How to Grep "\N" in File

How to grep \n in file

Use -F to match fixed strings:

$ grep -F "\n" file
echo "\nThis line has new line char." >> mno.txt

From man grep:

-F, --fixed-strings

Interpret PATTERN as a list of fixed strings, separated by newlines,
any of which is to be matched. (-F is specified by POSIX.)

How to give a pattern for new line in grep?

grep patterns are matched against individual lines so there is no way for a pattern to match a newline found in the input.

However you can find empty lines like this:

grep '^$' file
grep '^[[:space:]]*$' file # include white spaces 

grep behavior with \n in regex

grep strips the trailing newline from each line, just as if it were reading from a file. Its input is assumed to be a POSIX text file, not an arbitrary byte stream, so each input line will be newline-free.

grep, cut and remove \n from file

With your shown samples, please try following. Written and tested with shown samples in GNU awk.

awk '
match($0,/uid=[^,]*/){
  val1=substr($0,RSTART+4,RLENGTH-4)
  next
}
{
  val=""
  while($0){
    match($0,/LDAPresource=[^ ]*/)
    val=(val?val OFS:"")(val1 ";" substr($0,RSTART+13,RLENGTH-13))
    $0=substr($0,RSTART+RLENGTH)
  }
  print val
}' Input_file

Explanation: Adding detailed explanation for above.

awk '                                 ##Starting awk program from here.
match($0,/uid=[^,]*/){                ##Using match function to match regex uid= till comma comes in current line.
  val1=substr($0,RSTART+4,RLENGTH-4)  ##Creating val1 variable which has sub string of matched regex of above.
  next                                ##next will skip all further statements from here.
}
{
  val=""                              ##Nullifying val variable here.
  while($0){                          ##Running loop till current line value is not null.
    match($0,/LDAPresource=[^ ]*/)    ##using match to match regex from string LDAPresource= till space comes.
    val=(val?val OFS:"")(val1 ";" substr($0,RSTART+13,RLENGTH-13))  ##Creating val which has val1 ; and sub string of above matched regex.
    $0=substr($0,RSTART+RLENGTH)      ##Saving rest of line in current line.
  }
  print val                           ##Printing val here.
}' Input_file                         ##Mentioning Input_file name here.

How can I format my grep output to show line numbers at the end of the line, and also the hit count?

-n returns line number.

-i is for ignore-case. Only to be used if case matching is not necessary

$ grep -in null myfile.txt

2:example two null,
4:example four null,

Combine with awk to print out the line number after the match:

$ grep -in null myfile.txt | awk -F: '{print $2" - Line number : "$1}'

example two null, - Line number : 2
example four null, - Line number : 4

Use command substitution to print out the total null count:

$ echo "Total null count :" $(grep -ic null myfile.txt)

Total null count : 2

grep for special characters in Unix

Tell grep to treat your input as fixed string using -F option.

grep -F '*^%Q&$*&^@$&*!^@$*&^&^*&^&' application.log

Option -n is required to get the line number,

grep -Fn '*^%Q&$*&^@$&*!^@$*&^&^*&^&' application.log

grep results: how to format output file with one result for each line

Using awk for fixing the problem is an idea, after some adjustments.

You need \r (the Windows linefeed) and use double quotes.

grep 'ACCESSION' chrom_CDS_2.txt | awk '{print $0 "\r"}' > accession_out.txt

When you use awk, you do not need grep:

awk '/ACCESSION/ {print $0 "\r"}' chrom_CDS_2.txt > accession_out.txt

Another possibility is using sed: By default don't print lines. When ACCESSION is part of the line, replace the complete line with the complete line (&, matched part), followed by \r and use /p for printing it.

sed -n 's/.*ACCESSION.*/&\r/p' chrom_CDS_2.txt > accession_out.txt

How do you search for files containing DOS line endings (CRLF) with grep on Linux?

grep probably isn't the tool you want for this. It will print a line for every matching line in every file. Unless you want to, say, run todos 10 times on a 10 line file, grep isn't the best way to go about it. Using find to run file on every file in the tree then grepping through that for "CRLF" will get you one line of output for each file which has dos style line endings:

find . -not -type d -exec file "{}" ";" | grep CRLF

will get you something like:

./1/dos1.txt: ASCII text, with CRLF line terminators
./2/dos2.txt: ASCII text, with CRLF line terminators
./dos.txt: ASCII text, with CRLF line terminators

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