The regex:
^ *x *\=.*$
means "match a literal x preceded by an arbitrary count of spaces, followed by an arbitrary count of spaces, then an equal sign and then anything up to the end of line." Sed was invoked as:
sed -r -e 's|^ *x *\=.*$||g' file
However it doesn't find a single match, although it should. What's wrong with the regex?
To all: thanks for the answers and effort! It seems that the problem was in tabs present in input file, which are NOT matched by the space specifier ' '. However the solution with \s works regardless of present tabs!
^\s*x\s*=.*$
Maybe you must escape some chars, figure it out one by one.
BTW: Regex tags should really have three requirements:
what is the input string, what is the output string and your platform/language.
sed processes the file line-by-line, and executes the given program for each. The simplest program that does what you want is
sed -re '/^ *x *=.*$/!d' file
"/^ *x *=.*$/" selects each line that matches the pattern.
"!" negates the result.
"d" deletes the line.
sed will by default print all lines unless told otherwise. This effectively prints lines that matches the pattern.
One alternative way of writing it is:
sed -rne '/^ *x *=.*$/p' file
"/^ *x *=.*$/" selects each line that matches the pattern.
"p" prints the line.
The difference here is that I used the "-n" switch to suppress the automatic printing of lines, and instead print only the lines I have selected.
You can also use "grep" for this task:
grep -E '^ *x *=.*$' file
or maybe '^[ ]*x[ ]*='. It's a bit more compatible, but will not match tabs or the like. And, if you don't need groups, why bother about the rest of the line?
Related
I want to print lines that contains single word only.
For example:
this is a line
another line
one
more
line
last one
I want to get the ones with single word only
one
more
line
EDIT: Guys, thank you for answers. Almost all of the answers work for my test file. However I wanted to list single lines in bash history. When I try your answers like
history | your posted commands
all of them below fails. Some only prints some numbers (might line numbers?)
You want to get all those commands in history that contain just one word. Considering that history prints the number of the command as a first column, you need to match those lines consisting in two words.
For this, you can say:
history | awk 'NF==2'
If you just want to print the command itself, say:
history | awk 'NF==2 {print $2}'
To rehash your problem, any line containing a space or nothing should be removed.
grep -Ev '^$| ' file
Your problem statement is unspecific on whether lines containing only punctuation might also occur. Maybe try
grep -Ex '[A-Za-z]+' file
to only match lines containing only one or more alphabetics. (The -x option implicitly anchors the pattern -- it requires the entire line to match.)
In Bash, the output from history is decorated with line numbers; maybe try
history | grep -E '^ *[0-9]+ [A-Za-z]+$'
to match lines where the line number is followed by a single alphanumeric token. Notice that there will be two spaces between the line number and the command.
In all cases above, the -E selects extended regular expression matching, aka egrep (basic RE aka traditional grep does not support e.g. the + operator, though it's available as \+).
Try this:
grep -E '^\s*\S+\s*$' file
With the above input, it will output:
one
more
line
If your test strings are in a file called in.txt, you can try the following:
grep -E "^\w+$" in.txt
What it means is:
^ starting the line with
\w any word character [a-zA-Z0-9]
+ there should be at least 1 of those characters or more
$ line end
And output would be
one
more
line
Assuming your file as texts.txt and if grep is not the only criteria; then
awk '{ if ( NF == 1 ) print }' texts.txt
If your single worded lines don't have a space at the end you can also search for lines without an empty space :
grep -v " "
I think that what you're looking for could be best described as a newline followed by a word with a negative lookahead for a space,
/\n\w+\b(?! )/g
example
A csv file example.csv, it has
hello,world,wow
this,is,amazing
I want to get the first column elements, at the beginning I wrote a sed command like:
sed -n 's/\([^,]*\),*/\1/p' example.csv
output:
helloworld,now
thisis,amazing
Then I modified my command to the following and get what I want:
sed -n 's/\([^,]*\).*/\1/p' example.csv
output:
hello
this
command1 I used comma(,) and command2 I replaced comma with dot(.), and it works as expected, can anyone explain how sed really works to get the 1st output? What's the story behind? Is it because of the dot(.) or because of the substitution group & back-reference?
