cat myfile.sql
DELIMITER $$
USE `AA4`$$
DROP TRIGGER /*!50032 IF EXISTS */ `AT_Card_INSERT_Trigger`$$
CREATE
/*!50017 DEFINER='root'#'%' */
TRIGGER `AT_Card_INSERT_Trigger` AFTER INSERT ON `card`
FOR EACH ROW BEGIN
The following works as expected and removes the definer clause.
sed -e 's/DEFINER=[^*]*\*/\*/' myfile.sql
But it does not work with spaces after or before equal to sign. For e.g. if I have a line like this...
/*!50017 DEFINER = 'root'#'%' */
Then I need a sed statement something like this...
sed -e 's/DEFINER\ =\ [^*]*\*/\*/' myfile.sql
But there are 2 more possibilites with space (no space before, no space after). It is also possible that there can be more than 1 space before or after "=".
How do I handle it all?
You can just add * (a single space followed by *) to match zero or more spaces:
sed -e 's/DEFINER *= *[^*]*\*/\*/' myfile.sql
If you're on OSX, you could allow for more whitespace than just spaces using this:
# OSX
sed -E -e 's/DEFINER[[:space:]]*=[[:space:]]*[^*]*\*/\*/' myfile.sql
and the same will work with GNU sed(1) if you use -r in place of -E:
# GNU
sed -r -e 's/DEFINER[[:space:]]*=[[:space:]]*[^*]*\*/\*/' myfile.sql
GNU sed(1) also understands \s for whitespace:
# GNU
sed -r -e 's/DEFINER\s*=\s*[^*]*\*/\*/' myfile.sql
You already know how to use the Kleene star; it works for spaces too:
sed -e 's/DEFINER[ ]*=[ ]*[^*]*\*/\*/' myfile.sql
That handles spaces only. If you need tabs as well, drop them into the [ ] sections too. If you have an extended sed that knows about \s you can get even fancier.
Related
I have yet another list of subdomain. I want to remove any Wildcard subdomain which include these special characters:
()!&$#*+?
Mostly, the data are prefixly random. Also, could be middle. Here's some sample of output data
(www.imgur.com
***************diet.blogspot.com
*-1.gbc.criteo.com
------------------------------------------------------------i.imgur.com
This has been quite an inconvenience while scanning through the list. As always, I'm trying sed to fix it:
sed -i "/[!()#$&?+]/d" foo.txt ###Didn't work
sed -i "/[\!\(\)\#\$\&\?\+]/d" ###Escaping char didn't work
Performing commands above still result in an unchanged list and the file still on original state. I'm thinking that; to fix this is to pipe series of sed command in order to remove it one by one:
cat foo.txt | sed -e "/!/d" -e "/#/d" -e "/\*/d" -e "/\$/d" -e "/(/d" -e "/)/d" -e "/+/d" -e "/\'/d" -e "/&/d" >> foo2.txt
cat foo.txt | sed -e "/\!/d" | sed -e "/\#/d" | sed -e "/\*/d" | sed -e "/\$/d" | sed -e "/\+/d" | sed -e "/\'/d" | sed -e "/\&/d" >> foo2.txt
If escaping all special char doesn't work, it must've been my false logic. Also tried with /g still doesn't increase my luck.
As a side note: I don't want - to be deleted as some valid subdomain can have - character:
line-apps.com
line-apps-beta.com
line-apps-rc.com
line-apps-dev.com
Any help would be cherished.
Using sed
$ sed '/[[:punct:]]/d' input_file
This should delete all lines with special characters, however, it would help if you provided sample data.
To do what you're trying to do in your answer (which adds [ and ] and more to the set of characters in your question) would be:
sed '/[][!?+,#$&*() ]/d'
or just:
grep -v '[][!?+,#$&*() ]'
Per POSIX to include ] in a bracket expression it must be the first character otherwise it indicates the end of the bracket expression.
