I have a text file line:
110254,TRCN,"OJSC "Transcontainer",LSE,1,2014-05-21,2014-11-28,,,2,,,0,
I would like to change the middle qoute character from " into a \", but not the ones following or preceding a comma.
More generally, I would like to use sed to do the following:
Ignore /, *"/
Ignore /" *,/
Ignore /^ *"/
Ignore /" *$/
Substitute all other " characters for \"
How do I go about doing this with sed?
If I understood the requirements, I believe that you can use :
sed 's/\([^,]\+\)"\([^,]\+\)/\1\\"\2/g'
It matches any " not followed nor preceeded by a comma, but with any other text, and replaces it with \". For this I used ^ to negate the comma.
> echo '110254,TRCN,"OJSC "Transcontainer",LSE,1,2014-05-21,2014-11-28,,,2,,,0,' | sed 's/\([^,]\+\)"\([^,]\+\)/\1\\"\2/g'
110254,TRCN,"OJSC \"Transcontainer",LSE,1,2014-05-21,2014-11-28,,,2,,,0,
This might work for you (GNU sed):
sed ':a;s/\([^,\\]\)"\([^,\\]\|$\)/\1\\"\2/g;ta' file
Related
With the following text file, test.txt:
{\x22order\x22:\x22161332\x22,\x22name\x22:\x22chiller\x22,\x22code\x22:\x22chiller\x22}
{\x22order\x22:\x22161332\x22,\x22name\x22:\x22chiller\x22,\x22code\x22:\x22chiller\x22}
{\x22order\x22:\x22161332\x22,\x22name\x22:\x22chiller\x22,\x22code\x22:\x22chiller\x22}
How do I replace occurrences of \x22 with single quotation marks in Sed?
I've tried this with no avail: sed -i "s#\x22#'#g" test.txt
You need to escape the backslash to make it literal and you need to use '\'' to represent a single quote in a single-quote-delimited (which they all should be unless absolutely necessary to be otherwise) string or script
$ sed 's/\\x22/'\''/g' file
{'order':'161332','name':'chiller','code':'chiller'}
{'order':'161332','name':'chiller','code':'chiller'}
{'order':'161332','name':'chiller','code':'chiller'}
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($0,s+length(old))} 1' file
Even the above would need tweaked if old or new contained escape characters.
sed pattern replace, trailing spaces ending with + only with +
input
Summary of differences with Numeric +34
First-step changes +34
output
Summary of differences with Numeric+34
First-step changes+34
Did not find answer here
dos this work for you?
sed 's/ *+/+/'
add g if you want to multi-replacements in line.
sed -r 's|\s+([+]\S+)$|\1|' file
Output:
Summary of differences with Numeric+34
First-step changes+34
You can for example do:
$ sed -r 's/\s{2,}\+/+/g' file
Summary of differences with Numeric+34
First-step changes+34
This removes multiple spaces (at least 2) whenever they are followed by the + character. Note + has to be escaped to be interpreted as character and not as a regex symbol.
I am using the Unix sed command on a string that can contain all types of characters (&, |, !, /, ?, etc).
Is there a complex delimiter (with two characters?) that can fix the error:
sed: -e expression #1, char 22: unknown option to `s'
The characters in the input file are of no concern - sed parses them fine. There may be an issue, however, if you have most of the common characters in your pattern - or if your pattern may not be known beforehand.
At least on GNU sed, you can use a non-printable character that is highly improbable to exist in your pattern as a delimiter. For example, if your shell is Bash:
$ echo '|||' | sed s$'\001''|'$'\001''/'$'\001''g'
In this example, Bash replaces $'\001' with the character that has the octal value 001 - in ASCII it's the SOH character (start of heading).
Since such characters are control/non-printable characters, it's doubtful that they will exist in the pattern. Unless, that is, you are doing something weird like modifying binary files - or Unicode files without the proper locale settings.
Another way to do this is to use Shell Parameter Substitution.
