I'm using jQuery. I have to check if a given list of words are in a paragraph or not. I want the exact match of a word or a phrase(whole word match).ie, if i search for 'be' in 'Be a bee', only one match is there. I have done like this.
var searchText="tool,media,be,team";
var regexExactMatch = new RegExp('\^' + searchText.split(",").join("|") + '\$');
if (regexExactMatch.test(item.Name))
{
//Found
}
It is working for one search term, ie, without any comma (eg: media).
But for comma separated search, it will break.
How to do a exact match search for multiple search terms. I'm very very new to regex. Also I have to do the same search for integers and date (MM/dd/yyyy). Thanks in advance.
For full input string match use
new RegExp('^(?:' + searchText.split(",").join("|") + ')$');
^^^ ^
For a whole word search, replace ^ and $ with \b:
new RegExp('\\b(?:' + searchText.split(",").join("|") + ')\\b');
Otherwise, the anchors are applied respectively to the first and last alternatives only (i.e. your regex will look like /^tool|media|be|team$/ looking for tool at the beginning only, media and be anywhere in the string and team only at the end of the string).
Note I am using (?:...) non-capturing group since grouping is only necessary here, not capturing (no storing of the submatch). If you need to access the matched text, you can access the 0th group that equals the whole match.
Also, you do not need those \s before ^ and $, they are not necessary at all and are ignored in the constructor notation since there are no escape sequences like \^ and \$.
Remove the ^ from the beginning and $ from the end of the RegExp. Like this :
var regexExactMatch = new RegExp(searchText.split(",").join("|"));
Reason
^ will set the condition that the matched text need to be at the beginning of the string and $ set the condition that the matched text need to be at the end of the string, which can only happen if there is only that text in the string.
Related
I want to remove all of the symbols (The symbol depends on what I select at the time) after each word, without knowing what the word could be. But leave them in before each word.
A couple of examples:
!!hello! my! !!name!!! is !!bob!! should return...
!!hello my !!name is !!bob ; for !
and
$remove$ the$ targetted$# $$symbol$$# only $after$ a $word$ should return...
$remove the targetted# $$symbol# only $after a $word ; for $
You need to use capture groups and replace:
"!!hello! my! !!name!!! is !!bob!!".replace(/([a-zA-Z]+)(!+)/g, '$1');
Which works for your test string. To work for any generic character or group of characters:
var stripTrailing = trail => {
let regex = new RegExp(`([a-zA-Z0-9]+)(${trail}+)`, 'g');
return str => str.replace(regex, '$1');
};
Note that this fails on any characters that have meaning in a regular expression: []{}+*^$. etc. Escaping those programmatically is left as an exercise for the reader.
UPDATE
Per your comment I thought an explanation might help you, so:
First, there's no way in this case to replace only part of a match, you have to replace the entire match. So we need to find a pattern that matches, split it into the part we want to keep and the part we don't, and replace the whole match with the part of it we want to keep. So let's break up my regex above into multiple lines to see what's going on:
First we want to match any number of sequential alphanumeric characters, that would be the 'word' to strip the trailing symbol from:
( // denotes capturing group for the 'word'
[ // [] means 'match any character listed inside brackets'
a-z // list of alpha character a-z
A-Z // same as above but capitalized
0-9 // list of digits 0 to 9
]+ // plus means one or more times
)
The capturing group means we want to have access to just that part of the match.
Then we have another group
(
! // I used ES6's string interpolation to insert the arg here
+ // match that exclamation (or whatever) one or more times
)
Then we add the g flag so the replace will happen for every match in the target string, without the flag it returns after the first match. JavaScript provides a convenient shorthand for accessing the capturing groups in the form of automatically interpolated symbols, the '$1' above means 'insert contents of the first capture group here in this string'.
So, in the above, if you replaced '$1' with '$1$2' you'd see the same string you started with, if you did 'foo$2' you'd see foo in place of every word trailed by one or more !, etc.
I have this regular expression:
/([a-záäéěíýóôöúüůĺľŕřčšťžňď])-$\s*/gmi
This regex selects č- from my text:
sme! a Želiezovce 2015: Spoloíč-
ne pre Európu. Oslávili aj 940.
But I want to select only - (without č) (if some character from the list [a-záäéěíýóôöúüůĺľŕřčšťžňď] is before the -).
In other languages you would use a lookbehind
/(?<=[a-záäéěíýóôöúüůĺľŕřčšťžňď])-$\s*/gmi
This matches -$\s* only if it's preceded by one of the characters in the list.