Consider printing lines you want instead of deleting lines you do not want, though, e.g.:
grep '^[[:alnum:]_.-]$' file
to print lines that only contain letters, numbers, underscores, dashes, and/or periods.
I'm wondering whether it is possible to write a 100% reliable sed command to escape any regex metacharacters in an input string so that it can be used in a subsequent sed command. Like this:
#!/bin/bash
# Trying to replace one regex by another in an input file with sed
search="/abc\n\t[a-z]\+\([^ ]\)\{2,3\}\3"
replace="/xyz\n\t[0-9]\+\([^ ]\)\{2,3\}\3"
# Sanitize input
search=$(sed 'script to escape' <<< "$search")
replace=$(sed 'script to escape' <<< "$replace")
# Use it in a sed command
sed "s/$search/$replace/" input
I know that there are better tools to work with fixed strings instead of patterns, for example awk, perl or python. I would just like to prove whether it is possible or not with sed. I would say let's concentrate on basic POSIX regexes to have even more fun! :)
I have tried a lot of things but anytime I could find an input which broke my attempt. I thought keeping it abstract as script to escape would not lead anybody into the wrong direction.
Btw, the discussion came up here. I thought this could be a good place to collect solutions and probably break and/or elaborate them.
Note:
If you're looking for prepackaged functionality based on the techniques discussed in this answer:
bash functions that enable robust escaping even in multi-line substitutions can be found at the bottom of this post (plus a perl solution that uses perl's built-in support for such escaping).
#EdMorton's answer contains a tool (bash script) that robustly performs single-line substitutions.
Ed's answer now has an improved version of the sed command used below, corrected in calestyo's answer, which is needed if you want to escape string literals for potential use with other regex-processing tools, such as awk and perl. In short: for cross-tool use, \ must be escaped as \\ rather than as [\], which means: instead of the
sed 's/[^^]/[&]/g; s/\^/\\^/g' command used below, you must use
sed 's/[^^\]/[&]/g; s/[\^]/\\&/g;'
All snippets below assume bash as the shell (POSIX-compliant reformulations are possible):
SINGLE-line Solutions
Escaping a string literal for use as a regex in sed:
To give credit where credit is due: I found the regex used below in this answer.
Assuming that the search string is a single-line string:
search='abc\n\t[a-z]\+\([^ ]\)\{2,3\}\3' # sample input containing metachars.
searchEscaped=$(sed 's/[^^]/[&]/g; s/\^/\\^/g' <<<"$search") # escape it.
sed -n "s/$searchEscaped/foo/p" <<<"$search" # Echoes 'foo'
Every character except ^ is placed in its own character set [...] expression to treat it as a literal.
Note that ^ is the one char. you cannot represent as [^], because it has special meaning in that location (negation).
Then, ^ chars. are escaped as \^.
Note that you cannot just escape every char by putting a \ in front of it because that can turn a literal char into a metachar, e.g. \< and \b are word boundaries in some tools, \n is a newline, \{ is the start of a RE interval like \{1,3\}, etc.
The approach is robust, but not efficient.
The robustness comes from not trying to anticipate all special regex characters - which will vary across regex dialects - but to focus on only 2 features shared by all regex dialects:
the ability to specify literal characters inside a character set.
the ability to escape a literal ^ as \^
Escaping a string literal for use as the replacement string in sed's s/// command:
The replacement string in a sed s/// command is not a regex, but it recognizes placeholders that refer to either the entire string matched by the regex (&) or specific capture-group results by index (\1, \2, ...), so these must be escaped, along with the (customary) regex delimiter, /.
Assuming that the replacement string is a single-line string:
replace='Laurel & Hardy; PS\2' # sample input containing metachars.
replaceEscaped=$(sed 's/[&/\]/\\&/g' <<<"$replace") # escape it
sed -n "s/.*/$replaceEscaped/p" <<<"foo" # Echoes $replace as-is
MULTI-line Solutions
Escaping a MULTI-LINE string literal for use as a regex in sed:
Note: This only makes sense if multiple input lines (possibly ALL) have been read before attempting to match.
