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]+$/12/' file
You can add the -i option to edit in-place and/or using it as -i.bak to create a backup of the original with the .bak extension before modifying the original.
I have a file pattern.txt which is composed of one very long line of complicated code (~8200 chars).
This code can be found in multiple files inside multiple directories.
I can easily identify a list of these files using
grep -rli 'uniquepartofthecode' *
My concern is how do I replace it with the exact text from within the file ?
I tried to do:
var=$(cat pattern.txt)
sed -i "s/$var//g" targetfile.txt
but I got the following error :
sed: -e expression #1, char 96: unknown option to `s'
sed is interpreting my $var content as a regular expression, I would like it to just match the exact text.
The pattern.txt content could be more or less any combination of characters so I'm afraid I cannot escape every characters efficiently.
Is there a solution using sed ? Or should I use another tool for that ?
EDIT:
I tried using this solution to make a proper regex pattern from my text file.
Is it possible to escape regex metacharacters reliably with sed
the overall process is:
var=$(cat pattern.txt)
searchEscaped=$(sed 's/[^^]/[&]/g; s/\^/\\^/g' <<<"$var")
sed -n "s/$searchEscaped/foo/p" <<<"$var" # if ok, echoes 'foo'
This last command displays "foo". $searchEscaped seems to be properly escaped.
Though, this is not returning anything (it should display foo + the rest of the file without the matched part):
sed -n "s/$searchEscaped/foo/p" targetfile.txt
I think that the best solution is to not use regular expressions at all and resort to string replacement.
One way to do this is using perl:
$ echo "$string_to_replace"
some other stuff abc$^%!# some more
$ echo "$search"
abc$^%!#
$ perl -spe '$len = length $search;
while (($pos = index($_, $search, $n)) > -1) {
substr($_, $pos, $len) = "replacement";
$n = $pos + $len;
}' <<<"$string_to_replace" -- -search="$search"
some other stuff replacement some more
The -p switch tells perl to loop through each line of the variable $string_to_replace (which could easily be replaced by a file). -s allows options to be passed to the script - in this case, I've passed a shell variable containing the search string.
For each line of the file, the while loop runs through all of the matches of the search string. substr is used on the left hand of the assignment to replace a substring of $_, which refers to the current line being processed.
I have a tsv.-file and there are some lines which do not end with an '"'. So now I would like to remove every line break which is not directly after an '"'.
How could I accomplish that with sed? Or any other bash shell program...
Kind regards,
Snafu
This sed command should do it:
sed '/"$/!{N;s/\n//}' file
It says: on every line not matching "$ do:
read next line, append it to pattern space;
remove linebreak between the two lines.
Example:
$ cat file.txt
"test"
"qwe
rty"
foo
$ sed '/"$/!{N;s/\n//}' file.txt
"test"
"qwerty"
foo
To elaborate on #Lev's answer, the BSD (OSX) version of sed is less forgiving about the command syntax within the curly braces -- the semicolon command separator is required for both commands:
sed '/"$/!{N;s/\n//;}' file.txt
per the documentation here -- an excerpt:
Following an address or address range, sed accepts curly braces '{...}' so several commands may be applied to that line or to the lines matched by the address range. On the command line, semicolons ';' separate each instruction and must precede the closing brace.
give this awk one-liner a try:
awk '{printf "%s%s",$0,(/"$/?"\n":"")}' file
test
kent$ cat f
"foo"
"bar"
"a long
text with
many many
lines"
"lalala"
kent$ awk '{printf "%s%s",$0,(/"$/?"\n":"")}' f
"foo"
"bar"
"a longtext withmany manylines"
"lalala"
This might work for you (GNU sed):
sed ':a;/"$/!{N;s/\n//;ta}' file
This checks if the last character of the pattern space is a " and if not appends another line, removes a newline and repeats until the condition is met or the end-of-file is encountered.
An alternative is:
sed -r ':a;N;s/([^"])\n/\1/;ta;P;D' file
The mechanism is left for the reader to ponder.
I have pattern like below
hi
hello
hallo
greetings
salutations
no more hello for you
I am trying to replace all newlines with tab using the following command
sed -e "s_/\n_/\t_g"
but it's not working.
Could anybody please help? I'm looking for a solution in sed/awk.
tr is better here, I think:
tr "\n" "\t" < newlines
As Nifle suggested in a comment, newlines here is the name of the file holding the original text.
