Sed: How to Delete Lines Matching a Pattern That Contains Forward Slashes

sed: How to delete lines matching a pattern that contains forward slashes?

You were very close. When you use a nonstandard character for a pattern delimiter, such as |pattern|, the first use of that character must be escaped:

$ sed '\|^/dev/xvdb|d' /etc/fstab
/dev/xvda1 / ext4 defaults 1 1
/dev/md0    /mnt/ibsraid    xfs defaults,noatime    0   2
/mnt/ibsraid/varlog /var/log    none    bind    0   0

Similarly, one can use:

sed '\?^/dev/xvdb?d' /etc/fstab

Lastly, it is possible to use slashes inside of /pattern/ if they are escaped in the way that you showed in your answer.

using sed to delete lines containing slashes /

You can use still use alternate delimiter:

sed '\~//~d' file

Just escape the start of delimeter once.

sed: Replace lines matching a pattern that contains forward slashes?

In your command, you are missing s for substitution and have wrongly escaped \ character. Also as you replied to my comment, that you want to replace it from anywhere in the file, you don't have to use ^ character in your regex. And dot . in regex means any character so they need to be escaped too.

You can use this command,

sed -i 's/OPP\/com\.user\.opp\.orchest\.po\.services\.stub-npo\/npo-stub/OPP\/com.user.opp.orchest.po.services.stub-ica\/npo-ica/g' yourfilename

Delete lines containing pattern at the end of line

There are two approaches here, either print all non-matching lines with

sed -in '/LOCATION=$/!p' file

or delete all matching names with

sed -i '/LOCATION=$/d' file

The first uses the n command line option to suppress the default action of printing the line. We then test for lines that end in LOCATION= and invert the pattern (only keeping those that don't match). When we get a desirable line, we print it with the p option.

The second looks for lines matching the end of line pattern, and deletes those that do.

Your file contains blank lines, and both of these keep those. If we don't want to keep those, we can change the first option to

sed -in '/^$/!{/LOCATION=$/!p}' file

which first checks if a line is not empty, and only bothers checking if it should be printed if it isn't empty. We can modify the second option to

sed -i '/^$/d;/LOCATION=$/d' file

which deletes blank lines and then checks about deleting the other pattern.

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We can modify the options to work with different line ending by specifying the difference in the pattern. The difference between line endings on Unix/Linux (\n) and Windows (\r\n) is the presence of an extra carriage return on Windows. Modifying the four commands above to accept either, we get

1. sed -in '/LOCATION=\r\{0,1\}$/!p' file
2. sed -i '/LOCATION=\r\{0,1\}$/d' file
3. sed -in '/^\r\{0,1\}$/!{/LOCATION=\r\{0,1\}$/!p}' file
4. sed -i '/^\r\{0,1\}$/d;/LOCATION=\r\{0,1\}$/d' file

Note that in each of these we allow an optional \r before the end of line. We use the curly bracket notation, as sed does not support the question mark optional quantifier in normal mode (using the r option to GNU sed for enabling extended regular expressions, we can replace \{0,1\} with ?).

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On a Windows shell, all of the options above require double quotes instead of single quotes.

How can I remove the text between the first and last slash in a string using sed?

sed does not support the various Perl regex extensions you tried to use. But really all you need is

sed 's:/.*/:/:'

Regular expressions perform longest-leftmost matching, so /.*/ by definition matches from the first slash to the last.

"Why doesn't tool X support the regex dialect of tool Y" is a common FAQ; see e.g. Why are there so many different regular expression dialects? but really, these online regex testers shuld be more explicit about which tools they support.

SED: delete lines after pattern match while skipping a few lines, then delete the pattern line

Another approach:

sed '/($(ose),)/,/^endif/{/($(ose),)/d;/^else/,/^endif/d}' file

Output:

...
    dh_installdocs \
        $(archdir)/UserManual*.pdf $(archdir)/VirtualBox*.chm \
        $(addprefix $(archdir)/, LICENSE)
    rm $(addprefix $(archdir)/,UserManual*.pdf VirtualBox*.chm \
        LICENSE)
...

Add option -i to edit "in place".

How to use sed to delete lines which have certain pattern and at some specific line range?

sed "50,${/^\s*PUTS/d}" the_file

this line won't work, because you used double quotes, and you need escape the dollar: \$ or use single quote: '50,${/.../d}' file

sed "50,99999{/^\s*$pattern/d}" file

this line should work.

EDIT

wait, I just noticed that you set env var via set... this is not correct if you were with Bash. you should use export PAT="PUT" in your script.

check @Jonathan and @tripleee 's comments

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