Since tools such as sed and awk operate on a single line at a time by default, extra steps are needed to make them read more than one line at a time.
# Define sample multi-line literal.
search='/abc\n\t[a-z]\+\([^ ]\)\{2,3\}\3
/def\n\t[A-Z]\+\([^ ]\)\{3,4\}\4'
# Escape it.
searchEscaped=$(sed -e 's/[^^]/[&]/g; s/\^/\\^/g; $!a\'$'\n''\\n' <<<"$search" | tr -d '\n') #'
# Use in a Sed command that reads ALL input lines up front.
# If ok, echoes 'foo'
sed -n -e ':a' -e '$!{N;ba' -e '}' -e "s/$searchEscaped/foo/p" <<<"$search"
The newlines in multi-line input strings must be translated to '\n' strings, which is how newlines are encoded in a regex.
$!a\'$'\n''\\n' appends string '\n' to every output line but the last (the last newline is ignored, because it was added by <<<)
tr -d '\n then removes all actual newlines from the string (sed adds one whenever it prints its pattern space), effectively replacing all newlines in the input with '\n' strings.
-e ':a' -e '$!{N;ba' -e '}' is the POSIX-compliant form of a sed idiom that reads all input lines a loop, therefore leaving subsequent commands to operate on all input lines at once.
If you're using GNU sed (only), you can use its -z option to simplify reading all input lines at once:
sed -z "s/$searchEscaped/foo/" <<<"$search"
Escaping a MULTI-LINE string literal for use as the replacement string in sed's s/// command:
# Define sample multi-line literal.
replace='Laurel & Hardy; PS\2
Masters\1 & Johnson\2'
# Escape it for use as a Sed replacement string.
IFS= read -d '' -r < <(sed -e ':a' -e '$!{N;ba' -e '}' -e 's/[&/\]/\\&/g; s/\n/\\&/g' <<<"$replace")
replaceEscaped=${REPLY%$'\n'}
# If ok, outputs $replace as is.
sed -n "s/\(.*\) \(.*\)/$replaceEscaped/p" <<<"foo bar"
Newlines in the input string must be retained as actual newlines, but \-escaped.
-e ':a' -e '$!{N;ba' -e '}' is the POSIX-compliant form of a sed idiom that reads all input lines a loop.
's/[&/\]/\\&/g escapes all &, \ and / instances, as in the single-line solution.
s/\n/\\&/g' then \-prefixes all actual newlines.
IFS= read -d '' -r is used to read the sed command's output as is (to avoid the automatic removal of trailing newlines that a command substitution ($(...)) would perform).
${REPLY%$'\n'} then removes a single trailing newline, which the <<< has implicitly appended to the input.
bash functions based on the above (for sed):
quoteRe() quotes (escapes) for use in a regex
quoteSubst() quotes for use in the substitution string of a s/// call.
both handle multi-line input correctly
Note that because sed reads a single line at at time by default, use of quoteRe() with multi-line strings only makes sense in sed commands that explicitly read multiple (or all) lines at once.
Also, using command substitutions ($(...)) to call the functions won't work for strings that have trailing newlines; in that event, use something like IFS= read -d '' -r escapedValue <(quoteSubst "$value")
# SYNOPSIS
# quoteRe <text>
quoteRe() { sed -e 's/[^^]/[&]/g; s/\^/\\^/g; $!a\'$'\n''\\n' <<<"$1" | tr -d '\n'; }
# SYNOPSIS
# quoteSubst <text>
quoteSubst() {
IFS= read -d '' -r < <(sed -e ':a' -e '$!{N;ba' -e '}' -e 's/[&/\]/\\&/g; s/\n/\\&/g' <<<"$1")
printf %s "${REPLY%$'\n'}"
}
Example:
from=$'Cost\(*):\n$3.' # sample input containing metachars.
to='You & I'$'\n''eating A\1 sauce.' # sample replacement string with metachars.