However, Javascript doesn't have lookbehind, so the workaround is to use a capturing group for the part of the regular expression after it.
var match = /[a-záäéěíýóôöúüůĺľŕřčšťžňď](-$\s*)/gmi.match(string);
When you use this, match[1] will contain the part of the string beginning with the hyphen.
First, in regex everything you put in parenthesis will be broken down in the matching process, so that the matches array will contain the full matching string at it's 0 position, followed by all of the regex's parenthesis from left to right.
/[a-záäéěíýóôöúüůĺľŕřčšťžňď](-)$\s*/gmi
Would have returned the following matches for you string: ["č-", "-"] so you can extract the specific data you need from your match.
Also, the $ character indicates in regex the end of the line and you are using the multiline flag, so technically this part \s* is just being ignored as nothing can appear in a line after the end of it.
The correct regex should be /[a-záäéěíýóôöúüůĺľŕřčšťžňď](-)$/gmi
I want to find strings that contain words in a particular order, allowing non-standard characters in between the words but excluding a particular word or symbol.
I'm using javascript's replace function to find all instances and put into an array.
So, I want select...from, with anything except 'from' in between the words. Or I can separate select...from from select...from (, as long as I exclude nesting. I think the answer is the same for both, i.e. how do I write: find x and not y within the same regexp?
From the internet, I feel this should work: /\bselect\b^(?!from).*\bfrom\b/gi but this finds no matches.
This works to find all select...from: /\bselect\b[0-9a-zA-Z#\(\)\[\]\s\.\*,%_+-]*?\bfrom\b/gi but modifying it to exclude the parenthesis "(" at the end prevents any matches: /\bselect\b[0-9a-zA-Z#\(\)\[\]\s\.\*,%_+-]*?\bfrom\b\s*^\(/gi
Can anyone tell me how to exclude words and symbols within this regexp?
Many thanks
Emma
Edit: partial string input:
left outer join [stage].[db].[table14] o on p.Project_id = o.project_id
left outer join
(
select
different_id
,sum(costs) - ( sum(brushes) + sum(carpets) + sum(fabric) + sum(other) + sum(chairs)+ sum(apples) ) as overallNumber
from
(
select ace from [stage].db.[table18] J
Javascript:
sequel = stringInputAsAbove;
var tst = sequel.replace(/\bselect\b[\s\S]*?\bfrom\b/gi, function(a,b) { console.log('match: '+a); selects.push(b); return a; });
console.log(selects);
Console.log(selects) should print an array of numbers, where each number is the starting character of a select...from. This works for the second regexp I gave in my info, printing: [95, 251]. Your \s\S variation does the same, #stribizhev.
The first example ^(?!from).* should do likewise but returns [].
The third example \s*^\( should return 251 only but returns []. However I have just noticed that the positive expression \s*\( does give 95, so some progress! It's the negatives I'm getting wrong.
Your \bselect\b^(?!from).*\bfrom\b regex doesn't work as expected because:
^ means here beginning of a line, not negation of next part, so
the \bselect\b^ means, select word followed by beginning of a
line. After removal of ^ regex start to match something
(DEMO) but it is still invalid.
in multiline text .* without modification will not match new line,
so regex will match only select...from in single lines, but if you
change it for (.|\n)* (as a simple example) it will match
multiline, but still invalid
the * is greede quantifire, so it will match as much a possible,
but if you use reluctant quantifire *?, regex will match to first
occurance of from word, and int will start to return relativly
correct result.
\bselect\b(?!from) means match separate select word which is not
directly followed by separate from word, so it would be
selectfrom somehow composed of separate words (because
select\bfrom) so (?!from) doesn't work and it is redundant
In effect you will get regex very similar to what Stribizhev gave you: \bselect\b(.|\n)*?\bfrom\b
In third expression you meke same mistake: \bselect\b[0-9a-zA-Z#\(\)\[\]\s\.\*,%_+-]*?\bfrom\b\s*^\( using ^ as (I assume) a negation, not beginning of a line. Remove ^ and you will again get relativly valid result (match from select through from to closing parathesis ) ).
Your second regex works similar to \bselect\b(.|\n)*?\bfrom\b or \bselect\b[\s\S]*?\bfrom\b.
I wrote "relativly valid result", as I also think, that parsing SQL with regex could be very camplicated, so I am not sure if it will work in every case.