${parameter/pattern/replace} # substitute replace for pattern once
or
${parameter//pattern/replace} # substitute replace for pattern everywhere
Here is a quite complex example that is difficult with sed:
$ parameter="Common sed delimiters: [sed-del]"
$ pattern="\[sed-del\]"
$ replace="[/_%:\\#]"
$ echo "${parameter//$pattern/replace}"
result is:
Common sed delimiters: [/_%:\#]
However: This only work with bash parameters and not files where sed excel.
There is no such option for multi-character expression delimiters in sed, but I doubt
you need that. The delimiter character should not occur in the pattern, but if it appears in the string being processed, it's not a problem. And unless you're doing something extremely weird, there will always be some character that doesn't appear in your search pattern that can serve as a delimiter.
You need the nested delimiter facility that Perl offers. That allows to use stuff like matching, substituting, and transliterating without worrying about the delimiter being included in your contents. Since perl is a superset of sed, you should be able to use it for whatever you’re used sed for.
Consider this:
$ perl -nle 'print if /something/' inputs
Now if your something contains a slash, you have a problem. The way to fix this is to change delimiter, preferably to a bracketing one. So for example, you could having anything you like in the $WHATEVER shell variable (provided the backets are balanced), which gets interpolated by the shell before Perl is even called here:
$ perl -nle "print if m($WHATEVER)" /usr/share/dict/words
That works even if you have correctly nested parens in $WHATEVER. The four bracketing pairs which correctly nest like this in Perl are < >, ( ), [ ], and { }. They allow arbitrary contents that include the delimiter if that delimiter is balanced.
If it is not balanced, then do not use a delimiter at all. If the pattern is in a Perl variable, you don’t need to use the match operator provided you use the =~ operator, so:
$whatever = "some arbitrary string ( / # [ etc";
if ($line =~ $whatever) { ... }
With the help of Jim Lewis, I finally did a test before using sed :
if [ `echo $1 | grep '|'` ]; then
grep ".*$1.*:" $DB_FILE | sed "s#^.*$1*.*\(:\)## "
else
grep ".*$1.*:" $DB_FILE | sed "s|^.*$1*.*\(:\)|| "
fi
Thanks for help
Wow. I totally did not know that you could use any character as a delimiter.
At least half the time I use the sed and BREs its on paths, code snippets, junk characters, things like that. I end up with a bunch of horribly unreadable escapes which I'm not even sure won't die on some combination I didn't think of. But if you can exclude just some character class (or just one character even)
echo '#01Y $#1+!' | sed -e 'sa$#1+ashita' -e 'su#01YuHolyug'
> > > Holy shit!
That's so much easier.
Escaping the delimiter inline for BASH to parse is cumbersome and difficult to read (although the delimiter does need escaping for sed's benefit when it's first used, per-expression).
To pull together thkala's answer and user4401178's comment:
DELIM=$(echo -en "\001");
sed -n "\\${DELIM}${STARTING_SEARCH_TERM}${DELIM},\\${DELIM}${ENDING_SEARCH_TERM}${DELIM}p" "${FILE}"
This example returns all results starting from ${STARTING_SEARCH_TERM} until ${ENDING_SEARCH_TERM} that don't match the SOH (start of heading) character with ASCII code 001.
There's no universal separator, but it can be escaped by a backslash for sed to not treat it like separator (at least unless you choose a backslash character as separator).
Depending on the actual application, it might be handy to just escape those characters in both pattern and replacement.
If you're in a bash environment, you can use bash substitution to escape sed separator, like this:
safe_replace () {
sed "s/${1//\//\\\/}/${2//\//\\\/}/g"
}
It's pretty self-explanatory, except for the bizarre part.
Explanation to that:
${1//\//\\\/}
${ - bash expansion starts
1 - first positional argument - the pattern
// - bash pattern substitution pattern separator "replace-all" variant
\/ - literal slash
/ - bash pattern substitution replacement separator
\\ - literal backslash
\/ - literal slash
} - bash expansion ends
example use:
$ input="ka/pus/ta"
$ pattern="/pus/"
$ replacement="/re/"
$ safe_replace "$pattern" "$replacement" <<< "$input"
ka/re/ta
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