Im trying to print only lines that do not start with a letter from the file "main"
Ive tried sed -n '/^[a-z]/ /!w' main
and it gives me "w': event not found"
With sed as requested:
sed '/^[[:alpha:]]/d' main
or
sed -n '/^[^[:alpha:]]/p' main
or
sed -n '/^[[:alpha:]]/!p' main
Note: you could use [a-z] inplace of [[:alpha:]] but I prefer the latter because it is safe to use across different locales
there are many other ways to print lines
sed -n '/^[^a-zA-Z]/p' main
sed -n '/^[^a-z]/Ip' main
awk 'BEGIN{IGNORECASE=1}!/^[a-z]/' main
grep -vi "^[a-z]" main
ruby -ne 'print unless /^[a-z]/i' main
shell
while read -r line
do
case "$line" in
[^a-zA-Z]*) echo $line;;
esac
done < main
grep -v '^[a-z]' main
will do it.
Related
Hoping someone kind can help me pls!
I have an file input.list:
/scratch/user/IFS/IFS001/IFS003.GATK.recal.bam
/scratch/user/IFS/IFS002/IFS002.GATK.recal.bam
/scratch/user/EGS/ZFXHG22/ZFXHG22.GATK.recal.bam
and I want to extract the bit before .GATK.recal.bam - I have found a solution for this:
sed 's/\.GATK\.recal\.bam.*//' input.list | sed 's#.*/##'
I now want to incorporate this into a while loop but it's not working... please can someone take a look and guide me where I'm doing wrong. My attempt is below:
while read -r line; do ID=${sed 's/\.GATK\.recal\.bam.*//' $line | sed 's#.*/##'}; sbatch script.sh $ID; done < input.list
Apologies for the easy Q...
You can use the output of the sed command as input for the loop:
sed 'COMMAND' input.file | while read -r id ; do
some_command "${id}"
done
Instead of the loop, also xargs could be used:
sed 'COMMAND' input.file | xargs -n1 some_command
ps: GNU sed supports to execute the result of a s operation as a command. I wouldn't recommend to use this in production, for portability reasons at least, but it's worth mention probably:
sed 's/\(.*\)\.GATK\.recal\.bam.*/sbatch script.sh \1/e' input.file
You can do this in straight up bash (If you're using that shell; ksh93 and zsh will be very similar) no sed required:
while read -r line; do
id="${line##*/}" # Remove everything up to the last / in the string
id="${id%.GATK.recal.bam}" # Remove the trailing suffix
sbatch script.sh "$id"
done < input.list
At the very least you can use a single sed call per line:
id=$(sed -e 's/\.GATK\.recal\.bam$//' -e 's#.*/##' <<<"$line")
or with plain sh
id=$(printf "%s\n" "$line" | sed -e 's/\.GATK\.recal\.bam$//' -e 's#.*/##')
I need to modify an xml file using Sed to replace the line
url="jdbc:oracle:thin:#//ttpdbscan.axel.net:1521/axel.telco.net"
with
url="jdbc:oracle:thin:#//ttpdbscan.axeltelecom.net:1598/axelPRD.telco.net"
I have stored the lines like this
ACTUAL_DB=$(sed -n 's#^.*url="\(.*\).*"#\1#p' $FILE.xml)
and
NEW_DB="jdbc:oracle:thin:#//ttpdbscan.axeltelecom.net:1598/axelPRD.telco.net"
And the replacing method is this one
sed -i "s#$ACTUAL_DB#$NEW_DB#g" $File.xml
The problem is that when I run the script the file stays the same.
I have echoed the variables and all of them return the correct values.