Because sed is so line-oriented, it's more complicated to use in a case like this.
not sure about output you want
# awk -vRS="\n" -vORS="\t" '1' file
hi hello hallo greetings salutations no more hello for you
sed '$!{:a;N;s/\n/\t/;ta}' file
You can't replace newlines on a line-by-line basis with sed. You have to accumulate lines and replace the newlines between them.
text abc\n <- can't replace this one
text abc\ntext def\n <- you can replace the one after "abc" but not the one at the end
This sed script accumulates all the lines and eliminates all the newlines but the last:
sed -n '1{x;d};${H;x;s/\n/\t/g;p};{H}'
By the way, your sed script sed -e "s_/\n_/\t_g" is trying to say "replace all slashes followed by newlines with slashes followed by tabs". The underscores are taking on the role of delimiters for the s command so that slashes can be more easily used as characters for searching and replacing.
paste -s
-s Concatenate all of the lines of each separate input file in
command line order. The newline character of every line
except the last line in each input file is replaced with the
tab character, unless otherwise specified by the -d option.
You are almost there with your sed script, you'd just need to change it to:
sed -e "s/\n/\t/g"
The \ is enough for escape, you don't need to add _
And you need to add the / before g at the end to let sed know that this is the last part of the script.
How to insert a newline before a pattern within a line?
For example, this will insert a newline behind the regex pattern.
sed 's/regex/&\n/g'
How can I do the same but in front of the pattern?
Given this sample input file, the pattern to match on is the phone number.
some text (012)345-6789
Should become
some text
(012)345-6789
This works in bash and zsh, tested on Linux and OS X:
sed 's/regexp/\'$'\n/g'
In general, for $ followed by a string literal in single quotes bash performs C-style backslash substitution, e.g. $'\t' is translated to a literal tab. Plus, sed wants your newline literal to be escaped with a backslash, hence the \ before $. And finally, the dollar sign itself shouldn't be quoted so that it's interpreted by the shell, therefore we close the quote before the $ and then open it again.
Edit: As suggested in the comments by #mklement0, this works as well:
sed $'s/regexp/\\\n/g'
What happens here is: the entire sed command is now a C-style string, which means the backslash that sed requires to be placed before the new line literal should now be escaped with another backslash. Though more readable, in this case you won't be able to do shell string substitutions (without making it ugly again.)
Some of the other answers didn't work for my version of sed.
Switching the position of & and \n did work.
sed 's/regexp/\n&/g'
Edit: This doesn't seem to work on OS X, unless you install gnu-sed.
In sed, you can't add newlines in the output stream easily. You need to use a continuation line, which is awkward, but it works:
$ sed 's/regexp/\
&/'
Example:
$ echo foo | sed 's/.*/\
&/'
foo
See here for details. If you want something slightly less awkward you could try using perl -pe with match groups instead of sed:
$ echo foo | perl -pe 's/(.*)/\n$1/'
foo
$1 refers to the first matched group in the regular expression, where groups are in parentheses.
On my mac, the following inserts a single 'n' instead of newline:
sed 's/regexp/\n&/g'
This replaces with newline:
sed "s/regexp/\\`echo -e '\n\r'`/g"
echo one,two,three | sed 's/,/\
/g'
You can use perl one-liners much like you do with sed, with the advantage of full perl regular expression support (which is much more powerful than what you get with sed). There is also very little variation across *nix platforms - perl is generally perl. So you can stop worrying about how to make your particular system's version of sed do what you want.
In this case, you can do
perl -pe 's/(regex)/\n$1/'
-pe puts perl into a "execute and print" loop, much like sed's normal mode of operation.
' quotes everything else so the shell won't interfere
() surrounding the regex is a grouping operator. $1 on the right side of the substitution prints out whatever was matched inside these parens.
Finally, \n is a newline.
Regardless of whether you are using parentheses as a grouping operator, you have to escape any parentheses you are trying to match. So a regex to match the pattern you list above would be something like
\(\d\d\d\)\d\d\d-\d\d\d\d
\( or \) matches a literal paren, and \d matches a digit.
Better:
\(\d{3}\)\d{3}-\d{4}
I imagine you can figure out what the numbers in braces are doing.
Additionally, you can use delimiters other than / for your regex. So if you need to match / you won't need to escape it. Either of the below is equivalent to the regex at the beginning of my answer. In theory you can substitute any character for the standard /'s.
perl -pe 's#(regex)#\n$1#'
perl -pe 's{(regex)}{\n$1}'
A couple final thoughts.
using -ne instead of -pe acts similarly, but doesn't automatically print at the end. It can be handy if you want to print on your own. E.g., here's a grep-alike (m/foobar/ is a regex match):
perl -ne 'if (m/foobar/) {print}'
If you are finding dealing with newlines troublesome, and you want it to be magically handled for you, add -l. Not useful for the OP, who was working with newlines, though.
