I have a javascript regex against a text area in an html form. Here's the regex:
regex = /[EePp]+/
I would like to also use the regex to check the length of the string in the text area and to have it to be limited to a single character.
I tried
regex = /[EePp]{1}/
And it validates regex to only those characters, but still allows strings of more than one character in the text area:
<input type='text' onkeypress='validate(event)'>
Is it possible to do that through the regex?
If you add anchors, ^ and $ at the start and end of your regex, this will limit it to matching only the pattern and nothing else against the full extent of what is being searched.
So, /^[EePp]{1}$/ -- That says [EePP] at the very beginning of what is being searched,^, and there is nothing between it and the end, $, of what is being searched.
Turns out that in this case you don't need the {1}, because the anchors are telling it exactly how far the match extends. So:
/^[EePp]$/
should do it.
Related
I have tried this: but this isn't working as expected. However it restricts user to enter space at the beginning. where did I made the mistake?
Regex that I have tried to build so far: [^-\s][a-zA-Z\s]*$
Should Match: priya, Abc Xyz
Should not match: <spaces>binayak, $&&ay%%aac
First of all, you missed the beginning of string anchor ^.
Secondly, the inverse character class [^-\s] is a good attempt at not allowing spaces, but at the same time it does allow other special characters to be at the beginning of the string, which we do not want. Instead, we could just something very similar to the second character class, just without the \s: [a-zA-Z].
Full regex:
^[a-zA-Z][a-zA-Z\s]*$
I'm trying to create a regular expression in order to check some text inserted in a textarea. Basically, when typing I check the number of words inserted. As special chars, only commas and full stop are allowed.
The problem is, for example, when I type word,anotherword my regex recognises only one word instead of two. I cannot find a good regex for it.
My current regex is:
val.match(/\S+[A-Za-z]/g
What shall I add? Thanks a lot!
Since \S matches any non-whitespace characters, and a comma, too, you should be using
val.match(/\w+/g)
A \w matches word characters, those in A-Z, a-z, 0-9 ranges and a _ (underscore).
Do you want to extract the words, or check if the entered text is of the correct format?
If the first, use js Array.split(",") and check that each word is ok.
Otherwise, I think this refer should do it:
val.match(/^(\w+\,?)+$/);
