Regex match text followed by curly brackets - regex

I have a text like this:
"entity"
{
"id" "5040044"
"classname" "weapon_defibrillator_spawn"
"angles" "0 0 0"
"body" "0"
"disableshadows" "0"
"skin" "0"
"solid" "6"
"spawnflags" "3"
"origin" "449.47 5797.25 2856"
editor
{
"color" "0 0 200"
"visgroupshown" "1"
"visgroupautoshown" "1"
"logicalpos" "[-13268 14500]"
}
}
What would regex expression be to select only that part in Notepad++:
editor
{
"color" "0 0 200"
"visgroupshown" "1"
"visgroupautoshown" "1"
"logicalpos" "[-13268 14500]"
}
First word is always "editor", but the number of lines and content in curly brackets may vary.

editor\s*{\s*(?:\"[a-z]*\"\s*\".*\"\s*)*\}
Demo
Also tested it in Notepad++ it works fine

The simplest way to find everything between curly brackets would be \{[^{}]*\} (example 1).
You can prepend editor\s* on it so it limits the search to only that specific entry: editor\s*\{[^{}]*\} (example 2).
However... if any of the keys or value strings within editor {...} contain a { or }, you're going to have edge cases.
You'll need to find double-quoted values and essentially ignore them. This example shows how you would stop before the first double quote within the group, and this example shows how to match up through the first key-value pair.
You essentially want to repeatedly match those key-value pairs until no more remain.
If your keys or values can contain \" within them, such as "help" "this is \"quoted\" text", you need to look for that \ character as well.
If there are nested groups within this group, you'll need to recursively handle those. Most regex (Notepad++ included) don't handle recursion, though, so to get around this, you copy-paste what you have so far inside of the code if it happens to come across more nested { and }. This does not handle more than one level of nesting, though.
TL;DR
For Notepad++, this is a single line regex you could use.

Related

VS Code if/else conditions with regular expression in user defined snippet

I am pretty bad at regex and need some help implementing my idea with the already complicated if-else syntax being used for user-defined snippets in VS Code.
I want to achieve the following:
Whenever I enter a number for variable $1 I want the snippet to create the
text "MAXVALUE $1" at the placeholder positon
if anything else is entered, there should be printed nothing
My current line for this with $1 being the variable I enter is:
"\t ${1/([0-9])|([a-zA-Z])/${1:+MAXVALUE }${2:+ }/}"
At this state I can capture the entire number EXCEPT the FRIST CHARACTER being entered and I can print MAXVALUE _mynumber_minus_char_at_index_0, LOL?!
If I enter a text, MAXVALUE won't be printed, but again the value from $1 minus the character at index 0 is being printed on screen.
Any help would be highly appreciated. If you got some useful links that explain advanced snippet creation for those kinda cases, I would be thankful as well.
For RegEx, well, time to learn them, so why not starting with a crazy-ass example like this - at least for me it is like rocket-science atm :D
Thanks in advance and best regards.
Using this snippet:
"Maxvalue": {
"prefix": "cll",
"body": [
"\t ${1/([0-9]+)|([a-zA-Z]+)/${1:+MAXVALUE }$1${2:+ }/}",
],
"description": "maxvalue"
},
([0-9]+) captures all the numbers you type; or
([a-zA-Z]+) captures all the letters you type
You were using ([0-9]) which captures, but more importantly
matches only the first number. If you don't match something it will not be transformed by the snippet transform, it just remains
untouched. That is why you were seeing everything but the first
number in the output.
You weren't actually outputting $1 anywhere - you see I added it to the transform after the MAXVALUE conditional.
${1:+MAXVALUE } is a conditional which means if there is a capture group 1, do something, in this case output MAXVALUE. That 1 in ${1:+MAXVALUE } is not a reference to your $1 tabstop. It is only a reference to the first capture group of your regex.
So you correctly outputted MAXVALUE when you had a capture group 1, but you didn't follow that up by outputting capture group 1 anywhere.
{2:+ } is anther conditional where the 2 refers to a capture group 2, if any, here ([a-zA-Z]+). So if there is a capture group 2, a space will be output. If there is no capture group 2, the conditional will fail and provide no output of its own. If you want nothing printed if you type letters, then match it and do nothing with it. As in the following:
"\t ${1/([0-9]+)|[a-zA-Z]+/${1:+MAXVALUE }$1/}", this will match all the letters you type (before tabbing to complete the transform) and they will disappear because you matched them and then didn't output them in the transform part anywhere.
If you simply want those letters to remain, don't match them as in
"\t ${1/([0-9]+)/${1:+MAXVALUE }$1/}"
If there is something you don't understand let me know.
[By the way, your question title mentions if/else conditions but you are using only if conditionals.]