# Should print the unmodified value of $to
sed -e ':a' -e '$!{N;ba' -e '}' -e "s/$(quoteRe "$from")/$(quoteSubst "$to")/" <<<"$from"
Note the use of -e ':a' -e '$!{N;ba' -e '}' to read all input at once, so that the multi-line substitution works.
perl solution:
Perl has built-in support for escaping arbitrary strings for literal use in a regex: the quotemeta() function or its equivalent \Q...\E quoting.
The approach is the same for both single- and multi-line strings; for example:
from=$'Cost\(*):\n$3.' # sample input containing metachars.
to='You owe me $1/$& for'$'\n''eating A\1 sauce.' # sample replacement string w/ metachars.
# Should print the unmodified value of $to.
# Note that the replacement value needs NO escaping.
perl -s -0777 -pe 's/\Q$from\E/$to/' -- -from="$from" -to="$to" <<<"$from"
Note the use of -0777 to read all input at once, so that the multi-line substitution works.
The -s option allows placing -<var>=<val>-style Perl variable definitions following -- after the script, before any filename operands.
Building upon #mklement0's answer in this thread, the following tool will replace any single-line string (as opposed to regexp) with any other single-line string using sed and bash:
$ cat sedstr
#!/bin/bash
old="$1"
new="$2"
file="${3:--}"
escOld=$(sed 's/[^^\\]/[&]/g; s/\^/\\^/g; s/\\/\\\\/g' <<< "$old")
escNew=$(sed 's/[&/\]/\\&/g' <<< "$new")
sed "s/$escOld/$escNew/g" "$file"
To illustrate the need for this tool, consider trying to replace a.*/b{2,}\nc with d&e\1f by calling sed directly:
$ cat file
a.*/b{2,}\nc
axx/bb\nc
$ sed 's/a.*/b{2,}\nc/d&e\1f/' file
sed: -e expression #1, char 16: unknown option to `s'
$ sed 's/a.*\/b{2,}\nc/d&e\1f/' file
sed: -e expression #1, char 23: invalid reference \1 on `s' command's RHS
$ sed 's/a.*\/b{2,}\nc/d&e\\1f/' file
a.*/b{2,}\nc
axx/bb\nc
# .... and so on, peeling the onion ad nauseum until:
$ sed 's/a\.\*\/b{2,}\\nc/d\&e\\1f/' file
d&e\1f
axx/bb\nc
or use the above tool:
$ sedstr 'a.*/b{2,}\nc' 'd&e\1f' file
d&e\1f
axx/bb\nc
The reason this is useful is that it can be easily augmented to use word-delimiters to replace words if necessary, e.g. in GNU sed syntax:
sed "s/\<$escOld\>/$escNew/g" "$file"
whereas the tools that actually operate on strings (e.g. awk's index()) cannot use word-delimiters.
NOTE: the reason to not wrap \ in a bracket expression is that if you were using a tool that accepts [\]] as a literal ] inside a bracket expression (e.g. perl and most awk implementations) to do the actual final substitution (i.e. instead of sed "s/$escOld/$escNew/g") then you couldn't use the approach of:
sed 's/[^^]/[&]/g; s/\^/\\^/g'
to escape \ by enclosing it in [] because then \x would become [\][x] which means \ or ] or [ or x. Instead you'd need:
sed 's/[^^\\]/[&]/g; s/\^/\\^/g; s/\\/\\\\/g'
So while [\] is probably OK for all current sed implementations, we know that \\ will work for all sed, awk, perl, etc. implementations and so use that form of escaping.
It should be noted that the regular expression used in some answers above among this and that one:
's/[^^\\]/[&]/g; s/\^/\\^/g; s/\\/\\\\/g'
seems to be wrong:
Doing first s/\^/\\^/g followed by s/\\/\\\\/g is an error, as any ^ escaped first to \^ will then have its \ escaped again.