In both regexes, ([^,]*) will consume the same part of the string - all the symbols preceding the first encountered comma. Apparently the difference is how are the remaining parts of those regexes treated.
In the first one, it's ,* - zero or more comma symbols. Obviously all it might consume is
the comma itself - the rest of the line isn't covered by a pattern.
In the second one, it's .* - zero or more of any symbols. It's not a big surprise that'll cover the remaining string completely - as it has nothing to stop at; any is, well, any. )
In both cases the pattern-covered part of the string is replaced by the contents of the capturing group (and that's, as I said already, 'all the symbols before the first comma') - and what's covered by the remaining part of the regex is just removed. So in first case the very first comma is erased, in the second - the comma and the rest of the string.
The reason behind that is that the pattern matches only to the first part of the word, i.e. only the Hello, part is replaced. The part ,* takes arbitrary amount of commas, and then nothing is set to be next, i.e. nothing else matches the pattern. For example:
hello,,,,,,,,,,,,,,,,,,world
would be replaced to
helloworld
A good example would be
sed -n 's/\([^,]*\),*$/\1/p' example.csv
This will work if and only if all the commas are at the end of the line and will trim them, e.g.
hello,,,,,,
Hope this makes the problem a bit clearer.
On regex the . (dot) is a place holder for one, single character.
Can I suggest not using sed?
cut -d, -f1 example.csv
Personally, I'm a huge sed fan, but cut is much more appropriate in this instance.
If you like first word, why not use awk
awk -F, '{print $1}' file
hello
this
Using sed with back reference
sed -nr 's/([^,]*),.*/\1/p' file
hello
this
It seems that to make it work you need the .* so it get the whole line.
The r option make you not need to escape the parentheses \(
$cat file0
"basic/strong/bold"
" /""?basic""/strong/bold"
"^/))basic"
basic
I want unix sed command such that only basic that is not in quotes should be changed.[change basic to ring]
Expected output:
$cat file0
"basic/strong/bold"
" /""?basic""/strong/bold"
"^/))basic"
ring
If we disallow escaping quotes, then any basic that is not within " is preceded by an even number of ". So this should do the trick:
sed -r 's/^([^"]*("[^"]*){2}*)basic/\1ring/' file
And as ДМИТРИЙ МАЛИКОВ mentioned, adding the --in-place option will immediately edit the file, instead of returning the new contents.
How does this work?
We anchor the regular expression to the beginning of each line with ". Then we allow an arbitrary number of non-" characters (with [^"]*). Then we start a new subpattern "[^"]* that consists of one " and arbitrarily many non-" characters. We repeat that an even number of times (with {2}*). And then we match basic. Because we matched all of that stuff in the line before basic we would replace that as well. That's why this part is wrapped in another pair of parentheses, thus capturing the line and writing it back in the replacement with \1 followed by ring.
One caveat: if you have multiple basic occurrences in one line, this will only replace the last one that is not enclosed in double quotes, because regex matches cannot overlap. A solution would be a lookbehind, but since this would be a variable-length lookbehind, which is only supported by the .NET regex engine. So if that is the case in your actual input, run the command multiple times until all occurrences are replaced.
$> sed -r 's/^([^\"]*)(basic)([^\"]*)$/\1ring\3/' file0
"basic/strong/bold"
" /""?basic""/strong/bold"
"^/))basic"
ring
If you wanna edit file in place use --in-place option.