MongoDB: Match multiple values in string field

I have a collection of entities that contain a string field. I'm looking for a way to query the collection with a combined number of values, and get all entities that contain all of these values, with these specifications:
contain ALL provided query values, not just some of them
case-insensitive
regardless of order
'word' query values can be part of something bigger (for example separated by _ or any other character)
So as an example, if I provide these words as the query values:
i am spiderman
(I can separate them by whitespace, give an array, or whatever works..)
I expect these results:
- "i am_spiderMan" // should match
- "AM i spiderman?!" // should match
- "who am I? supermanspiderman" // should match
- "I am superman" // should not match
- "i am spider_man" // should not match
I hope this covers all the cases I tried to describe.
I tried regex, and also did some research with similar questions but could not get it to work.
You could use regular expr. This is working perfectly. When you pass the sentence, you need to put all worlds into array as I have shown below. Refer $all to include all words to find. Reg expr case insensitive
db.collection.find ({ key: { $all: [ /spiderman/i, /i/i, /am/i ] } })

Regex Multiple rows [duplicate]

I'm trying to get the list of all digits preceding a hyphen in a given string (let's say in cell A1), using a Google Sheets regex formula :
=REGEXEXTRACT(A1, "\d-")
My problem is that it only returns the first match... how can I get all matches?
Example text:
"A1-Nutrition;A2-ActPhysiq;A2-BioMeta;A2-Patho-jour;A2-StgMrktg2;H2-Bioth2/EtudeCas;H2-Bioth2/Gemmo;H2-Bioth2/Oligo;H2-Bioth2/Opo;H2-Bioth2/Organo;H3-Endocrino;H3-Génétiq"
My formula returns 1-, whereas I want to get 1-2-2-2-2-2-2-2-2-2-3-3- (either as an array or concatenated text).
I know I could use a script or another function (like SPLIT) to achieve the desired result, but what I really want to know is how I could get a re2 regular expression to return such multiple matches in a "REGEX.*" Google Sheets formula.
Something like the "global - Don't return after first match" option on regex101.com
I've also tried removing the undesired text with REGEXREPLACE, with no success either (I couldn't get rid of other digits not preceding a hyphen).
Any help appreciated!
Thanks :)
You can actually do this in a single formula using regexreplace to surround all the values with a capture group instead of replacing the text:
=join("",REGEXEXTRACT(A1,REGEXREPLACE(A1,"(\d-)","($1)")))
basically what it does is surround all instances of the \d- with a "capture group" then using regex extract, it neatly returns all the captures. if you want to join it back into a single string you can just use join to pack it back into a single cell:
You may create your own custom function in the Script Editor:
function ExtractAllRegex(input, pattern,groupId) {
return [Array.from(input.matchAll(new RegExp(pattern,'g')), x=>x[groupId])];
}
Or, if you need to return all matches in a single cell joined with some separator:
function ExtractAllRegex(input, pattern,groupId,separator) {
return Array.from(input.matchAll(new RegExp(pattern,'g')), x=>x[groupId]).join(separator);
}
Then, just call it like =ExtractAllRegex(A1, "\d-", 0, ", ").
Description:
input - current cell value
pattern - regex pattern
groupId - Capturing group ID you want to extract
separator - text used to join the matched results.
Edit
I came up with more general solution:
=regexreplace(A1,"(.)?(\d-)|(.)","$2")
It replaces any text except the second group match (\d-) with just the second group $2.
"(.)?(\d-)|(.)"
1 2 3
Groups are in ()
---------------------------------------
"$2" -- means return the group number 2
Learn regular expressions: https://regexone.com
Try this formula:
=regexreplace(regexreplace(A1,"[^\-0-9]",""),"(\d-)|(.)","$1")
It will handle string like this:
"A1-Nutrition;A2-ActPhysiq;A2-BioM---eta;A2-PH3-Généti***566*9q"
with output:
1-2-2-2-3-
I wasn't able to get the accepted answer to work for my case. I'd like to do it that way, but needed a quick solution and went with the following:
Input:
1111 days, 123 hours 1234 minutes and 121 seconds
Expected output:
1111 123 1234 121
Formula:
=split(REGEXREPLACE(C26,"[a-z,]"," ")," ")
The shortest possible regex:
=regexreplace(A1,".?(\d-)|.", "$1")
Which returns 1-2-2-2-2-2-2-2-2-2-3-3- for "A1-Nutrition;A2-ActPhysiq;A2-BioMeta;A2-Patho-jour;A2-StgMrktg2;H2-Bioth2/EtudeCas;H2-Bioth2/Gemmo;H2-Bioth2/Oligo;H2-Bioth2/Opo;H2-Bioth2/Organo;H3-Endocrino;H3-Génétiq".
Explanation of regex:
.? -- optional character
(\d-) -- capture group 1 with a digit followed by a dash (specify (\d+-) multiple digits)
| -- logical or
. -- any character
the replacement "$1" uses just the capture group 1, and discards anything else
Learn more about regex: https://twiki.org/cgi-bin/view/Codev/TWikiPresentation2018x10x14Regex
This seems to work and I have tried to verify it.
The logic is
(1) Replace letter followed by hyphen with nothing
(2) Replace any digit not followed by a hyphen with nothing
(3) Replace everything which is not a digit or hyphen with nothing
=regexreplace(A1,"[a-zA-Z]-|[0-9][^-]|[a-zA-Z;/é]","")
Result
1-2-2-2-2-2-2-2-2-2-3-3-
Analysis
I had to step through these procedurally to convince myself that this was correct. According to this reference when there are alternatives separated by the pipe symbol, regex should match them in order left-to-right. The above formula doesn't work properly unless rule 1 comes first (otherwise it reduces all characters except a digit or hyphen to null before rule (1) can come into play and you get an extra hyphen from "Patho-jour").
Here are some examples of how I think it must deal with the text
The solution to capture groups with RegexReplace and then do the RegexExctract works here too, but there is a catch.
=join("",REGEXEXTRACT(A1,REGEXREPLACE(A1,"(\d-)","($1)")))
If the cell that you are trying to get the values has Special Characters like parentheses "(" or question mark "?" the solution provided won´t work.
In my case, I was trying to list all “variables text” contained in the cell. Those “variables text “ was wrote inside like that: “{example_name}”. But the full content of the cell had special characters making the regex formula do break. When I removed theses specials characters, then I could list all captured groups like the solution did.
There are two general ('Excel' / 'native' / non-Apps Script) solutions to return an array of regex matches in the style of REGEXEXTRACT:
Method 1)
insert a delimiter around matches, remove junk, and call SPLIT
Regexes work by iterating over the string from left to right, and 'consuming'. If we are careful to consume junk values, we can throw them away.
(This gets around the problem faced by the currently accepted solution, which is that as Carlos Eduardo Oliveira mentions, it will obviously fail if the corpus text contains special regex characters.)
First we pick a delimiter, which must not already exist in the text. The proper way to do this is to parse the text to temporarily replace our delimiter with a "temporary delimiter", like if we were going to use commas "," we'd first replace all existing commas with something like "<<QUOTED-COMMA>>" then un-replace them later. BUT, for simplicity's sake, we'll just grab a random character such as  from the private-use unicode blocks and use it as our special delimiter (note that it is 2 bytes... google spreadsheets might not count bytes in graphemes in a consistent way, but we'll be careful later).
=SPLIT(
LAMBDA(temp,
MID(temp, 1, LEN(temp)-LEN(""))
)(
REGEXREPLACE(
"xyzSixSpaces:[ ]123ThreeSpaces:[ ]aaaa 12345",".*?( |$)",
"$1"
)
),
""
)
We just use a lambda to define temp="match1match2match3", then use that to remove the last delimiter into "match1match2match3", then SPLIT it.
