I have a string like this :
002_part1_part2_______________by_test
and I would like to stop the match at the second underscore character, like this :
002_part1_part2_
How can I do that with a Regular expression ?
Thanks
Create a pattern to match any character but not of an _ zero or more times followed by an underscore symbol. Put that pattern inside a capturing or non-capturing group and make it to repeat exactly 3 times by adding range quantifier {3} next to that group.
^(?:[^_]*_){3}
DEMO
You can use:
.*\d_
EXPLANATION:
Match any single character that is NOT a line break character (line feed) «.*»
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
Match a single character that is a “digit” (any decimal number in any Unicode script) «\d»
Match the character “_” literally «_»
https://regex101.com/r/uX0qD5/1
Related
I want to detect the two following circumstances, preferably with one regex:
This is a sentence ^that I wrote today.
And:
This is a sentence ^(that I wrote) today.
So basically, if there are parentheses after the caret, I want to match whatever is inside them. Otherwise, I just want to match just the next word.
I'm new to regex. Is this possible without making it too complicated?
\^(\w+|\([\w ]+\))
Options: case insensitive; ^ and $ match at line breaks
Match the character “^” literally «\^»
Match the regular expression below and capture its match into backreference number 1 «(\w+|\([\w ]+\))»
Match either the regular expression below (attempting the next alternative only if this one fails) «\w+»
Match a single character that is a “word character” (letters, digits, etc.) «\w+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Or match regular expression number 2 below (the entire group fails if this one fails to match) «\([\w ]+\)»
Match the character “(” literally «\(»
Match a single character present in the list below «[\w ]+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
A word character (letters, digits, etc.) «\w»
The character “ ” « »
Match the character “)” literally «\)»
Created with RegexBuddy
Is there any simple way to transform:
"<A[hello|home]>"
to:
"hello|home"
Thanks!
Apart from the clever advice in the comments to simply remove certain characters, if you are unable to remove these characters because they are present elsewhere in the text and do want to match that format, here is a way to do it with regex:
Search: <\w+\[([^|]*\|[^\]]*)\]>
Replace: \1 or $1 depending on editor or regex engine.
See the Substitution pane at the bottom of the demo.
Explanation
<\w+\[([^|]*\|[^\]]*)\]>
Match the character “<” literally <
Match a single character that is a “word character” (Unicode; any letter or ideograph, digit, connector punctuation) \w+
Between one and unlimited times, as many times as possible, giving back as needed (greedy) +
Match the character “[” literally \[
Match the regex below and capture its match into backreference number 1 ([^|]*\|[^\]]*)
Match any character that is NOT a “|” [^|]*
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) *
Match the character “|” literally \|
Match any character that is NOT a “]” [^\]]*
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) *
Match the character “]” literally \]
Match the character “>” literally >
\1
Insert the backslash character \
Insert the character “1” literally 1
I try to make regular expression which helps me filter strings like
blah_blah_suffix
where suffix is any string that has length from 2 to 5 characters. So I want accept strings
blah_blah_aa
blah_blah_abcd
but discard
blah_blah_a
blah_aaa
blah_blah_aaaaaaa
I use grepl in the following way:
samples[grepl("blah_blah_.{2,5}", samples)]
but it ignores upper bound for repetition (5). So it discards strings blah_blah_a,
blah_aaa, but accepts string blah_blah_aaaaaaa.
I know there is a way to filter strings without usage of regular expression but I want to understand how to use grepl correctly.
You need to bound the expression to the start and end of the line:
^blah_blah_.{2,5}$
The ^ matches beginning of line and $ matches end of line. See a working example here: Regex101
If you want to bound the expression to the beginning and end of a string (not multi-line), use \A and \Z instead of ^ and $.
