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Combining two sed commands
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Closed 1 year ago.
I made this script:
xrandr | grep '*' | sed 's/\S*\(*+\)\S*//g'| sed 's/ //g' | sed 's/x.*//'
How can I combine the three sed commands:
sed 's/\S*\(*+\)\S*//g'
sed 's/ //g'
sed 's/x.*//'
into a single command?
With -e:
xrandr | grep '*' | sed -e 's/\S*\(*+\)\S*//g' -e 's/ //g' -e 's/x.*//'
Note that the grep is not necessary:
xrandr | sed -e '/\*/!d' -e 's/\S*\(*+\)\S*//g' -e 's/ //g' -e 's/x.*//'
You could put those commands in a file called sedscr for example, one per line as such:
s/x.*//
s/ //g
s/\S*\(*+\)\S*//g
And then call:
xrandr | grep '*' | sed -f sedscr
I prefer this method, just in case I want to add more commands in the future.
You can simply consider all sed commands as script and add a ; between commands :
xrandr | grep '*' | sed 's/\S*\(*+\)\S*//g ; s/ //g ; s/x.*//'
With newlines :
echo coobas | sed 's:c:f:
s:s:r:'
Without sed but just grep :
$ xrandr | grep -oP '^\s+\K\d+(?=.*?\*)'
1440
or with perl :
$ xrandr | perl -lne 'print $1 if /^\s+(\d+)(?=.*?\*)/'
1440
Another thing to consider is to make a sed config file (call it config.sed) listing all replacement rules. E.g.:
1,/^END/{
s/x.*//
s/ //g
s/\S*\(*+\)\S*//g
}
and then run
sed -f config.sed filein.txt > fileout.txt
Related
sed -i 's/1.1.1.1/ `hostname -I | cut -f1 -d " "`/g' file.txt
Not able to overwrite IP address using sed command in a given file. How to run this (hostname -I | cut -f1 -d " ") command with sed command?
This might work for you (GNU sed):
sed -i '/1\.1\.1\.1/{s//$(hostname -I | cut -f1 -d " ")/;s/.*/echo "&"/e}' file
Place the commands between $(...) and then echo the line using the evaluate flag of the substitution command.
Just use a variable.
hn=$(hostname -I | cut -f1 -d" ") && sed -i "s/1\.1\.1\.1/$hn/g" file
Sed's e and s///e work on the entire pattern space, so it's just more trouble than it's worth.
I have a log file that is line separated by \n with multi-line logs (ones containing SQL) being separated by \r\n. In order to pull the SQL statements out of the file I need to convert the \r\n to \s. Not knowing sed I googled and found a good solution that works very well but it fails when I switch to tail -f.
eg. these work:
tail -1000 /mylogfile.log | sed -e ':a;N;$!ba;s/\r\n/ /g' | grep "Executing command"
cat /mylogfile.log | sed -e ':a;N;$!ba;s/\r\n/ /g' | grep "Executing command"
but this returns no data at all
tail -f /mylogfile.log | sed -e ':a;N;$!ba;s/\r\n/ /g' | grep "Executing command"
EDIT: For the person who added a "This question already has an answer here:", no that does not answer the question at all. First, that other question wasn't even resolved for the person who asked it. Second, it talks only about grep, the problem is with sed. I can have 10 greps and it still works fine if I switch from sed to perl. eg
tail -f /mylog.log | perl -ne 's/\r\n/ /g; print;' | grep "Executing command" | grep -vi ANALYZE | grep -vi DESCRIBE | grep -vi "SHOW PARTITION"
I have a script which searches source files that contain a "TODO"
note inside the comments. Furthermore I use a concatenation of grep, git blame, uniq and sort to get
the list ordered by the person who wrote the TODO comment.
The following works fine in bash and zsh:
#!/bin/bash
for FILE in $(grep -r -i "todo" apps/business | awk '{print $1}' | sed 's/://' | sed 's/\#//')
do
git blame $FILE | grep -i "todo"
done | sort -k2 | uniq
Now I want to count all the entries. Instead of calling the (time expensive)
grep/git blame again, I want to save everything into $MATCHES to count it
without evaluating it again.
MATCHES=$(for FILE in $(grep -r -i "todo" apps/business | awk '{print $1}' | sed 's/://' | sed 's/\#//')
do
git blame $FILE | grep -i "todo"
done | sort -k2 | uniq)
echo $MATCHES
That's where I experience different behaviour in bash/zsh:
zsh: Returns the same as the first script (as expected)
bash: Ignores the newlines of git blame, puts everything on one line. wc -l counts 1 line.
What am I missing here? Why is bash behaving differently here?
And how do I get bash to not-ignore the newline?
zsh doesn't perform word-splitting on the unquoted parameter expansion $MATCH by default. Use echo "$MATCHES" | wc -l, and bash should work as well.
Note this is the wrong way to iterate over the output of a command; use a while loop and the read command instead.
grep -ri "todo" apps/business | awk '{print $1}' | sed -e 's/://' -e 's/\#//' |
while IFS= read -r FILE; do
git blame "$FILE" | grep -i todo
done | sort -k2 | uniq
I have some html files and want to extract only lines with containing these tags:
head
p
I used sed to extract these parts of the files, as follows:
grep "<head>" myfile.html | sed -e 's%\(head\)\(.*\)\(/head\)%title\2\/title%'
grep "<p>" myfile.html | sed -e 's%\(<p>\)\(.*\)\(</p\)\(>\)%\2\\%'
Everything is Ok, but I get "\" character at the end of each line. How I can overcome this problem?
In this command, you're telling it to add a backslash by including the double backslash:
sed -e 's%\(<p>\)\(.*\)\(</p\)\(>\)%\2\\%'
Try removing the backslashes:
sed -e 's%\(<p>\)\(.*\)\(</p\)\(>\)%\2%'
Also, you don't need grep:
sed -ne '/<p>/{s%\(<p>\)\(.*\)\(</p\)\(>\)%\2%;p}'
Don't use \ at the end of the replacement string:
grep "<p>" myfile.html | sed -e 's%\(<p>\)\(.*\)\(</p\)\(>\)%\2%'
I have a product which has a command called db2level whose output is given below
I need to extract 8.1.1.64 out of it, so far i came up with,
db2level | grep "DB2 v" | awk '{print$5}'
which gave me an output v8.1.1.64",
Please help me to fetch 8.1.1.64. Thanks
grep is enough to do that:
db2level| grep -oP '(?<="DB2 v)[\d.]+(?=", )'
Just with awk:
db2level | awk -F '"' '$2 ~ /^DB2 v/ {print substr($2,6)}'
db2level | grep "DB2 v" | awk '{print$5}' | sed 's/[^0-9\.]//g'
remove all but numbers and dot
sed is your friend for general extraction tasks:
db2level | sed -n -e 's/.*tokens are "DB2 v\([0-9.]*\)".*/\1/p'
The sed line does print no lines (the -n) but those where a replacement with the given regexp can happen. The .* at the beginning and the end of the line ensure that the whole line is matched.
Try grep with -o option:
db2level | grep -E -o "[0-9]+\.[0-9]+\.[0-9]\+[0-9]+"
Another sed solution
db2level | sed -n -e '/v[0-9]/{s/.*DB2 v//;s/".*//;p}'
This one desn't rely on the number being in a particular format, just in a particular place in the output.
db2level | grep -o "v[0-9.]*" | tr -d v
Try s.th. like db2level | grep "DB2 v" | cut -d'"' -f2 | cut -d'v' -f2
cut splits the input in parts, seperated by delimiter -d and outputs field number -f