-Z0-9]+' testfile | sed 's/ fn0 / /'

share|improve this answer

answered 9 hours ago

markgrafmarkgraf

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0

$ sed 's/.*(tail[[:digit:]]*) fn[[:digit:]]*/1/' file
tail logfile
tail3 logfile1
tail4 logfile2

The sed expression is replaces everything on each line up to some substring matching tailXX fnYY with tailXX (where XX and YY are some positive integers).

share|improve this answer

answered 9 hours ago

Kusalananda♦Kusalananda

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4

You can use awk:

awk -v pattern="fn0" 'for (i=0;i<=NF;i++) if ($i==pattern) print $(i-1),$(i+1) ' file

or if you want to use a regex pattern:

awk -v pattern="^fn0$" 'for (i=0;i<=NF;i++) if ($i~pattern) print $(i-1),$(i+1) ' file

Output:

tail logfile
tail3 logfile1
tail4 logfile2

share|improve this answer

edited 9 hours ago

answered 9 hours ago

pLumopLumo

7,2081414 silver badges3232 bronze badges

add a comment | 

4

You can use awk:

awk -v pattern="fn0" 'for (i=0;i<=NF;i++) if ($i==pattern) print $(i-1),$(i+1) ' file

or if you want to use a regex pattern:

awk -v pattern="^fn0$" 'for (i=0;i<=NF;i++) if ($i~pattern) print $(i-1),$(i+1) ' file

Output:

tail logfile
tail3 logfile1
tail4 logfile2

share|improve this answer

edited 9 hours ago

answered 9 hours ago

pLumopLumo

7,2081414 silver badges3232 bronze badges

add a comment | 

You can use awk:

awk -v pattern="fn0" 'for (i=0;i<=NF;i++) if ($i==pattern) print $(i-1),$(i+1) ' file

or if you want to use a regex pattern:

awk -v pattern="^fn0$" 'for (i=0;i<=NF;i++) if ($i~pattern) print $(i-1),$(i+1) ' file

Output:

tail logfile
tail3 logfile1
tail4 logfile2

share|improve this answer

edited 9 hours ago

answered 9 hours ago

pLumopLumo

7,2081414 silver badges3232 bronze badges

awk -v pattern="fn0" 'for (i=0;i<=NF;i++) if ($i==pattern) print $(i-1),$(i+1) ' file

awk -v pattern="^fn0$" 'for (i=0;i<=NF;i++) if ($i~pattern) print $(i-1),$(i+1) ' file

tail logfile
tail3 logfile1
tail4 logfile2

share|improve this answer

edited 9 hours ago

answered 9 hours ago

pLumopLumo

7,2081414 silver badges3232 bronze badges

answered 9 hours ago

pLumopLumo

7,2081414 silver badges3232 bronze badges

answered 9 hours ago

pLumopLumo

7,2081414 silver badges3232 bronze badges

3

With perl

$ perl -lane '($i) = grep $F[$_] eq "fn0" 0..$#F;
 print "$F[$i-1] $F[$i+1]"' ip.txt
tail logfile
tail3 logfile1
tail4 logfile2

• 
  -l will remove newline character from input line, and add it back while printing


• 
  -a split input line based on whitespaces, @F array will have the data


• 
  ($i) = grep $F[$_] eq "fn0" 0..$#F get index of the element whose exact content is fn0
  


• 
  print "$F[$i-1] $F[$i+1]" print the required fields

With sed that supports ERE

$ sed -E 's/^(.* )?([^ ]+) fn0 ([^ ]+).*/2 3/' ip.txt
tail logfile
tail3 logfile1
tail4 logfile2

• 
  ^(.* )? optional fields at start of line


• 
  ([^ ]+) fn0 ([^ ]+) capture fields before and after fn0 (assuming single space as field separator)


• 
  .* rest of the line


• 
  2 3 required fields in output

share|improve this answer

answered 8 hours ago

SundeepSundeep

7,73611 gold badge1111 silver badges2828 bronze badges

add a comment | 

3

With perl

$ perl -lane '($i) = grep $F[$_] eq "fn0" 0..$#F;
 print "$F[$i-1] $F[$i+1]"' ip.txt
tail logfile
tail3 logfile1
tail4 logfile2

• 
  -l will remove newline character from input line, and add it back while printing


• 
  -a split input line based on whitespaces, @F array will have the data


• 
  ($i) = grep $F[$_] eq "fn0" 0..$#F get index of the element whose exact content is fn0
  


• 
  print "$F[$i-1] $F[$i+1]" print the required fields

With sed that supports ERE

$ sed -E 's/^(.* )?([^ ]+) fn0 ([^ ]+).*/2 3/' ip.txt
tail logfile
tail3 logfile1
tail4 logfile2

• 
  ^(.* )? optional fields at start of line


• 
  ([^ ]+) fn0 ([^ ]+) capture fields before and after fn0 (assuming single space as field separator)


