I'm trying to create a regex that will select an entire line where it contains a matching string.
I can't seem to get it to work. Here is the expression:
^.*?(\bEventname 2\b).*$
You can see the test case and what I've tried here:
https://www.regex101.com/r/mT5rZ3/1
here's what I use and it works perfectly for me
^.*substring.*$
This answer solves the question with 463 steps instead of 952 steps. Just ensure a new line at the end of the file.
.*Eventname 2.*\n
https://www.regex101.com/r/mT5rZ3/5
EDIT 4-9-2022
With .*Eventname 2.*\n? it also solves with 463 steps, but there is no need to ensure a new line at the end of the file.
If you are using the PHP regex . don't match newlines. So
.*(\bEventname 2\b).*
would be enough. If . matches newline you would need *? to make the dots non-greedy (so it just matches one line, instead of everything). You also need to be in multi-line mode to use ^ and $, but that shouldn't be necessary (since you only want to match one line anyway).
Try this:
(.*(?:Eventname 2).*)
explaination:
( ... ) : groups and captures the line
(?:...) : groups without capturing the string that the line needs to contain
.* : any characters
You are using a string containing several lines. By default, the ^ and $ operators will match the beginning and end of the whole string. The m modifier will cause them to match the beginning and end of a line.
Related
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
I have a list of items, such as:
this_thing.ety
other-stuff.ety
34-pairings.ety
I want to do this:
"At the beginning of every line, insert "images/"
so the result of search/replace with reg exp would yield:
images/this_thing.ety
images/other-stuff.ety
images/34-pairings.ety
I am using:
^.
as my anchor to find the beginning of each line but everything I've tried to add "images/" has resulted in actually replacing that first character. I am using Notepad ++, but can use anything.
I thought using ${foo} was on the right track but I'm missing something here.
In a regex ^.is matching begin of line and a character. If you replace this by 'image', first character, which matched, will be replaced. Empty line wont have 'image' but stay identical (they don't match ^.)
Just use ^ as regexp for begin of line
. is the any character symbol, but can only account for one character. You will want to use ^..*$ or ^.+$ if your version of regex allows so that every line that contains at least one character will be fully replaced. With replace, it would look like this
s/^(.+)$/images\/\1/
where the \1 re-inserts the part in parenthesis in the regex. In older versions of regex, try
s/^^\(..*\)$/\1/
I am attempting to edit a csv file, below is a sample line from this file.
|MIGRATE|;|10000|;|2ACC0003|;|30/09/13|;|Positive Adjmt.|;||;|MIGRATE|;|95004U
The beginning of the line |MIGRATE| needs to be modified without changing the second MIGRATE so the line would read
|MIGRATE|;|MIG_IN|;|10000|;|2ACC0003|;|30/09/13|;|Positive Adjmt.|;||;|MIGRATE|;|95004U
There are 7700 or so lines so if I am forced to do this manually I will probably cry a little.
Thanks in advance!
Just replace all the ones you want not changed with another word temporarily, then replace the rest with what you want. I'm not sure what you're asking here, but from what I can guess this might help.
It seems like you could just search for Just search for:
^\|MIGRATE\|
And replace with:
|MIGRATE|;|MIG_IN|
Make sure you've checked 'Regular expression' in the 'Search Mode' options.
Explanation: The ^ is a begin anchor; it will match the beginning of the line, ensuring that it does not match the second |MIGRATE|. The \ characters are required to escape the | characters since they normally have special meaning in regular expressions, and you want to match a literal |.
You can use beginning of line anchors:
Find:
^(\|MIGRATE\|)
Replace with:
$1;|MIG_IN|
regex101 demo
Just make sure that you are using the regular expression mode of the Search&Replace.
If you want to be a bit fancier, you can use a positive lookbehind:
Find:
(?<=^\|MIGRATE\|)
Replace with:
;|MIG_IN|
^ Will match only at the beginning of a line.
( ... ) is called a capture group, and will save the contents of the match in variable you can use (in the first regex, I accessed the variable using $1 in the replace. The first capture gets stored to $1, the second to $2, etc.)
| is a special character meaning 'or' in regex (to match a character or group of characters or another, e.g. a|b matches a or b. As such, you need to escape it with a backslash to make a regex match a literal |.
In my second regex, I used (?<= ... ) which is called a positive lookbehind. It makes sure that the part to be matched has what's inside before it. For instance, (?<=a)b matches a b only if it has an a before it. So that the b in ab matches but not in bb.
The website I linked also explains the details of the regex and you can try out some regex yourself!
I have a file containing lines like:
13
13-55
some text 11
I want to create a regex to match only first to type of lines, but not the last one.
Reges created by me is [0-9\-]+
You have to specify that you are testing from the beggining to the end of the string:
Try with following regex:
^[0-9-]+$
Try using anchors (^ and $ to denote beginning and end of string respectively) and use the multiline option (this one depends on the language/engine/environment of the regex).
^[0-9-]+$
Note, you can drop the backslash for the - if it's at the beginning or end of a character class.
If you want to match lines which start with number.
^[0-9-]+$
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*