So here is my input:
list toblock '123
456'
How will I use sed to replace list toblock with wildcard with space for this code didn't work
sed -i "s/list toblock '*.'/list maclist 'sampletoreplace'/g"
It's .* you needed, not *. (. means any character, * mean repeat any times including 0 time).
Also sed is a line-stream editor, so by default the match will be limited within the current line.
If you want to match multiple lines, the simplest way is to use -z switch (GNU sed):
echo "list toblock '123
456'"|sed -z "s/list toblock '.*'/list maclist 'sampletoreplace'/g"
list maclist 'sampletoreplace'
But since sed usually works with RegEx greedy mode, and not support .*?(stop greedy mode), so you might want to change to:
sed -z "s/list toblock '[^']*'/list maclist 'sampletoreplace'/g"
In which [^'] means any character that is not a '.
Related
I have one file named `config_3_setConfigPW.ldif? containing the following line:
{pass}
on terminal, I used following commands
SLAPPASSWD=Pwd&0011
sed -i "s#{pass}#$SLAPPASSWD#" config_3_setConfigPW.ldif
It should replace {pass} to Pwd&0011 but it generates Pwd{pass}0011.
The reason is that the SLAPPASSWD shell variable is expanded before sed sees it. So sed sees:
sed -i "s#{pass}#Pwd&0011#" config_3_setConfigPW.ldif
When an "&" is on the right hand side of a pattern it means "copy the matched input", and in your case the matched input is "{pass}".
The real problem is that you would have to escape all the special characters that might arise in SLAPPASSWD, to prevent sed doing this. For example, if you had character "#" in the password, sed would think it was the end of the substitute command, and give a syntax error.
Because of this, I wouldn't use sed for this. You could try gawk or perl?
eg, this will print out the modified file in awk (though it still assumes that SLAPPASSWD contains no " character
awk -F \{pass\} ' { print $1"'${SLAPPASSWD}'"$2 } ' config_3_setConfigPW.ldif
That's because$SLAPPASSWD contains the character sequences & which is a metacharacter used by sed and evaluates to the matched text in the s command. Meaning:
sed 's/{pass}/match: &/' <<< '{pass}'
would give you:
match: {pass}
A time ago I've asked this question: "Is it possible to escape regex metacharacters reliably with sed". Answers there show how to reliably escape the password before using it as the replacement part:
pwd="Pwd&0011"
pwdEscaped="$(sed 's/[&/\]/\\&/g' <<< "$pwd")"
# Now you can safely pass $pwd to sed
sed -i "s/{pass}/$pwdEscaped/" config_3_setConfigPW.ldif
Bear in mind that sed NEVER operates on strings. The thing sed searches for is a regexp and the thing it replaces it with is string-like but has some metacharacters you need to be aware of, e.g. & or \<number>, and all of it needs to avoid using the sed delimiters, / typically.
If you want to operate on strings you need to use awk:
awk -v old="{pass}" -v new="$SLAPPASSWD" 's=index($0,old){ $0 = substr($0,1,s-1) new substr($0,s+length(old))} 1' file
Even the above would need tweaked if old or new contained escape characters.
I am new to this forum and this is my first query, hope I am not duplicating stuff.
I need to replace the following strings in a C file and I am using Sed for this purpose.
Input:
#define N_MAX_ITEMS 20 // some comment
#define N_TOTAL_COUNT 10
Expected output:
N_MAX_ITEMS = 20
N_TOTAL_COUNT = 10
The inputs can have multiple or single spaces or tabs in between.
I am using the following sed command in MAKEfile, to be executed in Cygwin environment.
macros_$(HW_TYPE).ini: macros_temp.h
rm -f $#
sed -e "s/\(#define[ \t]*\) \(N_.*\) \([\s\t]*[A-Z0-9_(].*\)/\2=\3/" <$^ >>$#
but the output is not proper particularly if for the expression that have comments appended.
The generic idea is, that my sed should be able to select different words till the first space or tab encountered.
This might work for you (GNU sed):
sed -r 's/^#define\s+(\S+)\s+(\S+).*/\1 = \2/' file
Use the metacharacters \s and \S for whitespace and non-whitespace and the + metacharacter for one or more.
I'm having issues matching strings even if they start with any number of white spaces. It's been very little time since I started using regular expressions, so I need some help
Here is an example. I have a file (file.txt) that contains two lines
#String1='Test One'
String1='Test Two'
Im trying to change the value for the second line, without affecting line 1 so I used this
sed -i "s|String1=.*$|String1='Test Three'|g"
This changes the values for both lines. How can I make sed change only the value of the second string?
Thank you
With gnu sed, you match spaces using \s, while other sed implementations usually work with the [[:space:]] character class. So, pick one of these:
sed 's/^\s*AWord/AnotherWord/'
sed 's/^[[:space:]]*AWord/AnotherWord/'
Since you're using -i, I assume GNU sed. Either way, you probably shouldn't retype your word, as that introduces the chance of a typo. I'd go with:
sed -i "s/^\(\s*String1=\).*/\1'New Value'/" file
Move the \s* outside of the parens if you don't want to preserve the leading whitespace.