• 
  .* rest of the line


• 
  2 3 required fields in output

share|improve this answer

answered 8 hours ago

SundeepSundeep

7,73611 gold badge1111 silver badges2828 bronze badges

add a comment | 

With perl

$ perl -lane '($i) = grep $F[$_] eq "fn0" 0..$#F;
 print "$F[$i-1] $F[$i+1]"' ip.txt
tail logfile
tail3 logfile1
tail4 logfile2

• 
  -l will remove newline character from input line, and add it back while printing


• 
  -a split input line based on whitespaces, @F array will have the data


• 
  ($i) = grep $F[$_] eq "fn0" 0..$#F get index of the element whose exact content is fn0
  


• 
  print "$F[$i-1] $F[$i+1]" print the required fields

With sed that supports ERE

$ sed -E 's/^(.* )?([^ ]+) fn0 ([^ ]+).*/2 3/' ip.txt
tail logfile
tail3 logfile1
tail4 logfile2

• 
  ^(.* )? optional fields at start of line


• 
  ([^ ]+) fn0 ([^ ]+) capture fields before and after fn0 (assuming single space as field separator)


• 
  .* rest of the line


• 
  2 3 required fields in output

share|improve this answer

answered 8 hours ago

SundeepSundeep

7,73611 gold badge1111 silver badges2828 bronze badges

$ perl -lane '($i) = grep $F[$_] eq "fn0" 0..$#F;
 print "$F[$i-1] $F[$i+1]"' ip.txt
tail logfile
tail3 logfile1
tail4 logfile2

$ sed -E 's/^(.* )?([^ ]+) fn0 ([^ ]+).*/2 3/' ip.txt
tail logfile
tail3 logfile1
tail4 logfile2

share|improve this answer

answered 8 hours ago

SundeepSundeep

7,73611 gold badge1111 silver badges2828 bronze badges

answered 8 hours ago

SundeepSundeep

7,73611 gold badge1111 silver badges2828 bronze badges

answered 8 hours ago

SundeepSundeep

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0

a solution with grep and sed:

egrep -oh '[a-zA-Z0-9]+ fn0 [a-zA-Z0-9]+' testfile | sed 's/ fn0 / /'

share|improve this answer

answered 9 hours ago

markgrafmarkgraf

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markgraf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment | 

0

a solution with grep and sed:

egrep -oh '[a-zA-Z0-9]+ fn0 [a-zA-Z0-9]+' testfile | sed 's/ fn0 / /'

share|improve this answer

answered 9 hours ago

markgrafmarkgraf

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Check out our Code of Conduct.

add a comment | 

a solution with grep and sed:

egrep -oh '[a-zA-Z0-9]+ fn0 [a-zA-Z0-9]+' testfile | sed 's/ fn0 / /'

share|improve this answer

answered 9 hours ago

markgrafmarkgraf

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share|improve this answer

answered 9 hours ago

markgrafmarkgraf

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answered 9 hours ago

markgrafmarkgraf

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answered 9 hours ago

markgrafmarkgraf

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0

$ sed 's/.*(tail[[:digit:]]*) fn[[:digit:]]*/1/' file
tail logfile
tail3 logfile1
tail4 logfile2

The sed expression is replaces everything on each line up to some substring matching tailXX fnYY with tailXX (where XX and YY are some positive integers).

share|improve this answer

answered 9 hours ago

Kusalananda♦Kusalananda

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add a comment | 

0

$ sed 's/.*(tail[[:digit:]]*) fn[[:digit:]]*/1/' file
tail logfile
tail3 logfile1
tail4 logfile2

The sed expression is replaces everything on each line up to some substring matching tailXX fnYY with tailXX (where XX and YY are some positive integers).

share|improve this answer

answered 9 hours ago

Kusalananda♦Kusalananda

158k1818 gold badges313313 silver badges498498 bronze badges

add a comment | 

$ sed 's/.*(tail[[:digit:]]*) fn[[:digit:]]*/1/' file
tail logfile
tail3 logfile1
tail4 logfile2

The sed expression is replaces everything on each line up to some substring matching tailXX fnYY with tailXX (where XX and YY are some positive integers).

share|improve this answer

answered 9 hours ago

Kusalananda♦Kusalananda

158k1818 gold badges313313 silver badges498498 bronze badges

$ sed 's/.*(tail[[:digit:]]*) fn[[:digit:]]*/1/' file
tail logfile
tail3 logfile1
tail4 logfile2

share|improve this answer

answered 9 hours ago

Kusalananda♦Kusalananda

158k1818 gold badges313313 silver badges498498 bronze badges

answered 9 hours ago

Kusalananda♦Kusalananda

158k1818 gold badges313313 silver badges498498 bronze badges

answered 9 hours ago

Kusalananda♦Kusalananda

158k1818 gold badges313313 silver badges498498 bronze badges