Assuming the file you have is File.xml (if it is not a variable), you may use
sed -i "s#${ACTUAL_DB}#${NEW_DB}#g" File.xml
Try also with other delimiters:
sed -i "s~${ACTUAL_DB}~${NEW_DB}~g" File.xml
If your sed does not support -i use
sed "s~${ACTUAL_DB}~${NEW_DB}~g" File.xml 1<> File.xml
See sed edit file in place
So I saved the output into another file and found out that the string had an extra space
so it looked like this
ACTUAL_DB= "jdbc:oracle:thin:#//ttpdbscan.axel.net:1521/axel.telco.net "
I removed the extra space with
"$(echo -e "${ACTUAL_DB}" | tr -d '[:space:]')"
And now the sed is working as intended
I want to do the following substitution in a text file:
the original string: "---a---"
after substitution : "---\a---"
and I run the following command:
sed -r -e "s/-(a)-/-\\\1-/g" test.txt
but it doesn't give the right result. What command args should I use?
Remember that backslashes are significant in Bash's double-quoted strings as well as in sed itself. Either use single quotes:
sed -r -e 's/-(a)-/-\\\1-/g' test.txt
Or escape the backslashes again:
sed -r -e "s/-(a)-/-\\\\\\1-/g" test.txt
If you echo the strings, you'll see what's happening:
$ echo "s/-(a)-/-\\\1-/g"
s/-(a)-/-\\1-/g
$ echo 's/-(a)-/-\\\1-/g'
s/-(a)-/-\\\1-/g
$ echo "s/-(a)-/-\\\\\\1-/g"
s/-(a)-/-\\\1-/g
The first one (your original) just looks like a literal backslash followed by a literal 1 to sed.
Try sed -r -e "s/-(a)-/-\\\\\\1-/g" or sed -r -e 's/-(a)-/-\\\1-/g'
The problem is that \ is captured by bash if you use double quotes.
With " you will have to do something like:
[jaypal:~/Temp] echo "---a---" | sed -r "s/-(a)-/-\\\\\1-/g"
---\a---
You have to replace 1 with a
sed -r -e "s/-(a)-/-\\\a-/g" test.txt
so many answers ......
Kaizen ~/so_test
$ echo "---a---" | sed -n 's/a/\\a/p'
---\a---
since you have a text file the following should work :
sed -i 's/a/\\a/g' filename.txt ;
does this help ?
I am having a log file a.log and i need to extract a piece of information from it.
To locate the start and end line numbers of the pattern i am using the following.
start=$(sed -n '/1112/=' file9 | head -1)
end=$(sed -n '/true/=' file9 | head -1)
i need to use the variables (start,end) in the following command:
sed -n '16q;12,15p' orig-data-file > new-file
so that the above command appears something like:
sed -n '($end+1)q;$start,$end'p orig-data-file > new-file
I am unable to replace the line numbers with the variables. Please suggest the correct syntax.
Thanks,
Rosy
When I realized how to do it, I was looking for anyway to get line number into a file containing the requested info, and display the file from that line to EOF.
So, this was my way.
with
PATTERN="pattern"
INPUT_FILE="file1"
OUTPUT_FILE="file2"
line number of first match of $PATTERN into $INPUT_FILE can be retrieved with
LINE=`grep -n ${PATTERN} ${INPUT_FILE} | awk -F':' '{ print $1 }' | head -n 1`
and the outfile will be the text from that $LINE to EOF. This way:
sed -n ${LINE},\$p ${INPUT_FILE} > ${OUTPUT_FILE}
The point here, is the way how can variables be used with command sed -n:
first witout using variables
sed -n 'N,$p' <file name>
using variables
LINE=<N>; sed -n ${LINE},\$p <file name>
Remove the single quotes thus. Single quotes turn off the shell parsing of the string. You need shell parsing to do the variable string replacements.
sed -n '('$end'+1)q;'$start','$end''p orig-data-file > new-file
i need to redirect all of the stdout of a program except the first line into a file.
Is there a common unix program that removes lines from stdin and spits the rest out to stdout?
Others have already mentioned "tail". sed will also work:
sed 1d
As will Awk:
awk 'NR > 1'
tail -n +2 -f -
sed -e 1d < input > output