There are a couple of solutions you could use to go about your problem
If you want to ignore lines that begin with a comment character such as '#' you could use something like this:
sed -i "/^\s*#/! s|String1=.*$|String1='Test Three'|g" file.txt
which will only operate on lines that do not match the regular expression /.../! that begins ^ with optional whiltespace\s* followed by an octothorp #
The other option is to include the characters before 'String' as part of the substitution. Doing it this way means you'll need to capture \(...\) the group to include it in the output with \1
sed -i "s|^\(\s*\)String1=.*$|\1String1='Test Four'|g" file.txt
With GNU sed, try:
sed -i "s|^\s*String1=.*$|String1='Test Three'|" file
or
sed -i "/^\s*String1=/s/=.*/='Test Three'/" file
Using awk you could do:
awk '/String1/ && f++ {$2="Test Three"}1' FS=\' OFS=\' file
#String1='Test One'
String1='Test Three'
It will ignore first hits of string1 since f is not true.
I have a data file that needs to have several uniq identifiers stripped of hyphens.
So I have:
(Special_Section "data-values")
and I want to have it replaced with:
(Special_Section "datavalues")
I wanted to use a simple sed find/replace, but the data and values are different each time. Preferably, I'd run this in-place since the file has a lot of other information I want to keep in tact.
Does sed or awk have a way to remove the hyphen from the matched portion only?
Currently I can match with: sed -i 's/Special_Section "[a-zA-Z0-9]*-[a-zA-Z0-9]*"/&/g *myfiles*
But I would like to then run s/-// on & if it's possible.
You seems to be using GNU sed, so something like this might work:
sed -ri '
s/(Special_Section [^-]*)-([^)]*)/\1\2/g
' <your_filename_glob>
does this work?
sed -i '/(Special_Section ".*-.*")/{s/-//}' yourFile
Close - scan for the lines and then substitute on those that match:
sed -i '/Special_Section "[a-zA-Z0-9]*-[a-zA-Z0-9]*"/s/\( "[a-zA-Z0-9]*\)-\([a-zA-Z0-9]*\)"/\1\2/' *myfiles*
You can split that over several lines to avoid the scroll bar in SO:
sed -i '/Special_Section "[a-zA-Z0-9]*-[a-zA-Z0-9]*"/{
s/\( "[a-zA-Z0-9]*\)-\([a-zA-Z0-9]*\)"/\1\2/
}' *myfiles*
And on further thoughts, you can also do:
sed -i 's/\(Special_Section "[a-zA-Z0-9]*\)-\([a-zA-Z0-9]*"\)/\1\2/' *myfiles*
This is more compact. You can add the g qualifier if you need it. Both solutions use the special \(...\) notation to capture parts of the regular expression.
How can one replace a part of a line with sed?
The line
DBSERVERNAME xxx
should be replaced to:
DBSERVERNAME yyy
The value xxx can vary and there are two tabs between dbservername and the value. This name-value pair is one of many from a configuration file.
I tried with the following backreference:
echo "DBSERVERNAME xxx" | sed -rne 's/\(dbservername\)[[:blank:]]+\([[:alpha:]]+\)/\1 yyy/gip'
and that resulted in an error: invalid reference \1 on `s' command's RHS.
Whats wrong with the expression? Using GNU sed.
This works:
sed -rne 's/(dbservername)\s+\w+/\1 yyy/gip'
(When you use the -r option, you don't have to escape the parens.)
Bit of explanation:
-r is extended regular expressions - makes a difference to how the regex is written.
-n does not print unless specified - sed prints by default otherwise,
-e means what follows it is an expression. Let's break the expression down:
s/// is the command for search-replace, and what's between the first pair is the regex to match, and the second pair the replacement,
gip, which follows the search replace command; g means global, i.e., every match instead of just the first will be replaced in a line; i is case-insensitivity; p means print when done (remember the -n flag from earlier!),
The brackets represent a match part, which will come up later. So dbservername is the first match part,
\s is whitespace, + means one or more (vs *, zero or more) occurrences,
\w is a word, that is any letter, digit or underscore,
\1 is a special expression for GNU sed that prints the first bracketed match in the accompanying search.
Others have already mentioned the escaping of parentheses, but why do you need a back reference at all, if the first part of the line is constant?
You could simply do
sed -e 's/dbservername.*$/dbservername yyy/g'
You're escaping your ( and ). I'm pretty sure you don't need to do that. Try:
sed -rne 's/(dbservername)[[:blank:]]+\([[:alpha:]]+\)/\1 yyy/gip'
You shouldn't be escaping things when you use single quotes. ie.
echo "DBSERVERNAME xxx" | sed -rne 's/(dbservername[[:blank:]]+)([[:alpha:]]+)/\1 yyy/gip'
You shouldn't be escaping your parens. Try:
echo "DBSERVERNAME xxx" | sed -rne 's/(dbservername)[[:blank:]]+([[:alpha:]]+)/\1 yyy/gip'
This might work for you:
echo "DBSERVERNAME xxx" | sed 's/\S*$/yyy/'
DBSERVERNAME yyy
Try this
sed -re 's/DBSERVERNAME[ \t]*([^\S]+)/\yyy/ig' temp.txt
or this
awk '{if($1=="DBSERVERNAME") $2 ="YYY"} {print $0;}' temp.txt