Use the ^ and $ to make sure it starts and end around the found correct format.
I am trying to replace all invalid characters with a blank string, but my regex is also matching newlines (which I want to keep). Does anyone know where my regex is matching a newline?
var re = /[\0-\x1F\x7F-\x9F\xAD\u0378\u0379\u037F-\u0383\u038B\u038D\u03A2\u0528-\u0530\u0557\u0558\u0560\u0588\u058B-\u058E\u0590\u05C8-\u05CF\u05EB-\u05EF\u05F5-\u0605\u061C\u061D\u06DD\u070E\u070F\u074B\u074C\u07B2-\u07BF\u07FB-\u07FF\u082E\u082F\u083F\u085C\u085D\u085F-\u089F\u08A1\u08AD-\u08E3\u08FF\u0978\u0980\u0984\u098D\u098E\u0991\u0992\u09A9\u09B1\u09B3-\u09B5\u09BA\u09BB\u09C5\u09C6\u09C9\u09CA\u09CF-\u09D6\u09D8-\u09DB\u09DE\u09E4\u09E5\u09FC-\u0A00\u0A04\u0A0B-\u0A0E\u0A11\u0A12\u0A29\u0A31\u0A34\u0A37\u0A3A\u0A3B\u0A3D\u0A43-\u0A46\u0A49\u0A4A\u0A4E-\u0A50\u0A52-\u0A58\u0A5D\u0A5F-\u0A65\u0A76-\u0A80\u0A84\u0A8E\u0A92\u0AA9\u0AB1\u0AB4\u0ABA\u0ABB\u0AC6\u0ACA\u0ACE\u0ACF\u0AD1-\u0ADF\u0AE4\u0AE5\u0AF2-\u0B00\u0B04\u0B0D\u0B0E\u0B11\u0B12\u0B29\u0B31\u0B34\u0B3A\u0B3B\u0B45\u0B46\u0B49\u0B4A\u0B4E-\u0B55\u0B58-\u0B5B\u0B5E\u0B64\u0B65\u0B78-\u0B81\u0B84\u0B8B-\u0B8D\u0B91\u0B96-\u0B98\u0B9B\u0B9D\u0BA0-\u0BA2\u0BA5-\u0BA7\u0BAB-\u0BAD\u0BBA-\u0BBD\u0BC3-\u0BC5\u0BC9\u0BCE\u0BCF\u0BD1-\u0BD6\u0BD8-\u0BE5\u0BFB-\u0C00\u0C04\u0C0D\u0C11\u0C29\u0C34\u0C3A-\u0C3C\u0C45\u0C49\u0C4E-\u0C54\u0C57\u0C5A-\u0C5F\u0C64\u0C65\u0C70-\u0C77\u0C80\u0C81\u0C84\u0C8D\u0C91\u0CA9\u0CB4\u0CBA\u0CBB\u0CC5\u0CC9\u0CCE-\u0CD4\u0CD7-\u0CDD\u0CDF\u0CE4\u0CE5\u0CF0\u0CF3-\u0D01\u0D04\u0D0D\u0D11\u0D3B\u0D3C\u0D45\u0D49\u0D4F-\u0D56\u0D58-\u0D5F\u0D64\u0D65\u0D76-\u0D78\u0D80\u0D81\u0D84\u0D97-\u0D99\u0DB2\u0DBC\u0DBE\u0DBF\u0DC7-\u0DC9\u0DCB-\u0DCE\u0DD5\u0DD7\u0DE0-\u0DF1\u0DF5-\u0E00\u0E3B-\u0E3E\u0E5C-\u0E80\u0E83\u0E85\u0E86\u0E89\u0E8B\u0E8C\u0E8E-\u0E93\u0E98\u0EA0\u0EA4\u0EA6\u0EA8\u0EA9\u0EAC\u0EBA\u0EBE\u0EBF\u0EC5\u0EC7\u0ECE\u0ECF\u0EDA\u0EDB\u0EE0-\u0EFF\u0F48\u0F6D-\u0F70\u0F98\u0FBD\u0FCD\u0FDB-\u0FFF\u10C6\u10C8-\u10CC\u10CE\u10CF\u1249\u124E\u124F\u1257\u1259\u125E\u125F\u1289\u128E\u128F\u12B1\u12B6\u12B7\u12BF\u12C1\u12C6\u12C7\u12D7\u1311\u1316\u1317\u135B\u135C\u137D-\u137F\u139A-\u139F\u13F5-\u13FF\u169D-\u169F\u16F1-\u16FF\u170D\u1715-\u171F\u1737-\u173F\u1754-\u175F\u176D\u1771\u1774-\u177F\u17DE\u17DF\u17EA-\u17EF\u17FA-\u17FF\u180F\u181A-\u181F\u1878-\u187F\u18AB-\u18AF\u18F6-\u18FF\u191D-\u191F\u192C-\u192F\u193C-\u193F\u1941-\u1943\u196E\u196F\u1975-\u197F\u19AC-\u19AF\u19CA-\u19CF\u19DB-\u19DD\u1A1C\u1A1D\u1A5F\u1A7D\u1A7E\u1A8A-\u1A8F\u1A9A-\u