You can also try to use positive lookahead to match just position in text, like:
(?=\bselect\b(?:.|\n)*?\bfrom\b)
DEMO - the () was added to regex just to return beginning index of match in groups, so it would be easier to check it validity
Negation in regex
We use ^ as negation in character class, for example [^a-z] means match anything but not letter, so it will match number, symbol, whitespace, etc, but not letter from range a to z (Look here). But this negation is on a level of single character. I you use [^from] it will prevent regex from matching characters f,r,o and m (demo). Also the [^from]{4} will avoid matching from but also form, morf, etc.
To exlude the whole word from matching by regex, you need to use negative look ahead, like (?!from), which will fail to match, if there will be chosen word from fallowing given position. To avoid matching whole line containing from you could use ^(?!.*from.*).+$ (demo).
However in your case, you don't need to use this construction, because if you replace greedy quantifire .*\bfrom with .*?\bfrom it will match to first occurance of this word. Whats more it would couse problems. Take a look on this regex, it will not match anything because (?![\s\S]*from[\s\S]*) is not restricted by anything, so it will match only if there is no from after select, but we want to match also from! in effect this regex try to match and exclude from at once, and fail. so the (?!.*word.*) construction works much better to exclude matching line with given word.
So what to do if we don't what to match a word in a fragment of a match? I think select\b([^f]|f(?!rom))*?\bfrom\b is a good solution. With ([^f]|f(?!rom))*? it will match everything between select and from, but will not exclude from.
But if you would like to match only select...from not followed by ( then it is good idea to use (?!\() like. But in your regex (multiline, use of (.|\n)*? or [\s\S]*? it will cause to match up to next select...from part, because reluctant quantifire will chenge a plece where it need to match to make whole regex . In my opinion, good solution would be to use again:
select\b([^f]|f(?!rom))*?\bfrom\b(?!\s*?\()
which will not overlap additional select..from and will not match if there is \( after select...from - check it here
So, I'm writing a google script, and I've been using the regex101 tool to troubleshoot it, but I'm having absolutely no luck.
Here's the regex I need:
Is there a way in JS to isolate "randomstuff" from "myemail+randomstuff#gmail.com" without getting the + or # included in the returned result?
I've tried using my more limited regex skills to do so, but the end result always includes '+' and '#'.
You can use the match function with captured groups as
matched = "myemail+randomstuff#gmail.com".match(/\+(.*?)#/);
matched[1]
// Outputs
// randomstuff
\+ Matches +
(.*?) Matches anything, captured in group 1
# Matches #
matched[1] The match will return an Array containing the matched results. The contentes of the array will be the captured groups in the regex match along( which are indexed by the the capture group number. Here the group is one hence we used matched[1]) and the entire match( indexed by 0)
var originalString = "myemail+randomstuff#gmail.com";
/\+([^#]+)#/.exec(originalString )[1];
This will return
"randomstuff"
I used this regex http://regexr.com/3b1fa
Here's a fiddle: http://jsfiddle.net/rt4h215g/
/\w+(?=#)/g
the (?=#) uses positive look ahead to check and see when the # symbol presents itself without including it in the match. Since + isn't considered a word character you can simply use /w+ to get the characters before the # symbol and after the +.
I have this regular expression to match a valid name: /^['"\s\-.*0-9\u00BF-\u1FFF\u2C00-\uD7FF\w]+$/.test(name)
I'm having trouble figuring out how to transform this match style regex into one designed to filter out invalid characters using replace.
Ideally I would like to be able to take an invalid name in name, run it through the replace to replace any invalid characters, and then have the original test return true no matter what (as invalid characters will be filtered out).
Just use a negated character class by adding a ^ in front:
name.replace(/[^'"\s\-.*0-9\u00BF-\u1FFF\u2C00-\uD7FF\w]/g, "")
Example:
var name = "'41%!\u2000abc";
var sanitized = name.replace(/[^'"\s\-.*0-9\u00BF-\u1FFF\u2C00-\uD7FF\w]/g, "");
console.log(/^['"\s\-.*0-9\u00BF-\u1FFF\u2C00-\uD7FF\w]+$/.test(name)); // false
console.log(/^['"\s\-.*0-9\u00BF-\u1FFF\u2C00-\uD7FF\w]+$/.test(sanitized)); // true
/^['"\s\-.*0-9\u00BF-\u1FFF\u2C00-\uD7FF\w]+$/
The + at the end tells you to match a at least 1 or multiple characters of the types inside the brackets. The ^ at the beginning in combination with the $ at the end tells to match the whole input from its start to its end. So given regex matches a string consisting of only the characters of the set.
What you want is this:
/[^'"\s\-.*0-9\u00BF-\u1FFF\u2C00-\uD7FF\w]/g
[^] means to NOT match whatever is inside the brackets and is the opposite of [].