If I want to add content to the 10th line of a file, how do I do it?
This is what I came up with:
sed -ie '' "10s/^/new_content/g" file.txt
I keep getting the message "no such file or directory"
Also, if I want to replace 10 with N+1 and the new_content with a variable $VAR, would the format still be the same?
VAR= $(cat fileA.txt)
sed -ie '' "`expr $N +1`s/^/$VAR/g" fileB.txt
Thanks for your help!!!
Shell is fussy about spacing:
VAR= $(cat fileA.txt)
You do not want the space after the =. Actually, you don't want the assignment either, unless fileA.txt has only one line. You'd have to escape newlines with backslashes, etc, to get it to work, which is both hard and unnecessary.
Unless you're using an antique (non-POSIX) shell, use built-in arithmetic:
sed -i.bak -e "$(($N + 1))r fileA.txt" fileB.txt
The $(($N + 1)) has the shell evaluate the arithmetic expression. The r has sed read the file that is named.
[1addr]r file
Copy the contents of file to the standard output immediately before the next attempt to read a line of input. If file cannot be read for any reason, it is silently ignored and no error condition is set.
It turns out I needed to split -i and -e.
The final result that works was:
sed -i '' -e "10s/^/CONTENT/g" file.txt
Or to increase Ns by +1 each time it loops:
sed -i '' -e "$((N + 1))s/^/CONTENT/g" ~/dance/heatlist/prep.txt
I still have not figured out how to make CONTENT a variable $CONTENT
Related
I want to use grep to count the number of W occurrences in the file.txt, save it as the variable WATER_NUMBER. Then append that number to the end of file.txt.
Following here,
I tried
#!/bin/bash -l
WATER_NUMBER="$(grep -c W file.txt)"
sed -i -e '$a\"${WATER_NUMBER}"' file.txt
but I got "${WATER_NUMBER}" printed out, instead of the number. Can I ask how to modify it?
The command
sed -i '$a\"${WATER_NUMBER}"' file.txt
will simply add the line "${WATER_NUMBER}" at the end of the file. You could try
sed -i "$ a\$WATER_NUMBER" file.txt
but this will still add the line $WATER_NUMBER. The problem is that the variable WATER_NUMBER is not expanded in the sed script. In order to pass its value to sed, place it outside the quoting, like this
sed -i '$ a\'$WATER_NUMBER file.txt
Edit: I actually wrote my answer yesterday without really thinking about the reason as to why the variable is not expanded. This morning I wondered why this is the case even though the variable is in double quotes as opposed to single quotes. The reason is actually just the coincidence that the \ from the append command is in front of the $ from the variable, thus escaping it. To prevent this, you need to escape the \. On the other hand, a backslash is actually not needed to separate the a from the line you want to add, hence
sed -i "$ a $WATER_NUMBER" file.txt
will do the job.
How to save a number to a variable?
WATER_NUMBER=42
How to append the variable content to a file?
echo $WATER_NUMBER >> file.txt
I have a bash script which checks for a string pattern in file and delete entire line i same file but somehow its not deleting the line and no throwing any error .same command from command prompt deletes from file .
#array has patterns
for k in "${patternarr[#]}
do
sed -i '/$k/d' file.txt
done
sed version is >4
when this loop completes i want all lines matching string pattern in array to be deleted from file.txt
when i run sed -i '/pataern/d file.txt from command prompt then it works fine but not inside bash
Thanks in advance
Here:
sed -i '/$k/d' file.txt
The sed script is singly-quoted, which prevents shell variable expansion. It will (probably) work with
sed -i "/$k/d" file.txt
I say "probably" because what it will do depends on the contents of $k, which is just substituted into the sed code and interpreted as such. If $k contains slashes, it will break. If it comes from an untrustworthy source, you open yourself up to code injection (particularly with GNU sed, which can be made to execute shell commands).
Consider k=^/ s/^/rm -Rf \//e; #.
It is generally a bad idea to substitute shell variables into sed code (or any other code). A better way would be with GNU awk:
awk -i inplace -v pattern="$k" '!($0 ~ pattern)' file.txt
Or to just use grep -v and a temporary file.
first of all, you got an unclosed double quote around ${patternarr[#]} in your for statement.
Then your problem is that you use single quotes in the sed argument, making your shell not evaluate the $k within the quotes:
% declare -a patternarr=(foo bar fu foobar)
% for k in ${patternarr[#]}; do echo sed -i '/$k/d' file.txt; done
sed -i /$k/d file.txt
sed -i /$k/d file.txt
sed -i /$k/d file.txt
sed -i /$k/d file.txt
if you replace them with double quotes, here it goes:
% for k in ${patternarr[#]}; do echo sed -i "/$k/d" file.txt; done
sed -i /foo/d file.txt
sed -i /bar/d file.txt
sed -i /fu/d file.txt
sed -i /foobar/d file.txt
Any time you write a loop in shell just to manipulate text you have the wrong approach. This is probably closer to what you really should be doing (no surrounding loop required):
awk -v ks="${patternarr[#]}" 'BEGIN{gsub(/ /,")|(",ks); ks="("ks")} $0 !~ ks' file.txt
but there may be even better approaches still (e.g. only checking 1 field instead of the whole line, or using word boundaries, or string comparison or....) if you show us some sample input and expected output.
You need to use double quotes to interpolate shell variables inside the sed command, like:
for k in ${patternarr[#]}; do
sed -i "/$k/d" file.txt
done
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 "$file2": if one of those variables contain spaces, this will prevent your command from breaking.
Some further remarks:
As pointed by jim, you may want to use the GNU sed -i option.