1A9F\u1AAE-\u1AFF\u1B4C-\u1B4F\u1B7D-\u1B7F\u1BF4-\u1BFB\u1C38-\u1C3A\u1C4A-\u1C4C\u1C80-\u1CBF\u1CC8-\u1CCF\u1CF7-\u1CFF\u1DE7-\u1DFB\u1F16\u1F17\u1F1E\u1F1F\u1F46\u1F47\u1F4E\u1F4F\u1F58\u1F5A\u1F5C\u1F5E\u1F7E\u1F7F\u1FB5\u1FC5\u1FD4\u1FD5\u1FDC\u1FF0\u1FF1\u1FF5\u1FFF\u200B-\u200F\u202A-\u202E\u2060-\u206F\u2072\u2073\u208F\u209D-\u209F\u20BB-\u20CF\u20F1-\u20FF\u218A-\u218F\u23F4-\u23FF\u2427-\u243F\u244B-\u245F\u2700\u2B4D-\u2B4F\u2B5A-\u2BFF\u2C2F\u2C5F\u2CF4-\u2CF8\u2D26\u2D28-\u2D2C\u2D2E\u2D2F\u2D68-\u2D6E\u2D71-\u2D7E\u2D97-\u2D9F\u2DA7\u2DAF\u2DB7\u2DBF\u2DC7\u2DCF\u2DD7\u2DDF\u2E3C-\u2E7F\u2E9A\u2EF4-\u2EFF\u2FD6-\u2FEF\u2FFC-\u2FFF\u3040\u3097\u3098\u3100-\u3104\u312E-\u3130\u318F\u31BB-\u31BF\u31E4-\u31EF\u321F\u32FF\u4DB6-\u4DBF\u9FCD-\u9FFF\uA48D-\uA48F\uA4C7-\uA4CF\uA62C-\uA63F\uA698-\uA69E\uA6F8-\uA6FF\uA78F\uA794-\uA79F\uA7AB-\uA7F7\uA82C-\uA82F\uA83A-\uA83F\uA878-\uA87F\uA8C5-\uA8CD\uA8DA-\uA8DF\uA8FC-\uA8FF\uA954-\uA95E\uA97D-\uA97F\uA9CE\uA9DA-\uA9DD\uA9E0-\uA9FF\uAA37-\uAA3F\uAA4E\uAA4F\uAA5A\uAA5B\uAA7C-\uAA7F\uAAC3-\uAADA\uAAF7-\uAB00\uAB07\uAB08\uAB0F\uAB10\uAB17-\uAB1F\uAB27\uAB2F-\uABBF\uABEE\uABEF\uABFA-\uABFF\uD7A4-\uD7AF\uD7C7-\uD7CA\uD7FC-\uF8FF\uFA6E\uFA6F\uFADA-\uFAFF\uFB07-\uFB12\uFB18-\uFB1C\uFB37\uFB3D\uFB3F\uFB42\uFB45\uFBC2-\uFBD2\uFD40-\uFD4F\uFD90\uFD91\uFDC8-\uFDEF\uFDFE\uFDFF\uFE1A-\uFE1F\uFE27-\uFE2F\uFE53\uFE67\uFE6C-\uFE6F\uFE75\uFEFD-\uFF00\uFFBF-\uFFC1\uFFC8\uFFC9\uFFD0\uFFD1\uFFD8\uFFD9\uFFDD-\uFFDF\uFFE7\uFFEF-\uFFFB\uFFFE\uFFFF]/g;
Just at the beginning.... in the segment in between \0 and \x1f there are two characters: \x0d and \x0a that are being included there (those are the guilty ones). just change the beginning from /[\0-\x1f to /\0-\x09\x0b\x0c\x0e-\x1f If you want also to conserve tabs, then don't include \x09 also.
This works for me
https://regex101.com/r/nH2rY4/1
However, in regex101 if you go to the end of the expression and accidentally add a carriage return, it will match it. Compare it with this:
https://regex101.com/r/nH2rY4/2
Based on this question here (not a duplicate I think, there is no linebreak issue): Replace unicode space characters, this is a good start: var re = /(?![ \r\n])\s/. It'll matches all whitespace except for "real" space and new line characters. Now, you'll need to add all the accepted characters (letters, numbers, punctuation...) and negate them so anything that is not in the accepted list will match.
Newlines and carriage returns are both included in \0-\x1F.
Newline \n is \x0A.
Carriage return \r is \x0D.
