I have a csv file, with student name and marks. I want to update "marks" of a student with name "jack"(the only person in the csv). the data in csv file looks as below.
student,marks
jack,10
peter,20
rick,10
I found this awk '$1 == "Audrey" {print $2}' numbers.txt, but iam not sure on how to modify the file.
awk 'BEGIN{FS=OFS=","} $1=="jack"{$2=27} 1' foo.csv > tmp && mv tmp foo.csv
It worked for me with
sed -ir "s/^\(jack\),.*/\1,$new_grade/"
input.csv. with argument "r" or else i get the "error sed: 1: "input.csv": command i expects \ followed by text".
ed is usually better for in-place editing of files than sed:
printf "%s\n" "/^jack,/c" "jack,${new_grade}" "." w | ed -s input.csv
or using a heredoc to make it easier to read:
ed -s input.csv <<EOF
/^jack,/c
jack,${new_grade}
.
w
EOF
At the first line starting with jack,, change it to jack,XX where XX is the value of the new_grade variable, and write the new contents of the file.
You could use sed:
new_grade=9
sed -i'' "s/^\(jack\),.*/\1,$new_grade/"
The pattern ^\(jack\),.* matches the beginning of the line ^ followed by jack by a comma and the rest of the line .*. The replacement string \1,$new_mark contains the first captured group \1 (in this case jack) followed by a comma and the new mark.
Alternatively you could loop over the file and use a pattern substitution:
new_grade=9
while read -s line; do
echo ${line/jack,*/jack,$new_grade}
done < grades.txt > grades2.txt
Another approach with sed is to anchor the replacement to the digits at the end of the line with:
sed '/^jack,/s/[0-9][0-9]*$/12/' file
This uses the form sed '/find/s/match/replace' where find locates at the beginning of the line '^' the word "jack," eliminating all ambiguity with, e.g. jackson,33. Then the normal substitution form of 's/match/replace/' where match locates at least one digit at the end of the line (anchored by '$') and replaces it with the 12 (or whatever you choose).
Example Use/Output
With your example file in file, you would have:
$ sed '/^jack,/s/[0-9][0-9]*$/12/' file
student,marks
jack,12
peter,20
rick,10
(note: the POSIX character class of [[:digit:]] is equivalent to [0-9] which is another alternative)
The equivalent expression using the POSIX character class would be:
sed '/^jack,/s/[[:digit:]][[:digit:]]*$/12/' file
You can also use Extended Regular Expression which provides the '+' repetition operator to indicate one-or-more compared to the basic repetition designator of '*' to indicate zero-or-more. The equivalent ERE would be sed -E '/^jack,/s/[0-9]+$/12/' file
You can add the -i option to edit in-place and/or using it as -i.bak to create a backup of the original with the .bak extension before modifying the original.
Related
I have a file file.txt and it has the lines below. I want the queuename to be converted to uppercase, like this: queuename=SP00245B
# Queue name
#
queuename=sp00245b
awk '$1 == "queuename" {$2 = toupper($2)}1' FS== OFS== input-file
Note that this will fail if there are 2 = in the line, and only the values between the first 2 = will be uppercased. If that's an issue, it's an easy fix (left as an exercise for the reader).
A simple Perl solution:
perl -i -pe 's/^\s*queuename=\K(.*)/\U$1/' file.txt
(Remove -i if you don't want to modify the file in place.)
With GNU sed:
sed -i 's/\(^[[:blank:]]*queuename=\)\(.*\)/\1\U\2/' file.txt
This uses two captures groups and the \U sequence to toggle uppercase substitution for the second group.
You can also use the sed conversion \U to convert the portions of the matched pattern with the substitution command to uppercase. To covert everything following the '=' sign you could use, e.g.
sed '/^queuename=/s/=.*$/\U&/' filename
To edit the file in-place, include the -i option, e.g.
sed -i '/^queuename=/s/=.*$/\U&/' filename
Example Use/Output
$ echo "queuename=sp00245b" | sed '/^queuename=/s/=.*$/\U&/'
queuename=SP00245B
I have a text file and I would like to only delete the first character of the text file, is there a way to do this in shell script?