Bonus tip - if you have the pcre package installed, it comes with pcregrep, which uses full perl-compatible regexes.
In this case, I do not use sed. I use tr.
cat Somefile |tr ',' '\012'
This takes the comma and replaces it with the carriage return.
To insert a newline to output stream on Linux, I used:
sed -i "s/def/abc\\\ndef/" file1
Where file1 was:
def
Before the sed in-place replacement, and:
abc
def
After the sed in-place replacement. Please note the use of \\\n. If the patterns have a " inside it, escape using \".
Hmm, just escaped newlines seem to work in more recent versions of sed (I have GNU sed 4.2.1),
dev:~/pg/services/places> echo 'foobar' | sed -r 's/(bar)/\n\1/;'
foo
bar
echo pattern | sed -E -e $'s/^(pattern)/\\\n\\1/'
worked fine on El Captitan with () support
In my case the below method works.
sed -i 's/playstation/PS4/' input.txt
Can be written as:
sed -i 's/playstation/PS4\nplaystation/' input.txt
PS4
playstation
Consider using \\n while using it in a string literal.
sed : is stream editor
-i : Allows to edit the source file
+: Is delimiter.
I hope the above information works for you 😃.
in sed you can reference groups in your pattern with "\1", "\2", ....
so if the pattern you're looking for is "PATTERN", and you want to insert "BEFORE" in front of it, you can use, sans escaping
sed 's/(PATTERN)/BEFORE\1/g'
i.e.
sed 's/\(PATTERN\)/BEFORE\1/g'
You can also do this with awk, using -v to provide the pattern:
awk -v patt="pattern" '$0 ~ patt {gsub(patt, "\n"patt)}1' file
This checks if a line contains a given pattern. If so, it appends a new line to the beginning of it.
See a basic example:
$ cat file
hello
this is some pattern and we are going ahead
bye!
$ awk -v patt="pattern" '$0 ~ patt {gsub(patt, "\n"patt)}1' file
hello
this is some
pattern and we are going ahead
bye!
Note it will affect to all patterns in a line:
$ cat file
this pattern is some pattern and we are going ahead
$ awk -v patt="pattern" '$0 ~ patt {gsub(patt, "\n"patt)}1' d
this
pattern is some
pattern and we are going ahead
sed -e 's/regexp/\0\n/g'
\0 is the null, so your expression is replaced with null (nothing) and then...
\n is the new line
On some flavors of Unix doesn't work, but I think it's the solution to your problem.
echo "Hello" | sed -e 's/Hello/\0\ntmow/g'
Hello
tmow
This works in MAC for me
sed -i.bak -e 's/regex/xregex/g' input.txt sed -i.bak -e 's/qregex/\'$'\nregex/g' input.txt
Dono whether its perfect one...
After reading all the answers to this question, it still took me many attempts to get the correct syntax to the following example script:
#!/bin/bash
# script: add_domain
# using fixed values instead of command line parameters $1, $2
# to show typical variable values in this example
ipaddr="127.0.0.1"
domain="example.com"
# no need to escape $ipaddr and $domain values if we use separate quotes.
sudo sed -i '$a \\n'"$ipaddr www.$domain $domain" /etc/hosts
The script appends a newline \n followed by another line of text to the end of a file using a single sed command.
In vi on Red Hat, I was able to insert carriage returns using just the \r character. I believe this internally executes 'ex' instead of 'sed', but it's similar, and vi can be another way to do bulk edits such as code patches. For example. I am surrounding a search term with an if statement that insists on carriage returns after the braces:
:.,$s/\(my_function(.*)\)/if(!skip_option){\r\t\1\r\t}/
Note that I also had it insert some tabs to make things align better.
Just to add to the list of many ways to do this, here is a simple python alternative. You could of course use re.sub() if a regex were needed.
python -c 'print(open("./myfile.txt", "r").read().replace("String to match", "String to match\n"))' > myfile_lines.txt
sed 's/regexp/\'$'\n/g'
works as justified and detailed by mojuba in his answer .
However, this did not work:
sed 's/regexp/\\\n/g'
It added a new line, but at the end of the original line, a \n was added.