If you want to exclude them from your character class:
var re = /[\0-\x09\x0B\x0C\x0E-\x1F\x7F-\x9F\xAD\u0378\u0379\u037F-\u0383\u038B\u038D\u03A2\u0528-\u0530\u0557\u0558\u0560\u0588\u058B-\u058E\u0590\u05C8-\u05CF\u05EB-\u05EF\u05F5-\u0605\u061C\u061D\u06DD\u070E\u070F\u074B\u074C\u07B2-\u07BF\u07FB-\u07FF\u082E\u082F\u083F\u085C\u085D\u085F-\u089F\u08A1\u08AD-\u08E3\u08FF\u0978\u0980\u0984\u098D\u098E\u0991\u0992\u09A9\u09B1\u09B3-\u09B5\u09BA\u09BB\u09C5\u09C6\u09C9\u09CA\u09CF-\u09D6\u09D8-\u09DB\u09DE\u09E4\u09E5\u09FC-\u0A00\u0A04\u0A0B-\u0A0E\u0A11\u0A12\u0A29\u0A31\u0A34\u0A37\u0A3A\u0A3B\u0A3D\u0A43-\u0A46\u0A49\u0A4A\u0A4E-\u0A50\u0A52-\u0A58\u0A5D\u0A5F-\u0A65\u0A76-\u0A80\u0A84\u0A8E\u0A92\u0AA9\u0AB1\u0AB4\u0ABA\u0ABB\u0AC6\u0ACA\u0ACE\u0ACF\u0AD1-\u0ADF\u0AE4\u0AE5\u0AF2-\u0B00\u0B04\u0B0D\u0B0E\u0B11\u0B12\u0B29\u0B31\u0B34\u0B3A\u0B3B\u0B45\u0B46\u0B49\u0B4A\u0B4E-\u0B55\u0B58-\u0B5B\u0B5E\u0B64\u0B65\u0B78-\u0B81\u0B84\u0B8B-\u0B8D\u0B91\u0B96-\u0B98\u0B9B\u0B9D\u0BA0-\u0BA2\u0BA5-\u0BA7\u0BAB-\u0BAD\u0BBA-\u0BBD\u0BC3-\u0BC5\u0BC9\u0BCE\u0BCF\u0BD1-\u0BD6\u0BD8-\u0BE5\u0BFB-\u0C00\u0C04\u0C0D\u0C11\u0C29\u0C34\u0C3A-\u0C3C\u0C45\u0C49\u0C4E-\u0C54\u0C57\u0C5A-\u0C5F\u0C64\u0C65\u0C70-\u0C77\u0C80\u0C81\u0C84\u0C8D\u0C91\u0CA9\u0CB4\u0CBA\u0CBB\u0CC5\u0CC9\u0CCE-\u0CD4\u0CD7-\u0CDD\u0CDF\u0CE4\u0CE5\u0CF0\u0CF3-\u0D01\u0D04\u0D0D\u0D11\u0D3B\u0D3C\u0D45\u0D49\u0D4F-\u0D56\u0D58-\u0D5F\u0D64\u0D65\u0D76-\u0D78\u0D80\u0D81\u0D84\u0D97-\u0D99\u0DB2\u0DBC\u0DBE\u0DBF\u0DC7-\u0DC9\u0DCB-\u0DCE\u0DD5\u0DD7\u0DE0-\u0DF1\u0DF5-\u0E00\u0E3B-\u0E3E\u0E5C-\u0E80\u0E83\u0E85\u0E86\u0E89\u0E8B\u0E8C\u0E8E-\u0E93\u0E98\u0EA0\u0EA4\u0EA6\u0EA8\u0EA9\u0EAC\u0EBA\u0EBE\u0EBF\u0EC5\u0EC7\u0ECE\u0ECF\u0EDA\u0EDB\u0EE0-\u0EFF\u0F48\u0F6D-\u0F70\u0F98\u0FBD\u0FCD\u0FDB-\u0FFF\u10C6\u10C8-\u10CC\u10CE\u10CF\u1249\u124E\u124F\u1257\u1259\u125E\u125F\u1289\u128E\u128F\u12B1\u12B6\u12B7\u12BF\u12