Your regex will currently replace everything after !cats! in matching lines. If they were several occurences of !cats! on your line, only one will remain. If instead you just want to replace the value between two ! delimiters, you may consider use following sed command instead:
sed 's/!cats![^!]*/!cats!300/g'
I'm using sed to replace a character in a file with a variable. This variable is basically reading the contents of a file or a webpage which contains multiple hash-like strings like below, which are randomly generated:
define('AUTH_KEY', 'CVo|BO;Qt1B|+GE}+h2/yU7h=5`/wRV{>%h.b_;s%S8-p|>qpf]|/Vf#`&[g~*:&');
define('SECURE_AUTH_KEY', '{G2-<^jWRd7]2,?]6hhM^*asg.2C.+k=gf33-m+ZK_{Mt|q*<ELF4|gPjyxtTh!)');
define('LOGGED_IN_KEY', 'jSNo9Z;5d]tzZoh-QQ`{M-&~y??$R({:*m`0={67=+mF?L.e+R{;)+4}qCAAHz=C');
define('NONCE_KEY', '19Vt4=%8j/Z-&~ni0S<]9)J^~sy9dh|h9M_RX2#K0]F9+.v+[BP1d&B&}-FTKIJ,');
define('AUTH_SALT', 'jr7f?T|#Cbo]XVAo}N^ilkvD>dC-rr]5{al64|?_Hz }JG$yEi:_aU )Olp YAD+');
define('SECURE_AUTH_SALT', 'hm#Z%O!X_mr?lM|>>~r-?F%wi R($}|9R[):4^NTsj+gS[qnv}7|+0<9e-$DJjju');
define('LOGGED_IN_SALT', 'tyPHBOCkXZh_4H;G|^.&|^#JPB/f;{}y_Orj!6AH?#wovx+KKtTZ A[HMS9SZJ|N');
define('NONCE_SALT', 'Eb-/t 5D-vPV9I--8F<[^lcherGv.g+|7p6;+xP|5g6P}tup1K.vuHAQ=uWZ#}H^');
The variable is defined like so:
KEY=$(cat keyfile)
I used the following sed syntax:
sed -i "s/stringtoreplace/$KEY/" /path/to/file
I've also tried different variations, using single-quotes, quotes around the variables
sed -i "s/stringtoreplace/"$KEY"/" /path/to/file
sed -i "s/stringtoreplace/"${KEY}"/" /path/to/file
sed -i "s/stringtoreplace/'$KEY'/" /path/to/file
I think I "brute-forced" every way possible to put quotes and I don't know how to escape randomly generated characters like those. I keep getting the following error:
unterminated s command
Any suggestions on how to replace the string with the weird-hash-like variable?
sed is an excellent tool for simple substitutions on a single line but for anything else you should use awk:
awk -v key="$KEY" '{sub(/stringtoreplace/,key)}1' file
That will work no matter what characters "$KEY" contains, except for "&".
One possibility is to use bash instead of sed. That makes the substitution easy, but you'll have to emulate the -i option.
Something like this:
TMPFILE=$(mktemp)
KEY=$(cat keyfile)
while IFS= read -r LINE; do
echo "${LINE//stringtoreplace/$KEY}"
done </path/to/file >$TMPFILE
mv $TMPFILE /path/to/file
Try this:
KEY=$(cat keyfile | sed -e 's/[]\/()$*.^|[]/\\&/g')
sed -i "s/stringtoreplace/$KEY/" /path/to/file
The problem is that the $KEY variable contains slashes, which are the delimiters for your s command.
sed can do more than search and replace:
sed '/stringtoreplace/{
d
r keyfile
}' /path/to/file
I'm assuming "stringtoreplace" occurs by itself on a line.
I have a large number of words in a text file to replace.
This script is working up until the sed command where I get:
sed: 1: "*.js": invalid command code *
PS... Bash isn't one of my strong points - this doesn't need to be pretty or efficient
cd '/Users/xxxxxx/Sites/xxxxxx'
echo `pwd`;
for line in `cat myFile.txt`
do
export IFS=":"
i=0
list=()
for word in $line; do
list[$i]=$word
i=$[i+1]
done
echo ${list[0]}
echo ${list[1]}
sed -i "s/{$list[0]}/{$list[1]}/g" *.js
done
You're running BSD sed (under OS X), therefore the -i flag requires an argument specifying what you want the suffix to be.
Also, no files match the glob *.js.
This looks like a simple typo:
sed -i "s/{$list[0]}/{$list[1]}/g" *.js
Should be:
sed -i "s/${list[0]}/${list[1]}/g" *.js
(just like the echo lines above)
So myFile.txt contains a list of from:to substitutions, and you are looping over each of those. Why don't you create a sed script from this file instead?
cd '/Users/xxxxxx/Sites/xxxxxx'
sed -e 's/^/s:/' -e 's/$/:/' myFile.txt |
# Output from first sed script is a sed script!
# It contains substitutions like this:
# s:from:to:
# s:other:substitute:
sed -f - -i~ *.js
Your sed might not like the -f - which means sed should read its script from standard input. If that is the case, perhaps you can create a temporary script like this instead;
sed -e 's/^/s:/' -e 's/$/:/' myFile.txt >script.sed
sed -f script.sed -i~ *.js
Another approach, if you don't feel very confident with sed and think you are going to forget in a week what the meaning of that voodoo symbols is, could be using IFS in a more efficient way:
IFS=":"
cat myFile.txt | while read PATTERN REPLACEMENT # You feed the while loop with stdout lines and read fields separated by ":"
do
sed -i "s/${PATTERN}/${REPLACEMENT}/g"
done
The only pitfall I can see (it may be more) is that if whether PATTERN or REPLACEMENT contain a slash (/) they are going to destroy your sed expression.
You can change the sed separator with a non-printable character and you should be safe.
Anyway, if you know whats on your myFile.txt you can just use any.