Is it possible to make a regex match only the first line of a text? So if I have the text:
This is the first line.
This is the second line.
...
It would match "This is the first line.", whatever the first line is.
that's sounds more like a job for the filehandle buffer.
You should be able to match the first line with:
/^(.*)$/m
(as always, this is PCRE syntax)
the /m modifier makes ^ and $ match embedded newlines. Since there's no /g modifier, it will just process the first occurrence, which is the first line, and then stop.
If you're using a shell, use:
head -n1 file
or as a filter:
commandmakingoutput | head -n1
Please clarify your question, in case this is not wat you're looking for.
In case you need the very first line no matter what, here you go:
\A.*
It will select the first line, no matter what.
Yes, you can.
Example in javascript:
"This is the first line.\n This is the second line.".match(/^.*$/m)[0];
Returns
"This is the first line."
EDIT
Explain regex:
match(/^.*$/m)[0]
^: begin of line
.*: any char (.), 0 or more times (*)
$: end of line.
m: multiline mode (. acts like a \n too)
[0]: get first position of array of results
There is also negative lookbehind function (PowerGREP or Perl flavor). It works perfectly for my purposes.
Regex:
(?<!\s+)^(.+)$
where
(?<!\s+) is negative lookbehind - regex matches only strings that
are not preceded by a whitespace(s) (\s also stands for a line break)
^ is start of a string
(.+) is a string
$ is end of string
Related
I made this regex demo (working) here: https://regex101.com/r/WSwEbY/6
When I use it in notepad ++, it doesn't work with multiple lines:
hello ladies how are you Today
hello ladies how are you Today
-> result is on a single line:
helloLadiesHowAreYouTodayHelloLadiesHowAreYouToday
Informations:
search: [^\w]+(\w)
replaceby: \U$1
n++version: 7.5.8
I also try to check 'multiline' or add '$' to en of the search.
Here, you tried to match everything that is not a word character:
[^\w]
However, the new line character \n is also not a word character so it will also be matched by [^\w] and replaced.
You should exclude \n from the character class as well:
[^\w\n]+(\w)
Demo
How about matching just the space or the start(^) with multiline flag?
(?:^| +)(\w)
sub:
\U$1
In addition to not matching newlines in the repeated character set, you should also alternate with a check for if you're at the start of a line - that way the first word on a line will be capitalized as well. Use the m flag so that ^ matches the start of a line:
(?:^|[^\w\n]+)(\w)
Replace with:
\U$1
Output:
HelloLadiesHowAreYouToday
IAmFineThankYou
https://regex101.com/r/dsOcOD/1
I would like to add some custom text to the end of all lines in my document opened in Notepad++ that start with 10 and contain a specific word (for example "frog").
So far, I managed to solve the first part.
Search: ^(10)$
Replace: \1;Batteries (to add ;Batteries to the end of the line)
What I need now is to edit this regex pattern to recognize only those lines that also contain a specific word.
For example:
Before: 1050;There is this frog in the lake
After: 1050;There is this frog in the lake;Batteries
You can use the regex to match your wanted lines:
(^(10).*?(frog).*)
the .*? is a lazy quantifier to get the minimum until frog
and replace by :
$1;Battery
Hope it helps,
You should allow any characters between the number and the end of line:
^10.*frog.*
And replacement will be $0;Batteries. You do not even need a $ anchor as .* matches till the end of a line since . matches any character but a line break char.
NOTE: There is no need to wrap the whole pattern with capturing parentheses, the $0 placeholder refers to the whole match value.
More details:
^ - start of a line
10 - a literal 10 text
.* - zero or more chars other than line break chars as many as possible
frog - a literal string
.* - zero or more chars other than line break chars as many as possible
try this
find with: (^(10).*(frog).*)
replace with: $1;Battery
Use ^(10.*frog.*)$ as regex. Replace it with something like $1;Batteries
In my quest to learn flex I'm having a scanner echo input adding line numbers.
After every line I display a counter and increment it.
Trouble is there is always a lone line number at the end of the display.
I need a regex that will ignore all line breaks except for the last one.
I tried [\n/<<EOF>>] to no avail.
Any thoughts?
I don't know what regex engine uses Flex but you can use this regex:
\z
Working demo
\z assert position at the very end of the string.
Matches the end of a string only. Unlike $, this is not affected by
multiline mode, and, in contrast to \Z, will not match before a
trailing newline at the end of a string.
If above regex doesn't work then you can use this one:
(?<=[\S\s])$
Working demo
Edit: since flex seems to work slightly different than other regex engines you could use this regex:
[\s\S]$
To get the latest character of each line. Then you can iterated over all lines until get the last one. Here you have an online flex regex engine tool:
http://ryanswanson.com/regexp/#start
Try below regex, It will search for a new line character at the end of the line.