A better way seems to be: 's/[^\^]/[&]/g; s/[\^]/\\&/g;'.
[^^\\] with sed (BRE/ERE) should be just [^\^] (or [^^\]). \ has no special meaning inside a bracket expression and needs not to be quoted.
Bash parameter expansion can be used to escape a string for use as a Sed replacement string:
# Define a sample multi-line literal. Includes a trailing newline to test corner case
replace='a&b;c\1
d/e
'
# Escape it for use as a Sed replacement string.
: "${replace//\\/\\\\}"
: "${_//&/\\\&}"
: "${_//\//\\\/}"
: "${_//$'\n'/\\$'\n'}"
replaceEscaped=$_
# Output should match "$replace"
sed -n "s/.*/$replaceEscaped/p" <<<''
In bash 5.2+, it can be simplified further:
# Define a sample multi-line literal. Includes a trailing newline to test corner case
replace='a&b;c\1
d/e
'
# Escape it for use as a Sed replacement string.
shopt -s extglob
shopt -s patsub_replacement # An & in the replacement will expand to what matched. bash 5.2+
: "${replace//#(&|\\|\/|$'\n')/\\&}"
replaceEscaped=$_
# Output should match "$replace"
sed -n "s/.*/$replaceEscaped/p" <<<''
Encapsulate it in a bash function:
##
# escape_replacement -v var replacement
#
# Escape special characters in _replacement_ so that it can be
# used as the replacement part in a sed substitute command.
# Store the result in _var_.
escape_replacement() {
if ! [[ $# = 3 && $1 = '-v' ]]; then
echo "escape_replacement: invalid usage" >&2
echo "escape_replacement: usage: escape_replacement -v var replacement" >&2
return 1
fi
local -n var=$2 # nameref (requires Bash 4.3+)
# We use the : command (true builtin) as a dummy command as we
# trigger a sequence of parameter expansions
# We exploit that the $_ variable (last argument to the previous command
# after expansion) contains the result of the previous parameter expansion
: "${3//\\/\\\\}" # Backslash-escape any existing backslashes
: "${_//&/\\\&}" # Backslash-escape &
: "${_//\//\\\/}" # Backslash-escape the delimiter (we assume /)
: "${_//$'\n'/\\$'\n'}" # Backslash-escape newline
var=$_ # Assign to the nameref
# To support Bash older than 4.3, the following can be used instead of nameref
#eval "$2=\$_" # Use eval instead of nameref https://mywiki.wooledge.org/BashFAQ/006
}
# Test the function
# =================
# Define a sample multi-line literal. Include a trailing newline to test corner case
replace='a&b;c\1
d/e
'
escape_replacement -v replaceEscaped "$replace"
# Output should match "$replace"
sed -n "s/.*/$replaceEscaped/p" <<<''
So I have a bash script which is working perfectly except for one issue with sed.
full=$(echo $full | sed -e 's/\b'$first'\b/ /' -e 's/ / /g')
This would work great except there are instances where the variable $first is preceeded immediately by a period, not a blank space. In those instances, I do not want the variable removed.
Example:
full="apple.orange orange.banana apple.banana banana";first="banana"
full=$(echo $full | sed -e 's/\b'$first'\b/ /' -e 's/ / /g')
echo $first $full;
I want to only remove the whole word banana, and not make any change to orange.banana or apple.banana, so how can I get sed to ignore the dot as a delimiter?
You want "banana" that is preceded by beginning-of-string or a space, and followed by a space or end-of-string
$ sed -r 's/(^|[[:blank:]])'"$first"'([[:blank:]]|$)/ /g' <<< "$full"
apple.orange orange.banana apple.banana
Note the use of -r option (for bsd sed, use -E) that enables extended regular expressions -- allow us to omit a lot of backslashes.