Taking COLUMNS of the result will prove that the correct result is returned, i.e. {" ", " ", " "}.
This is a particularly good function to turn into a Named Function, and call it something like REGEXGLOBALEXTRACT(text,regex) or REGEXALLEXTRACT(text,regex), e.g.:
=SPLIT(
LAMBDA(temp,
MID(temp, 1, LEN(temp)-LEN(""))
)(
REGEXREPLACE(
text,
".*?("&regex&"|$)",
"$1"
)
),
""
)
Method 2)
use recursion
With LAMBDA (i.e. lets you define a function like any other programming language), you can use some tricks from the well-studied lambda calculus and function programming: you have access to recursion. Defining a recursive function is confusing because there's no easy way for it to refer to itself, so you have to use a trick/convention:
trick for recursive functions: to actually define a function f which needs to refer to itself, instead define a function that takes a parameter of itself and returns the function you actually want; pass in this 'convention' to the Y-combinator to turn it into an actual recursive function
The plumbing which takes such a function work is called the Y-combinator. Here is a good article to understand it if you have some programming background.
For example to get the result of 5! (5 factorial, i.e. implement our own FACT(5)), we could define:
Named Function Y(f)=LAMBDA(f, (LAMBDA(x,x(x)))( LAMBDA(x, f(LAMBDA(y, x(x)(y)))) ) ) (this is the Y-combinator and is magic; you don't have to understand it to use it)
Named Function MY_FACTORIAL(n)=
Y(LAMBDA(self,
LAMBDA(n,
IF(n=0, 1, n*self(n-1))
)
))
result of MY_FACTORIAL(5): 120
The Y-combinator makes writing recursive functions look relatively easy, like an introduction to programming class. I'm using Named Functions for clarity, but you could just dump it all together at the expense of sanity...
=LAMBDA(Y,
Y(LAMBDA(self, LAMBDA(n, IF(n=0,1,n*self(n-1))) ))(5)
)(
LAMBDA(f, (LAMBDA(x,x(x)))( LAMBDA(x, f(LAMBDA(y, x(x)(y)))) ) )
)
How does this apply to the problem at hand? Well a recursive solution is as follows:
in pseudocode below, I use 'function' instead of LAMBDA, but it's the same thing:
// code to get around the fact that you can't have 0-length arrays
function emptyList() {
return {"ignore this value"}
}
function listToArray(myList) {
return OFFSET(myList,0,1)
}
function allMatches(text, regex) {
allMatchesHelper(emptyList(), text, regex)
}
function allMatchesHelper(resultsToReturn, text, regex) {
currentMatch = REGEXEXTRACT(...)
if (currentMatch succeeds) {
textWithoutMatch = SUBSTITUTE(text, currentMatch, "", 1)
return allMatches(
{resultsToReturn,currentMatch},
textWithoutMatch,
regex
)
} else {
return listToArray(resultsToReturn)
}
}
Unfortunately, the recursive approach is quadratic order of growth (because it's appending the results over and over to itself, while recreating the giant search string with smaller and smaller bites taken out of it, so 1+2+3+4+5+... = big^2, which can add up to a lot of time), so may be slow if you have many many matches. It's better to stay inside the regex engine for speed, since it's probably highly optimized.
You could of course avoid using Named Functions by doing temporary bindings with LAMBDA(varName, expr)(varValue) if you want to use varName in an expression. (You can define this pattern as a Named Function =cont(varValue) to invert the order of the parameters to keep code cleaner, or not.)
Whenever I use varName = varValue, write that instead.
to see if a match succeeds, use ISNA(...)
It would look something like:
Named Function allMatches(resultsToReturn, text, regex):
UNTESTED:
LAMBDA(helper,
OFFSET(
helper({"ignore"}, text, regex),
0,1)
)(
Y(LAMBDA(helperItself,
LAMBDA(results, partialText,
LAMBDA(currentMatch,
IF(ISNA(currentMatch),
results,
LAMBDA(textWithoutMatch,
helperItself({results,currentMatch}, textWithoutMatch)
)(
SUBSTITUTE(partialText, currentMatch, "", 1)
)
)
)(
REGEXEXTRACT(partialText, regex)
)
)
))
)