\n$
Have you tried simply doing:
\n$
Debuggex Demo
The \n matches the newline, the $ matches end of string.
I am trying to make simple regex that will check if a line is blank or not.
Case;
" some" // not blank
" " //blank
"" // blank
The pattern you want is something like this in multiline mode:
^\s*$
Explanation:
^ is the beginning of string anchor.
$ is the end of string anchor.
\s is the whitespace character class.
* is zero-or-more repetition of.
In multiline mode, ^ and $ also match the beginning and end of the line.
References:
regular-expressions.info/Anchors, Character Classes, and Repetition.
A non-regex alternative:
You can also check if a given string line is "blank" (i.e. containing only whitespaces) by trim()-ing it, then checking if the resulting string isEmpty().
In Java, this would be something like this:
if (line.trim().isEmpty()) {
// line is "blank"
}
The regex solution can also be simplified without anchors (because of how matches is defined in Java) as follows:
if (line.matches("\\s*")) {
// line is "blank"
}
API references
String String.trim()
Returns a copy of the string, with leading and trailing whitespace omitted.
boolean String.isEmpty()
Returns true if, and only if, length() is 0.
boolean String.matches(String regex)
Tells whether or not this (entire) string matches the given regular expression.
Actually in multiline mode a more correct answer is this:
/((\r\n|\n|\r)$)|(^(\r\n|\n|\r))|^\s*$/gm
The accepted answer: ^\s*$ does not match a scenario when the last line is blank (in multiline mode).
Try this:
^\s*$
Full credit to bchr02 for this answer. However, I had to modify it a bit to catch the scenario for lines that have */ (end of comment) followed by an empty line. The regex was matching the non empty line with */.
New: (^(\r\n|\n|\r)$)|(^(\r\n|\n|\r))|^\s*$/gm
All I did is add ^ as second character to signify the start of line.
The most portable regex would be ^[ \t\n]*$ to match an empty string (note that you would need to replace \t and \n with tab and newline accordingly) and [^ \n\t] to match a non-whitespace string.
Here Blank mean what you are meaning.
A line contains full of whitespaces or a line contains nothing.
If you want to match a line which contains nothing then use '/^$/'.
Somehow none of the answers from here worked for me when I had strings which were filled just with spaces and occasionally strings having no content (just the line terminator), so I used this instead:
if (str.trim().isEmpty()) {
doSomethingWhenWhiteSpace();
}
Well...I tinkered around (using notepadd++) and this is the solution I found
\n\s
\n for end of line (where you start matching) -- the caret would not be of help in my case as the beginning of the row is a string
\s takes any space till the next string
hope it helps
This regex will delete all empty spaces (blank) and empty lines and empty tabs from file
\n\s*
According the Perl documentation on regexes:
By default, the "^" character is guaranteed to match only the beginning of the string ... Embedded newlines will not be matched by "^" ... You may, however, wish to treat a string as a multi-line buffer, such that the "^" will match after any newline within the string ... you can do this by using the /m modifier on the pattern match operator.
The "after any newline" part means that it will only match at the beginning of the 2nd and subsequent lines. What if I want to match at the beginning of any line (1st, 2nd, etc.)?
EDIT: OK, it seems that the file has BOM information (3 chars) at the beginning and that's what's messing me up. Any way to get ^ to match anyway?
EDIT: So in the end it works (as long as there's no BOM), but now it seems that the Perl documentation is wrong, since it says "after any newline"
The ^ does match the 1st line with the /m flag:
~:1932$ perl -e '$a="12\n23\n34";$a=~s/^/:/gm;print $a'
:12
:23
:34
To match with BOM you need to include it in the match.
~:1939$ perl -e '$a="12\n23\n34";$a=~s/^(\d)/<\1>:/mg;print $a'
12
<2>:3
<3>:4
~:1940$ perl -e '$a="12\n23\n34";$a=~s/^(?:)?(\d)/<\1>:/mg;print $a'
<1>:2
<2>:3
<3>:4
You can use the /^(?:\xEF\xBB\xBF)?/mg regex to match at the beginning of the line anyway, if you want to preserve the BOM.
Conceptually, there's assumed to be a newline before the beginning of the string. Consequently, /^a/ will find a letter 'a' at the beginning of a string.
Put a empty line at the beginning of the file, this cool things down, and avoid to make regex hard to read.
Yes, the BOM. It might appear at the beginning of the file, so put an empty at the beginning of the file. The BOM will not be \s, or something can be seen by bare eye. It kills my hours when a BOM make my regex fail.