I have an input of the following schema
10,0,'string1_string2,_string3','',8,0,0,0.59,'20140101205216','20140128074836',584266915,5934
and I would like to replace all comma "," characters with tabs using sed. The constraint is to not replace "," inside text strings (i.e the comma in 'string1_string2,_string3' should not be replaced with tab). A regex to do this is ,(?!,_).
However the following sed does not work. I've tried all escaping permutations too.
sed s/",\(\?\!,_\)"/"\t"/g
Is there a way to do this?
On Mac OS X 10.9.1, you can use:
sed -E -e "s/('[^']*'|[^,]*),/\1X/g"
except that you'd replace the X with an actual tab. For your input line, that yields:
10X0X'string1_string2,_string3'X''X8X0X0X0.59X'20140101205216'X'20140128074836'X584266915X5934
which has X's where you want tabs. With GNU sed, you can use -r in place of -E (though it also recognizes -E). Mac sed will not expand \t to a tab; GNU sed will. With Bash, you can use the ANSI-C Quoting mechanism to have the shell embed a tab in the string passed to sed:
sed -E -e "s/('[^']*'|[^,]*),/\1"$'\t'"/g"
Without the extended regular expressions (activated by -r or -E), it isn't worth trying in sed; use awk instead.
The regex looks for either a single quote followed by zero or more non-quotes and a single quote or zero or more non-commas, followed by a comma, and replaces it with what was remembered as the either/or string and a 'tab' (using X to represent tab because it is more visible).
devnull points out that the answer above replaces the comma in a string at the end of a line. There's a workaround for that:
sed -E -e "s/('[^']*'|[^,]*)(,|$)/\1"$'\t'"/g; s/"$'\t'"$//"
The s///g before the semicolon adds a tab to the end of each line; the s/// after the semicolon removes the tab that was just added.
You could use Text::ParseWords:
perl -MText::ParseWords -n -l -e 'print join("\t", parse_line(",", 1, $_));' filename
For your input, it'd result in:
10 0 'string1_string2,_string3' '' 8 0 0 0.59 '20140101205216' '20140128074836' 584266915 5934
I would suggest take Perl's help if available because of availability of lookarounds:
s="10,0,'string1_string2,_string3','',8,0,0,0.59,'20140101205216','20140128074836',584266915,5934"
perl -pe "s/,(?=(([^']*'){2})*[^']*$)/\t/g" <<< "$s"
10\t0\t'string1_string2,_string3'\t''\t8\t0\t0\t0.59\t'20140101205216'\t'20140128074836'\t584266915\t5934
PS: Showing \t only for readability purpose.
This seems to work if I understand your question correctly:
sed -E 's/,([^_])/\t\1/g'
Output:
10 0 'string1_string2,_string3' '' 8 0 0 0.59 '20140101205216' '20140128074836' 584266915 5934
I am trying to vocab list for a Greek text we are translating in class. I want to replace every space or tab character with a paragraph mark so that every word appears on its own line. Can anyone give me the sed command, and explain what it is that I'm doing? I’m still trying to figure sed out.
For reasonably modern versions of sed, edit the standard input to yield the standard output with
$ echo 'τέχνη βιβλίο γη κήπος' | sed -E -e 's/[[:blank:]]+/\n/g'
τέχνη
βιβλίο
γη
κήπος
If your vocabulary words are in files named lesson1 and lesson2, redirect sed’s standard output to the file all-vocab with
sed -E -e 's/[[:blank:]]+/\n/g' lesson1 lesson2 > all-vocab
What it means:
The character class [[:blank:]] matches either a single space character or
a single tab character.
Use [[:space:]] instead to match any single whitespace character (commonly space, tab, newline, carriage return, form-feed, and vertical tab).
The + quantifier means match one or more of the previous pattern.
So [[:blank:]]+ is a sequence of one or more characters that are all space or tab.
The \n in the replacement is the newline that you want.
The /g modifier on the end means perform the substitution as many times as possible rather than just once.
