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 "$file2": if one of those variables contain spaces, this will prevent your command from breaking.
Some further remarks:
As pointed by jim, you may want to use the GNU sed -i option.
Your regex will currently replace everything after !cats! in matching lines. If they were several occurences of !cats! on your line, only one will remain. If instead you just want to replace the value between two ! delimiters, you may consider use following sed command instead:
sed 's/!cats![^!]*/!cats!300/g'
Related
how to remove comment lines (as # bal bla ) and empty lines (lines without charecters) from file with one sed command?
THX
lidia
If you're worried about starting two sed processes in a pipeline for performance reasons, you probably shouldn't be, it's still very efficient. But based on your comment that you want to do in-place editing, you can still do that with distinct commands (sed commands rather than invocations of sed itself).
You can either use multiple -e arguments or separate commands with a semicolon, something like (just one of these, not both):
sed -i 's/#.*$//' -e '/^$/d' fileName
sed -i 's/#.*$//;/^$/d' fileName
The following transcript shows this in action:
pax> printf 'Line # with a comment\n\n# Line with only a comment\n' >file
pax> cat file
Line # with a comment
# Line with only a comment
pax> cp file filex ; sed -i 's/#.*$//;/^$/d' filex ; cat filex
Line
pax> cp file filex ; sed -i -e 's/#.*$//' -e '/^$/d' filex ; cat filex
Line
Note how the file is modified in-place even with two -e options. You can see that both commands are executed on each line. The line with a comment first has the comment removed then all is removed because it's empty.
In addition, the original empty line is also removed.
#paxdiablo has a good answer but it can be improved.
(1) The '/^$/d' clause only matches 100% blank lines.
If you want to also match lines that are entirely whitespace (spaces, tabs etc.) use this instead:
'/^\s*$/d'
(2) The 's/#.*$//' clause only matches lines that start with the # character in column 0.
If you want to also match lines that have only whitespace before the first # use this instead:
'/^\s*#.*$/d'
The above criteria may not be universal (e.g. within a HEREDOC block, or in a Python multi-line string the different approaches could be significant), but in many cases the conventional definition of "blank" lines include whitespace-only, and "comment" lines include whitespace-then-#.
(3) Lastly, on OSX at least, the #paxdiablo solution in which the first clause turns comment lines into blank lines, and the second clause strips blank lines (including what were originally comments) doesn't work. It seems to be more portable to make both clauses /d delete actions as I've done.
The revised command incorporating the above is:
sed -e '/^\s*#.*$/d' -e '/^\s*$/d' inputFile
This tiny jewel removes all # comments, no matter where they begin in a line (see caution below):
sed -e 's/\s*#.*$//'
Example:
text="
this is a # test
#this is a test
#this is a #test
this is # another #test
"
$echo "$text" | sed -e 's/\s*#.*$//'
this is a
this is
Next this removes any resulting blank lines:
$echo "$text" | sed -e 's/\s*#.*$//' | sed -e '/^\s*$/d'
Caution: Depending on the syntax and/or interpretation of the lines your processing, this might not be an appropriate solution, as it just stupidly removes end of lines, even if the '#' is part of your data or code. However, for use cases where you'll never use a hash except for as an end of line comment then it works fine. So just as with all coding, context must be taken into consideration.
Alternative variant, using grep:
cat file.txt | grep -Ev '(#.*$)|(^$)'
you can use awk
awk 'NF{gsub(/^[ \t]*#/,"");print}' file
First example(paxdiablo) is very good except its not change file, just output result. If you want to change it inline:
sudo sed -i 's/#.*$//;/^$/d' inputFile
On (one of) my linux boxes, sed understands extended regular expressions with the -r option, so:
sed -r '/(^\s*#)|(^\s*$)/d' squid.conf.installed
is very useful for showing all non-blank, non comment lines.
The regex matches either start of line followed by zero or more spaces or tabs followed by either a hash or end of line, and deletes those matching lines from the input.
I have a file pattern.txt which is composed of one very long line of complicated code (~8200 chars).
This code can be found in multiple files inside multiple directories.
I can easily identify a list of these files using
grep -rli 'uniquepartofthecode' *
My concern is how do I replace it with the exact text from within the file ?
I tried to do:
var=$(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.
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
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}/${REPLACEMENT}/g"
done
The only pitfall I can see (it may be more) is that if whether PATTERN or REPLACEMENT contain a slash (/) they are going to destroy your sed expression.
You can change the sed separator with a non-printable character and you should be safe.
Anyway, if you know whats on your myFile.txt you can just use any.
All of the lines with comments in a file begin with #. How can I delete all of the lines (and only those lines) which begin with #? Other lines containing #, but not at the beginning of the line should be ignored.
This can be done with a sed one-liner:
sed '/^#/d'
This says, "find all lines that start with # and delete them, leaving everything else."
I'm a little surprised nobody has suggested the most obvious solution:
grep -v '^#' filename
This solves the problem as stated.
But note that a common convention is for everything from a # to the end of a line to be treated as a comment:
sed 's/#.*$//' 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