Print all lines that don’t have numbers, using sed

numeric datased

I need to print all the lines containing non numeric characters using regex in sed.
The input is a csv, and some of it's lines has text and numbers. I'm interested in printing only those without numbers. This is what I tried:

sed -E -n '/^\D*$/p' direcciones.csv

Why it doesn't find anything?

Best Answer

@steeldriver already explained why your attempt didn't work (should work with GNU sed, though).

But why not keep it simple? Printing all lines with only non-numeric characters is the same as dropping all lines with numeric characters:

sed '/[0-9]/d' direcciones.csv

Easier to write and easier to read, isn't it?

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If I am understanding the question correctly, you can accomplish this with one grep command.

For Q1a, your grep output can suppress the filename using -h, e.g.:

grep -hnr "the_string" /media/slowly/DATA/lots_of_files > output.txt

For Q1b, your grep output can include lines preceding and following matched lines using -A and -B, e.g.:

grep -hnr -A2 -B2 "the_string" /media/slowly/DATA/lots_of_files > output.txt

The output will contain a separator between matches, which you can suppress with --no-group-separator, e.g.:

grep -hnr -A2 -B2 --no-group-separator "the_string" /media/slowly/DATA/lots_of_files > output.txt

Note that the output uses a different delimiter for matching lines (:) and context lines (-).

Linux sed – Print Lines Between Start and End Pattern

cat input |
sed '/\*\*\*\*\* BEGIN \*\*\*\*\*/,/\*\*\*\*\* END *\*\*\*\*/ p;d' | 
tac |
sed '/\*\*\*\*\* END \*\*\*\*\*/,/\*\*\*\*\* BEGIN *\*\*\*\*/ p;d' |
tac

It works by having tac reverse the lines so that sed can find both delimiters in both orders.

Best Answer

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How to find lines containing a string and then printing those specific lines and something else

Linux sed – Print Lines Between Start and End Pattern

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