Create an If statement comparing a custom field MS Word

I'm trying to create an if statement (in MS Word) that looks at a custom field.
The custom field is DocProperty Client_ABV
I want it to print a line of text if client_abv matches a certain value else be completely blank (or delete the empty line if possible)
I believe it needs to look something like this:
{IF DocProperty.Client_ABV="Test" "Print this line if Test",""}
I've very little experience with this function in Word but I have some with conditional programming.
Can anyone shed any light. I've been googling it for the last 45 minutes and have had little success with the example pages I've found.
Use Ctrl+F9 to insert the field code { brackets }. They look like wavy brackets, but these are actually special "escape codes" that tell Word this is a field code.
You need a pair of brackets for both the IF and the DocProperty fields.
When performing a string comparison it's a good idea to put "quotes" around the field code as well as around the literal string.
There is no punctuation in the DocProperty field code (no period). And no comma between the true/false evaluation, only a space between the closing " and opening ".
If a paragraph mark should be part of the true/false evaluation (for example, you want to suppress the paragraph mark if the comparison is false) include it inside the "quotes" for the evaluation result. The field code will look a bit odd, but that does work.
For example:
{ IF "{ DocProperty Client_ABV }"="Test" "Print this line if Test¶
" ""}

Select until next dot followed by \s?

I could use some help writing a regex. I have the following text:
DEFINE BROWSE BW_SC20SDAN
&ANALYZE-SUSPEND _UIB-CODE-BLOCK _DISPLAY-FIELDS BW_SC20SDAN C-Win _FREEFORM
QUERY BW_SC20SDAN NO-LOCK DISPLAY
ZTYACC.prime COLUMN-LABEL "" FORMAT "X(35)"
ZUNACT.sec COLUMN-LABEL " " FORMAT "X(30)"
INFDON.sep COLUMN-LABEL "" FORMAT "99/99/9999"
IF INFDON.top THEN "S" ELSE (IF INFDON.REPORT THEN "R" ELSE (IF INFDON.prime <> "" THEN INFDON.prime ELSE "")) COLUMN-LABEL "R" FORMAT "X(1)"
/* _UIB-CODE-BLOCK-END */
&ANALYZE-RESUME
WITH SEPARATORS SIZE 83.57 BY 5.08
BGCOLOR 15 FGCOLOR 1 FONT 6 FIT-LAST-COLUMN.
I have to find this whole block in a text file, so far I have this regex:
(?:DEFINE|DEF)\s([\w\s]*)BROWSE\s+([\w-]+)\s+([^.]*)\.
My problem is that it selects only this :
DEFINE BROWSE BW_SC20SDAN
&ANALYZE-SUSPEND _UIB-CODE-BLOCK _DISPLAY-FIELDS BW_SC20SDAN C-Win _FREEFORM
QUERY BW_SC20SDAN NO-LOCK DISPLAY
ZTYACC.
When I want to select until the final point. Basically, the rule I want to apply is "until next dot followed by \s".
But I can't figure out how to write this regex.
Allow "non-dot" [^.] OR "dots not followed by space" \.(?!\s):
DEF(INE)?\s([\w\s]*)BROWSE\s+([\w-]+)\s+(([^.]|\.(?!\s))*)\.
Note also the simplification of the leading term.
Probably the most readable way to do that is
(?:DEFINE|DEF)\s([\w\s]*)BROWSE[\S\s]+?\.\s
You turn the + operator lazy with ?, meaning by default it matches everything until it hits the first period followed by a space.
If you have the option to use an ungreedy regex library, the simplest yet closest to what you specified would be
DEFINE\s+BROWSE.*?\.\s
Note, however, that the trailing whitespace may not be there at the end of your input text, leaving the last statement unmatched.
You may find it useful to have a lexer (scanner) like flex or ANTLR tokenize your string. This approach has the advantage that the lexer takes care of the white space and lets you specify the form of the block of interest in more detail.