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I am trying to use \d in regex in sed but it doesn't work:
sed -re 's/\d+//g' But this is working:
sed -re 's/[0-9]+//g' 
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It means that sed will read the next line and start processing it. Your test script doesn't do what you think. It matches the empty lines and applies the delete command to them.
A regular expression is a string that can be used to describe several sequences of characters. Regular expressions are used by several different Unix commands, including ed, sed, awk, grep, and to a more limited extent, vi.
sed does not support "non greedy" operator. You have to use "[]" operator to exclude "/" from match. P.S. there is no need to backslash "/".
r is used to read a file and append it at the current point. The point in your example is the address /EOF/ which means this script will find the line containing EOF and then append the file specified by $thingToAdd after that point. Then it will process the rest of the file.
\d is a switch not a regular expression macro. If you want to use some predefined "constant" instead of [0-9] expression just try run this code:
s/[[:digit:]]+//g There is no such special character group in sed. You will have to use [0-9].
In GNU sed, \d introduces a decimal character code of one to three digits in the range 0-255. As indicated in this comment.
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