I need to remove a specific data from xml file along with data, I tried lot but couldn't get the right approach. Please help me out.
Example Input:
<isOurAccount>false</isOurAccount><maturityDate/><openedDate/><valuationAmount>0<valuationAmount><value>0</value>
Expected output:
<isOurAccount>false</isOurAccount><valuationAmount>0<valuationAmount><value>0</value>
Similarly for rest of the elements for pattern <somevalue/>
Couldn't get the specific regular expression.
Thanks
Using a simple RegEx replace action to substitute with empty "" all occurrences of <[a-zA-Z]+ *\/> should suffice.
RegEx description retrieved from https://regex101.com
< matches the characters < literally
[a-zA-Z]+ match a single character present in the list below
Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
a-z a single character in the range between a and z (case sensitive)
A-Z a single character in the range between A and Z (case sensitive)
(space)* matches the space character literally
Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
\/ matches the escaped character / literally
> matches the characters > literally
Related
I have some simple sales figures in notepad++ of the form
($12 000)
($9 000)
etc. etc.
I would like to change them from this form to
-120000
-90000
etc.
I'm sure this is possible with regex/find-replace somehow. What is the best way for me to accomplish this within notepad++?
Find : (\d)
Replace : -\d
doesn't get me anywhere.
Any help much appreciated.
Use this regular expression:
\((\$\d+\s\d+)\)
Use this as the replacement:
-\1
Make sure the Regular expression radio button is checked.
RegexBuddy generates the following explanation for the regex:
Explanation
\((\$\d+\s\d+)\)
Match the character "(" literally «\(»
Match the regular expression below and capture its match into backreference number 1 «(\$\d+\s\d+)»
Match the character "$" literally «\$»
Match a single digit 0..9 «\d+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Match a single character that is a "whitespace character" (spaces, tabs, line breaks, etc.) «\s»
Match a single digit 0..9 «\d+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Match the character ")" literally «\)»
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 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 saw the phrase
^(?=.*[A-Z])(?=.*[a-z])(?=.*[0-9])[A-Za-z0-9_##%\*\-]{8,24}$
in regex, which was password checking mechanism. I read few courses about regular expressions, but I never saw combination ?=. explained.
I want know how it works. In the example it is searching for at least one capital letter, one small letter and one number. I guess it's something like "if".
(?=regex_here) is a positive lookahead. It is a zero-width assertion, meaning that it matches a location that is followed by the regex contained within (?= and ). To quote from the linked page:
lookaround actually matches characters, but then gives up the match,
returning only the result: match or no match. That is why they are
called "assertions". They do not consume characters in the string, but
only assert whether a match is possible or not. Lookaround allows you
to create regular expressions that are impossible to create without
them, or that would get very longwinded without them.
The . is not part of the lookahead, because it matches any single character that is not a line terminator.
Although i am a newbie to regex but what i understand about the above regex is
1- ?= is positive lookahead i.e. it matches the expression by looking ahead and sees if there is any pattern that matches your search paramater like [A-Z]
2- .* makes sure that they can be 0 or more number of characters before your matching expression i.e. it makes sure that u can lookahead till the end of the input string to find a match.
In short * is a quantifier which says 0 or more so if:
For instance u changed * with ? for [A-Z] part then your expression will only return true if ur 1st or 2nd letter is capital. OR if u changed it with + then ur expression will return true if any letter other than the first is a capital letter
^ asserts position at start of the string
Positive Lookahead (?=\D*\d)
Assert that the Regex below matches
\D matches any character that's not a digit (equivalent to [^0-9])
matches the previous token between zero and unlimited times, as many times as possible, giving back as needed (greedy)
\d matches a digit (equivalent to [0-9])
Positive Lookahead (?=[^a-z]*[a-z])
Assert that the Regex below matches
Match a single character not present in the list below [^a-z]
matches the previous token between zero and unlimited times, as many times as possible, giving back as needed (greedy)
a-z matches a single character in the range between a (index 97) and z (index 122) (case sensitive)
Match a single character present in the list below [a-z]
a-z matches a single character in the range between a (index 97) and z (index 122) (case sensitive)
Positive Lookahead (?=[^A-Z]*[A-Z])
Assert that the Regex below matches
Match a single character not present in the list below [^A-Z]
matches the previous token between zero and unlimited times, as many times as possible, giving back as needed (greedy)
A-Z matches a single character in the range between A (index 65) and Z (index 90) (case sensitive)
Match a single character present in the list below [A-Z]
A-Z matches a single character in the range between A (index 65) and Z (index 90) (case sensitive)
. matches any character (except for line terminators)
{8,30} matches the previous token between 8 and 30 times, as many times as possible, giving back as needed (greedy)
$ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)
I want to extract some 50 characters to the left and right of a certain word, but to make sure the outermost characters are not split the last character has to be either space, beginning of row or end of row. I tried something like this without success:
^.*(\s{0,50}(word)\s{0,50}).*$
This matches "word", but ends abruptly just before and after.
For example, using "... test test word test test ...", it matches " word ".
By using \s{0,50} you are effectively trying to match 0-50 spaces. You might want to change \s to either the characters you want (e.g [a-zA-Z\s.] or . to match all).
My suggestion is the following:
((\b.{0,50})?(word)(.{0,50}\b)?)
Note that I had to create two new groups and make them optional, so that the boundary would be matched. You might also want to add \b inside the groups to separate your word from the rest, like so:
((\b.{0,50}\b)?(word)(\b.{0,50}\b)?)
You can use this regex to extract upto 50 characters to the left and right of a certain word:
(.{0,50}\bword\b.{0,50})
Online Demo: http://regex101.com/r/uV8pL6
Explanation:
1st Capturing group (.{0,50}\bword\b.{0,50})
.{0,50} matches any character (except newline)
Quantifier: Between 0 and 50 #, as many times as possible, giving back as needed [greedy]
\b assert position at a word boundary (^\w|\w$|\W\w|\w\W)
word matches the characters word literally (case sensitive)
\b assert position at a word boundary (^\w|\w$|\W\w|\w\W)
.{0,50} matches any character (except newline)
Quantifier: Between 0 and 50 #, as many times as possible, giving back as needed [greedy]