I'm new to writing scripts so I really don't know where to start. I understand that the main command most people use is "sed" but I can only find how to use that as a find and replace tool.
All help is appreciated.
You can use the tail command, telling it to start from character 2:
tail -c +2 infile > outfile
You can use sed
sed '1s/^.//' startfile > endfile
1s means match line 1, in substitution mode (s)
^. means at the beginning of the line (^), match any character (.)
There's nothing between the last slashes, which means substitute with nothing (remove)
I used to use cut command to do this.
For example:
cat file|cut -c2-80
Will show characters from column 2 to 80 only.
In your case you can use:
cat file|cut -c2-10000 > newfile
I hope this help you.
[]s
You can also use the 0,addr2 address-range to limit replacements to the first substitution, e.g.
sed '0,/./s/^.//' file
That will remove the 1st character of the file and the sed expression will be at the end of its range -- effectively replacing only the 1st occurrence.
To edit the file in place, use the -i option, e.g.
sed -i '0,/./s/^.//' file
or simply redirect the output to a new file:
sed '0,/./s/^.//' file > newfile
A few other ideas:
awk '{print (NR == 1 ? substr($0,2) : $0)}' file
perl -0777 -pe 's/.//' file
perl -pe 's/.// unless $done; $done = 1' file
ed file <<END
1s/.//
w
q
END
dd allows you to specify an offset at which to start reading:
dd ibs=1 seek=1 if="$input" of="$output"
(where the variables are set to point to your input and output files, respectively)
I am having trouble using sed to substitute values and write to a new file. It writes to a new file, but fails to change any values. Here is my code:
cd/mydirectory
echo "Enter file name:"
read file_input
file1= "$file_input"
file1= "$file1.b"
file2= "$file_input"
file2= "${file2}Ins.b"
sed "/\!cats!/s/\!cats!.*/cats!300!/g $file1>$file2
I simply want to substitute whatever text was after cats with the value 300. Whenever I run this script it doesn't overwrite the previous value with 300. Any suggestions?
Try changing
sed "/\!cats!/s/\!cats!.*/cats!300!/g $file1>$file2
to
sed "s/cats.*/cats300/g" $file1 > $file2
To replace text, you often have to use sed like sed "s/foo/bar/g" file_in > file_out, to change all occurrences of foo with bar in file_in, redirecting the output to file_out.
Edit
I noticed that you are redirecting the output to the same file - you can't do that. You have 2 options:
Redirect the results to another file, with a different filename. e.g.:
sed "s/cats.*/cats300/g" $file1 > $file2.tmp
Note the .tmp after $file2
Use the -i flag (if using GNU sed):
sed -i "s/cats.*/cats300/g" $file1
The i stands for inline replacement.
I think this modified version of your script should work:
echo "Enter file name:"
read file_input
file1="$file_input" # No space after '='
file1="$file1.b" # No space after '='
file2="$file_input" # No space after '='
file2="${file2}Ins.b" # No space after '='
sed 's/!cats!.*/!cats!300!/g' "$file1" > "$file2"
Note the single quotes around sed expression: with them, there's no need to escape the !s in your expression. Note also the double quotes around "$file1" and "$file2": if one of those variables contain spaces, this will prevent your command from breaking.
Some further remarks:
As pointed by jim, you may want to use the GNU sed -i option.
Your regex will currently replace everything after !cats! in matching lines. If they were several occurences of !cats! on your line, only one will remain. If instead you just want to replace the value between two ! delimiters, you may consider use following sed command instead:
sed 's/!cats![^!]*/!cats!300/g'
I am new to this platform. Just had a requirement which I have been working over sometime but not able to find it.
If this pattern was to occur in the middle of a line. How to handle it. Suppose the line is like. aaaa ---- bbbb. If i want to erase the ----bbbb part how to do it. But I want to keep the aaaa part as it is in the file.