This might work for you (GNU sed):
sed -r 's/^/\n/;ta;:a;s/\n$//;t;s/\n("[^"]*")/\1\n/;ta;s/\nbasic/ring\n/;ta;s/\n([^"]*)/\1\n/;ta' file
Not a sed solution, but it substitutes words not in quotes
Assuming that there is no escaped quotes in strings, i.e. "This is a trap \" hehe", awk might be able to solve this problem
awk -F\" 'BEGIN {OFS=FS}
{
for(i=1; i<=NF; i++){
if(i%2)
gsub(/basic/,"ring",$i)
}
print
}' inputFile
Basically the words that are not in quotes are in odd-numbered fields, and the word "basic" is replaced by "ring" in these fields.
This can be written as a one-liner, but for clarity's sake I've written it in multiple lines.
If basic is at the beginning of line:
sed -e 's/^basic/ring/' file0
I have a file that looks like this:
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
I want it to look like this
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
I thought I could use sed to do this but I can't figure out how to store something in a buffer and then modify it.
Am I even using the right tool?
Thanks
You don't have to get tricky with regular expressions and replacement strings: use sed's p command to print the line intact, then modify the line and let it print implicitly
sed 'p; s/\.png//'
Glenn jackman's response is OK, but it also doubles the rows which do not match the expression.
This one, instead, doubles only the rows which matched the expression:
sed -n 'p; s/\.png//p'
Here, -n stands for "print nothing unless explicitely printed", and the p in s/\.png//p forces the print if substitution was done, but does not force it otherwise
That is pretty easy to do with sed and you not even need to use the hold space (the sed auxiliary buffer). Given the input file below:
$ cat input
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
you should use this command:
sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
The result:
$ sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
This commands is just a replacement command (s///). It matches anything starting with #" followed by non-period chars ([^.]*) and then by .png",. Also, it matches all non-period chars before .png", using the group brackets \( and \), so we can get what was matched by this group. So, this is the to-be-replaced regular expression:
#"\([^.]*\)\.png",
So follows the replacement part of the command. The & command just inserts everything that was matched by #"\([^.]*\)\.png", in the changed content. If it was the only element of the replacement part, nothing would be changed in the output. However, following the & there is a newline character - represented by the backslash \ followed by an actual newline - and in the new line we add the #" string followed by the content of the first group (\1) and then the string ",.
This is just a brief explanation of the command. Hope this helps. Also, note that you can use the \n string to represent newlines in some versions of sed (such as GNU sed). It would render a more concise and readable command:
sed 's/#"\([^.]*\)\.png",/&\n#"\1",/' input
I prefer this over Carles Sala and Glenn Jackman's:
sed '/.png/p;s/.png//'
Could just say it's personal preference.
or one can combine both versions and apply the duplication only on lines matching the required pattern
sed -e '/^#".*\.png",/{p;s/\.png//;}' input
How to look for lines which don't end with a ."
description="This has a full stop."
description="This has a full stop."
description="This line doesn't have a full stop"
You can use a character class to describe the occurrence of any character except .:
[^\n.](\n|$)
This will match any character that is neither a . nor new line character, and that is either followed by a new line character or by the end of the string. If multiline mode is supported, you can also use just $ instead of (\n|$).
Depends on your environent. On Linux/Unix/Cygwin you would do something like this:
grep -n -v '\."$' <file.txt
or
grep -n -v '\."[[:space:]]*$' <file.txt
if trailing whitespace is fine.
I guess the regular expression pattern you are looking for is the following:
\."$
\. means a real dot. (compared to . which means any character except \n)
" is the double quote that ends the line in your example.
$ means end of line.
The way you will use this pattern depends on the environment you are using, so give us more precision for a more precise answer :-)
In general, regular expression matches. It is not easy to do a don't match. The general solution for this kind of thing is to invert the truth value. For example:
grep: grep -v '\.$'
Perl: $line !~ /\.$/
Tcl: ![regexp {\.$} $line]
In this specific case, since it is just a character, you can use the character class syntax, [], since it accepts a ^ modifier to signify anything that is not the specified characters:
[^.]$
so, in Perl it would be something like:
$line =~ /[^.]$/