The -E option tells sed to use POSIX extended regex syntax and in particular for this case the + quantifier. Without -E, your sed command becomes sed -e 's/[[:blank:]]\+/\n/g'. (Note the use of \+ rather than simple +.)
Perl Compatible Regexes
For those familiar with Perl-compatible regexes and a PCRE-capable sed, use \s+ to match runs of at least one whitespace character, as in
sed -E -e 's/\s+/\n/g' old > new
or
sed -e 's/\s\+/\n/g' old > new
These commands read input from the file old and write the result to a file named new in the current directory.
Maximum portability, maximum cruftiness
Going back to almost any version of sed since Version 7 Unix, the command invocation is a bit more baroque.
$ echo 'τέχνη βιβλίο γη κήπος' | sed -e 's/[ \t][ \t]*/\
/g'
τέχνη
βιβλίο
γη
κήπος
Notes:
Here we do not even assume the existence of the humble + quantifier and simulate it with a single space-or-tab ([ \t]) followed by zero or more of them ([ \t]*).
Similarly, assuming sed does not understand \n for newline, we have to include it on the command line verbatim.
The \ and the end of the first line of the command is a continuation marker that escapes the immediately following newline, and the remainder of the command is on the next line.
Note: There must be no whitespace preceding the escaped newline. That is, the end of the first line must be exactly backslash followed by end-of-line.
This error prone process helps one appreciate why the world moved to visible characters, and you will want to exercise some care in trying out the command with copy-and-paste.
Note on backslashes and quoting
The commands above all used single quotes ('') rather than double quotes (""). Consider:
$ echo '\\\\' "\\\\"
\\\\ \\
That is, the shell applies different escaping rules to single-quoted strings as compared with double-quoted strings. You typically want to protect all the backslashes common in regexes with single quotes.
The portable way to do this is:
sed -e 's/[ \t][ \t]*/\
/g'
That's an actual newline between the backslash and the slash-g. Many sed implementations don't know about \n, so you need a literal newline. The backslash before the newline prevents sed from getting upset about the newline. (in sed scripts the commands are normally terminated by newlines)
With GNU sed you can use \n in the substitution, and \s in the regex:
sed -e 's/\s\s*/\n/g'
GNU sed also supports "extended" regular expressions (that's egrep style, not perl-style) if you give it the -r flag, so then you can use +:
sed -r -e 's/\s+/\n/g'
If this is for Linux only, you can probably go with the GNU command, but if you want this to work on systems with a non-GNU sed (eg: BSD, Mac OS-X), you might want to go with the more portable option.
All of the examples listed above for sed break on one platform or another. None of them work with the version of sed shipped on Macs.
However, Perl's regex works the same on any machine with Perl installed:
perl -pe 's/\s+/\n/g' file.txt
If you want to save the output:
perl -pe 's/\s+/\n/g' file.txt > newfile.txt
If you want only unique occurrences of words:
perl -pe 's/\s+/\n/g' file.txt | sort -u > newfile.txt
option 1
echo $(cat testfile)
Option 2
tr ' ' '\n' < testfile
This should do the work:
sed -e 's/[ \t]+/\n/g'
[ \t] means a space OR an tab. If you want any kind of space, you could also use \s.
[ \t]+ means as many spaces OR tabs as you want (but at least one)
s/x/y/ means replace the pattern x by y (here \n is a new line)
The g at the end means that you have to repeat as many times it occurs in every line.
You could use POSIX [[:blank:]] to match a horizontal white-space character.
sed 's/[[:blank:]]\+/\n/g' file
or you may use [[:space:]] instead of [[:blank:]] also.
Example:
$ echo 'this is a sentence' | sed 's/[[:blank:]]\+/\n/g'
this
is
a
sentence
You can also do it with xargs:
cat old | xargs -n1 > new
or
xargs -n1 < old > new
Using gawk:
gawk '{$1=$1}1' OFS="\n" file