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I have multiple files with the following text:

20~02~19~05-01-52
2249\\\2249\\\2249\\\2249\\\2249\\\2249\\\2248\\\

I'd like to use sed or another Linux command to replace \\\ with a newline.

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7 Answers 7

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13

sed 's|\\\\\\|\n|g' filename does it, if you're using GNU sed.

If you want POSIX sed, then this should work (quite a lot of escaping!):

sed 's|\\\\\\|\
|g' filename
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8

I'd use Perl here since it has more advanced regexes that can easily support repetition. So, something like this:

$ perl -pe 's/\\{3}/\n/g' file 
20~02~19~05-01-52
2249
2249
2249
2249
2249
2249
2248

The -pe means "print each input line after applying the script given by -e to it". The script itself simply replaces every occurrence of exactly 3 \ with a newline. We need \\ instead of \ because the \ needs to be escaped.

If your sed supports it (GNU sed does) you can use the same approach:

$ sed -E 's/\\{3}/\n/g' file 
20~02~19~05-01-52
2249
2249
2249
2249
2249
2249
2248
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1
  • 1
    You can use \\\{3\} in standard BRE in sed as well. Actually, the \{x,y\} BRE operator is more portable than the {x,y} ERE operator.
    – Stéphane Chazelas
    Commented Feb 20, 2020 at 19:00
4

With awk:

awk '{gsub(/\\{3}/, RS)}1' file
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2
  • zsh
printf '%s\n' ${(s:\\\:)"$(<FILE)"}
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2

Using sed: sed -n 's/\\\\\\/\n/g; p' filename.txt.

It is important to have the 'p' after the semicolon, otherwise it would not print the first line where no substitution takes place. 6 backslashes because 3 backslashes used as escape characters for the 3 backslashes you have. 'g' for global. '-n' to avoid automatic printing.

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3
  • 1
    I don't understand the bit about p and the first line. If I do the command without -n and without p, the output is the same, including the first line, at least with GNU's sed. Automatic printing does not depend on whether substitutions were made on a line.
    – JoL
    Commented Feb 20, 2020 at 18:55
  • Oh sorry, my bad! Yes, in this particular case, it works without the -n and p options. But usually I work with more complicated problems when using sed and automatic printing only gets in the way. It's normally easier when I always suppress automatic printing and print only when I want to, rather than having to think if this particular problem requires automatic suppression of printing.
    – srgk26
    Commented Feb 22, 2020 at 8:16
  • For example. for this very same problem, if I want to instead print only where substitutions were made, then I would have my p command come in without the semicolon in between, like this: sed -n 's/\\\\\\/\n/gp' filename.txt. Or normally I don't want to print a certain range of files, and perform substitutions on the included lines in a file. So in almost all my use cases, I would need to suppress automatic printing, so I developed the habit. Otherwise you're right, don't need it in this particular instance.
    – srgk26
    Commented Feb 22, 2020 at 8:19
0

Just a different approach, assuming that the file has no other patterns like this, using GNU sed:

sed -E 's;([0-9]{4}).{3};\1\n;g' < file
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0

I rewrote the code that receives the Serial data and used the following to generate linefeeds. Thank you so much for the posts . This will keep me studying for a while.

sed  -i 's/\\n/\n/g' *.txt
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