Can anybody help me please?
grep " 287 " file.txt | grep "HI" | sed -i 's/HIS/HID/g'
sed: no input files
Tried also xargs
grep " 287 " file.txt | grep HI | xargs sed -i 's/HIS/HID/g'
sed: invalid option -- '6'
This works fine
grep " 287 " file.txt | grep HI
If you want to keep your pipeline:
f=file.txt
tmp=$(mktemp)
grep " 287 " "$f" | grep "HI" | sed 's/HIS/HID/g' > "$tmp" && mv "$tmp" "$f"
Or, simplify:
sed -i -n '/ 287 / {/HI/ s/HIS/HID/p}' file.txt
That will filter out any line that does not contain " 287 " and "HI" -- is that what you want? I suspect you really want this:
sed -i '/ 287 / {/HI/ s/HIS/HID/}' file.txt
For lines that match / 287 /, execute the commands in braces. In there, for lines that match /HI/, search for the first "HIS" and replace with "HID". sed implicitly prints all lines if -n is not specified.
Other commands that do the same thing:
awk '/ 287 / && /HI/ {sub(/HIS/, "HID")} {print}' file.txt > new.txt
perl -i -pe '/ 287 / and /HI/ and s/HIS/HID/' file.txt
awk does not have an "in-place" option (except gawk -i inplace for recent gawk versions)
Related
I'm using this code to get all titles from urls with http://something.txt:
#!/usr/bin/perl -w
$output = `cat source.html | grep -o '<a .*href=.*>' | grep -E 'txt' | sed -e 's/<a /\n<a /g' | sed -e 's/<a .*title="//' | cut -f1 -d '"'`;
print("$output");
When i run this on perl i get the error:
sed: -e expression #1, char 6: unterminated `s' command
The error is related with this portion of code:
sed -e 's/<a /\n<a /g'
In backquotes, Perl uses the same rules as in double quotes. Therefore, \n corresponds to a newline; you have to backslash the backslash to pass literal \ to the shell:
`sed -e 's/<a /\\n<a /g'`
I have a example cut down from a log file.
112 172.172.172.1#50912 (ssl.bing.com):
I would like some how to remove the # and numbers after and (): from the url.
Would like the result.
112 172.172.172.1 ssl.bing.com
Here is the sed oneliner I have been working on.
cat newdns.log | sed -e 's/.*query: //' | cut -f 1 -d' ' | sort | uniq -c | sort -k2 > old.log
Thanks
Using sed, you could say:
sed 's/#[0-9]*//;s/(\(.*\)):$/\1/' filename
or, in a single substitution:
sed 's/#[0-9]* *(\(.*\)):$/ \1/' filename
Another sed:
sed -r 's/#[^ ]+|[():]//g'
$ echo '112 172.172.172.1#50912 (ssl.bing.com):' | sed -r 's/#[^ ]+|[():]//g'
112 172.172.172.1 ssl.bing.com
How do I extract 68 from v1+r0.68?
Using awk, returns everything after the last '.'
echo "v1+r0.68" | awk -F. '{print $NF}'
Using sed to get the number after the last dot:
echo 'v1+r0.68' | sed 's/.*[.]\([0-9][0-9]*\)$/\1/'
grep is good at extracting things:
kent$ echo " v1+r0.68"|grep -oE "[0-9]+$"
68
Match the digit string before the end of the line using grep:
$ echo 'v1+r0.68' | grep -Eo '[0-9]+$'
68
Or match any digits after a .
$ echo 'v1+r0.68' | grep -Po '(?<=\.)\d+'
68
Print everything after the . with awk:
echo "v1+r0.68" | awk -F. '{print $NF}'
68
Substitute everything before the . with sed:
echo "v1+r0.68" | sed 's/.*\.//'
68
type man grep
and you will see
...
-o, --only-matching
Show only the part of a matching line that matches PATTERN.
then type echo 'v1+r0.68' | grep -o '68'
if you want it any where special do:
echo 'v1+r0.68' | grep -o '68' > anyWhereSpecial.file_ending
Before:
eng-vshakya:scripts vshakya$ ls
American Samoa.png Faroe Islands.png Saint Barthelemy.png
After:
eng-vshakya:scripts vshakya$ ls
AmericanSamoa.png FaroeIslands.png SaintBarthelemy.png
Tried below prototype, but it does not work :( Sorry, not very good when it comes to awk/sed :(
ls *.png | sed 's/\ /\\\ /g' | awk '{print("mv "$1" "$1)}'
[ Above is prototype, real command, I guess, would be:
ls *.png | sed 's/\ /\\\ /g' | awk '{print("mv "$1" "$1)}' | sed 's/\ //g'
]
No need to use awk or sed when you can do this in pure bash.
[ghoti#pc ~/tmp1]$ ls -l
total 2
-rw-r--r-- 1 ghoti wheel 0 Aug 1 01:19 American Samoa.png
-rw-r--r-- 1 ghoti wheel 0 Aug 1 01:19 Faroe Islands.png
-rw-r--r-- 1 ghoti wheel 0 Aug 1 01:19 Saint Barthelemy.png
[ghoti#pc ~/tmp1]$ for name in *\ *; do mv -v "$name" "${name// /}"; done
American Samoa.png -> AmericanSamoa.png
Faroe Islands.png -> FaroeIslands.png
Saint Barthelemy.png -> SaintBarthelemy.png
[ghoti#pc ~/tmp1]$
Note that the ${foo/ /} notation is bash, and does not work in classic Bourne shell.
ghoti's solution is the right thing to do. Since you ask how to do it in sed, here's one way:
for file in *; do newfile=$( echo "$file" | tr -d \\n | sed 's/ //g' );
test "$file" != "$newfile" && mv "$file" "$newfile"; done
The tr is there to remove newlines in the filename, and is necessary to ensure that sed sees the entire filename in one line.
I'm very new to some of the command line utilities and have been looking for a while for a command that would accomplish my goal.
The goal is to find files that contain a string of text, replace it with a new string, and then write the results to a file that is named the same as the original, but in a different directory.
Obviously this is not working, so I am asking how you who know about this stuff would go about it.
grep -rl 'stringToFind' *.* | sed 's|oldString|newString|g' < fileNameFromGrep > ./new/fileNameFromGrep
Thanks for your input!
John
for f in "`find /YOUR/SEARCH/DIR/ROOT -type f -exec fgrep -l 'stirngToFind' \{\} \;`" ; do
sed 's|oldString|newString|g' < "${f} > ./new/"${f}
done
Will do it for you.
If you have spaces in filenames:
OLDIFS=$IFS
IFS=''
find /PATH -print0 -type f | while read -r -d $'' file
do
fgrep -l 'stirngToFind' "$file" && \
sed 's|oldString|newString|g' < "${file} > ./new/"${file}
done
IFS=$OLDIFS
#!/bin/bash
for file in *; do
if grep -qF 'stringToFind' "$file"; then
sed 's/oldString/newString/g' "$file" > "./new/$file"
fi
done
for file in path/to/dir/*
do
grep -q 'pattern' "$file" > /dev/null
if [ $? == 0 ]; then
sed 's/oldString/newString/g' "$file" > /path/to/newdir/"$file"
fi
done
You try:
sed -ie "s/oldString/newString/g" \
$(grep -Rsi 'pattern' path/to/dir/ | cut -d: -f1)
sed:
i in_place
e exec other command or script
grep:
R recursive
s Suppress error messages
i ignore case sensitive