Anchors Tutorial
/^[\w]+_[\w]+_[\w]{2,5}$/
DEMO
Options: dot matches newline; case insensitive; ^ and $ match at line breaks
Assert position at the beginning of a line (at beginning of the string or after a line break character) «^»
Match a single character that is a “word character” (letters, digits, and underscores) «[\w]+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Match the character “_” literally «_»
Match a single character that is a “word character” (letters, digits, and underscores) «[\w]+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Match the character “_” literally «_»
Match a single character that is a “word character” (letters, digits, and underscores) «[\w]{2,5}»
Between 2 and 5 times, as many times as possible, giving back as needed (greedy) «{2,5}»
Assert position at the end of a line (at the end of the string or before a line break character) «$»
I am building a RegEx that needs to find lines that have either:
DateTime.Now
or
Date.Now
But cannot have the literal "SystemDateTime" on the same line.
I started with this (DateTime\.Now|Date\.Now) but now I am stuck with where to put the "SystemDateTime"
Use this. Assuming you are not using /s modifier(or DOTALL) which takes newline characters under the dot(.)
(?!.*SystemDateTime)(DateTime\.Now|Date\.Now)
(?!.*SystemDateTime) means there is no SystemDateTime in front.
You could use negative lookahead like this:
(?!.*SystemDateTime)\bDate(?:Time)?\.Now\b
/(?!.*SystemDateTime)Date(?:Time)?\.Now/
DEMO
EXPLANATION:
Assert that it is impossible to match the regex below starting at this position (negative lookahead) «(?!.*SystemDateTime)»
Match any single character that is not a line break character «.*»
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
Match the characters “SystemDateTime” literally «SystemDateTime»
Match the characters “Date” literally «Date»
Match the regular expression below «(?:Time)?»
Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
Match the characters “Time” literally «Time»
Match the character “.” literally «\.»
Match the characters “Now” literally «Now»
I have to change a url that looks like
http://my-assets.s3.amazonaws.com/uploads/2011/10/PiaggioBeverly-001-106x106.jpg
into this format
http://my-assets.s3.amazonaws.com/uploads/2011/10/106x106/PiaggioBeverly-001.jpg
I understand I need to create a regular expression pattern that will divide the initial url into three groups:
http://my-assets.s3.amazonaws.com/uploads/
2011/10/
PiaggioBeverly-001-106x106.jpg
and then cut off the resolution string (106x106) from the third group, get rid of the hyphen at the end and move the resolution next to the second. Any idea how to get it done using something like preg_replace?
search this : (.*\/)(\w+-\d+)-(.*?)\.
and replace with : \1\3/\2.
demo here : http://regex101.com/r/fX7gC2
The pattern will be as follow(for input uploads/2011/10/PiaggioBeverly-001-106x106.jpg)
^(.*/)(.+?)(\d+x\d+)(\.jpg)$
And the groups will be holding as follows:
$1 = uploads/2011/10/
$2 = PiaggioBeverly-001-
$3 = 106x106
$4 = .jpg
Now rearrange as per your need. You can check this example from online.
As you have mentioned about preg_replace(), so if its in PHP, you can use preg_match() for this.
<?php
$oldurl = "http://my-assets.s3.amazonaws.com/uploads/2011/10/PiaggioBeverly-001-106x106.jpg";
$newurl = preg_replace('%(.*?)/(\w+)-(\w+)-(\w+)\.(\w+)%sim', '$1/$4/$2-$3.jpg', $oldurl);
echo $newurl;
#http://my-assets.s3.amazonaws.com/uploads/2011/10/106x106/PiaggioBeverly-001.jpg
?>
DEMO
EXPLANATION:
Options: dot matches newline; case insensitive; ^ and $ match at line breaks
Match the regular expression below and capture its match into backreference number 1 «(.*?)»
Match any single character «.*?»
Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character “/” literally «/»
Match the regular expression below and capture its match into backreference number 2 «(\w+)»
Match a single character that is a “word character” (letters, digits, and underscores) «\w+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Match the character “-” literally «-»
Match the regular expression below and capture its match into backreference number 3 «(\w+)»
Match a single character that is a “word character” (letters, digits, and underscores) «\w+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Match the character “-” literally «-»
Match the regular expression below and capture its match into backreference number 4 «(\w+)»
Match a single character that is a “word character” (letters, digits, and underscores) «\w+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Match the character “.” literally «\.»
Match the regular expression below and capture its match into backreference number 5 «(\w+)»
Match a single character that is a “word character” (letters, digits, and underscores) «\w+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»