Thanks
You can do it easily with sed:
sed -r 's/^--.*//' inputfile > outputfile
Or in place:
sed -r -i.bak 's/^--.*//' inputfile
This will create an inputfile.bak as a backup before modifying the file
Here is a good old bash solution:
while read -r line; do
echo "${line/#--*/}"
done < inputFile > outputFile
One way using awk:
awk '/^--/{$0=" ";}1' file
This will repalce the line with a space when it begins with --
Its not clear from your problem statement what the criteria (limitations) of the solution is.
What you are looking for is something that will support regular expressions. There are a lot of UNIX/Linux tools that can be used to solve this problem.
One simple solution is:
# cat file.txt | sed -e "{s/^--.*/ /}"
The regular expression "^--." will match any line beginning "^" with "--" followed
by any number of characters ".". "s" is the sed substitution command.
so "s/^--.*/ /" means, substitute all lines that start with -- and are followed by any
number of characters with a single space.
All of the lines with comments in a file begin with #. How can I delete all of the lines (and only those lines) which begin with #? Other lines containing #, but not at the beginning of the line should be ignored.
This can be done with a sed one-liner:
sed '/^#/d'
This says, "find all lines that start with # and delete them, leaving everything else."
I'm a little surprised nobody has suggested the most obvious solution:
grep -v '^#' filename
This solves the problem as stated.
But note that a common convention is for everything from a # to the end of a line to be treated as a comment:
sed 's/#.*$//' filename
though that treats, for example, a # character within a string literal as the beginning of a comment (which may or may not be relevant for your case) (and it leaves empty lines).
A line starting with arbitrary whitespace followed by # might also be treated as a comment:
grep -v '^ *#' filename
if whitespace is only spaces, or
grep -v '^[ ]#' filename
where the two spaces are actually a space followed by a literal tab character (type "control-v tab").
For all these commands, omit the filename argument to read from standard input (e.g., as part of a pipe).
The opposite of Raymond's solution:
sed -n '/^#/!p'
"don't print anything, except for lines that DON'T start with #"
you can directly edit your file with
sed -i '/^#/ d'
If you want also delete comment lines that start with some whitespace use
sed -i '/^\s*#/ d'
Usually, you want to keep the first line of your script, if it is a sha-bang, so sed should not delete lines starting with #!. also it should delete lines, that just contain only a hash but no text. put it all together:
sed -i '/^\s*\(#[^!].*\|#$\)/d'
To be conform with all sed variants you need to add a backup extension to the -i option:
sed -i.bak '/^\s*#/ d' $file
rm -Rf $file.bak
You can use the following for an awk solution -
awk '/^#/ {sub(/#.*/,"");getline;}1' inputfile
This answer builds upon the earlier answer by Keith.
egrep -v "^[[:blank:]]*#" should filter out comment lines.
egrep -v "^[[:blank:]]*(#|$)" should filter out both comments and empty lines, as is frequently useful.
For information about [:blank:] and other character classes, refer to https://en.wikipedia.org/wiki/Regular_expression#Character_classes.
If you want to delete from the file starting with a specific word, then do this:
grep -v '^pattern' currentFileName > newFileName && mv newFileName currentFileName
So we have removed all the lines starting with a pattern, writing the content into a new file, and then copy the content back into the source/current file.
You also might want to remove empty lines as well
sed -E '/(^$|^#)/d' inputfile
Delete all empty lines and also all lines starting with a # after any spaces:
sed -E '/^$|^\s*#/d' inputfile
For example, see the following 3 deleted lines (including just line numbers!):
1. # first comment
2.
3. # second comment
After testing the command above, you can use option -i to edit the input file in place.
Just this!
Here is it with a loop for all files with some extension:
ll -ltr *.filename_extension > list.lst
for i in $(cat list.lst | awk '{ print $8 }') # validate if it is the 8 column on ls
do
echo $i
sed -i '/^#/d' $i
done