C1\u12C6\u12C7\u12D7\u1311\u1316\u1317\u135B\u135C\u137D-\u137F\u139A-\u139F\u13F5-\u13FF\u169D-\u169F\u16F1-\u16FF\u170D\u1715-\u171F\u1737-\u173F\u1754-\u175F\u176D\u1771\u1774-\u177F\u17DE\u17DF\u17EA-\u17EF\u17FA-\u17FF\u180F\u181A-\u181F\u1878-\u187F\u18AB-\u18AF\u18F6-\u18FF\u191D-\u191F\u192C-\u192F\u193C-\u193F\u1941-\u1943\u196E\u196F\u1975-\u197F\u19AC-\u19AF\u19CA-\u19CF\u19DB-\u19DD\u1A1C\u1A1D\u1A5F\u1A7D\u1A7E\u1A8A-\u1A8F\u1A9A-\u1A9F\u1AAE-\u1AFF\u1B4C-\u1B4F\u1B7D-\u1B7F\u1BF4-\u1BFB\u1C38-\u1C3A\u1C4A-\u1C4C\u1C80-\u1CBF\u1CC8-\u1CCF\u1CF7-\u1CFF\u1DE7-\u1DFB\u1F16\u1F17\u1F1E\u1F1F\u1F46\u1F47\u1F4E\u1F4F\u1F58\u1F5A\u1F5C\u1F5E\u1F7E\u1F7F\u1FB5\u1FC5\u1FD4\u1FD5\u1FDC\u1FF0\u1FF1\u1FF5\u1FFF\u200B-\u200F\u202A-\u202E\u2060-\u206F\u2072\u2073\u208F\u209D-\u209F\u20BB-\u20CF\u20F1-\u20FF\u218A-\u218F\u23F4-\u23FF\u2427-\u243F\u244B-\u245F\u2700\u2B4D-\u2B4F\u2B5A-\u2BFF\u2C2F\u2C5F\u2CF4-\u2CF8\u2D26\u2D28-\u2D2C\u2D2E\u2D2F\u2D68-\u2D6E\u2D71-\u2D7E\u2D97-\u2D9F\u2DA7\u2DAF\u2DB7\u2DBF\u2DC7\u2DCF\u2DD7\u2DDF\u2E3C-\u2E7F\u2E9A\u2EF4-\u2EFF\u2FD6-\u2FEF\u2FFC-\u2FFF\u3040\u3097\u3098\u3100-\u3104\u312E-\u3130\u318F\u31BB-\u31BF\u31E4-\u31EF\u321F\u32FF\u4DB6-\u4DBF\u9FCD-\u9FFF\uA48D-\uA48F\uA4C7-\uA4CF\uA62C-\uA63F\uA698-\uA69E\uA6F8-\uA6FF\uA78F\uA794-\uA79F\uA7AB-\uA7F7\uA82C-\uA82F\uA83A-\uA83F\uA878-\uA87F\uA8C5-\uA8CD\uA8DA-\uA8DF\uA8FC-\uA8FF\uA954-\uA95E\uA97D-\uA97F\uA9CE\uA9DA-\uA9DD\uA9E0-\uA9FF\uAA37-\uAA3F\uAA4E\uAA4F\uAA5A\uAA5B\uAA7C-\uAA7F\uAAC3-\uAADA\uAAF7-\uAB00\uAB07\uAB08\uAB0F\uAB10\uAB17-\uAB1F\uAB27\uAB2F-\uABBF\uABEE\uABEF\uABFA-\uABFF\uD7A4-\uD7AF\uD7C7-\uD7CA\uD7FC-\uF8FF\uFA6E\uFA6F\uFADA-\uFAFF\uFB07-\uFB12\uFB18-\uFB1C\uFB37\uFB3D\uFB3F\uFB42\uFB45\uFBC2-\uFBD2\uFD40-\uFD4F\uFD90\uFD91\uFDC8-\uFDEF\uFDFE\uFDFF\uFE1A-\uFE1F\uFE27-\uFE2F\uFE53\uFE67\uFE6C-\uFE6F\uFE75\uFEFD-\uFF00\uFFBF-\uFFC1\uFFC8\uFFC9\uFFD0\uFFD1\uFFD8\uFFD9\uFFDD-\uFFDF\uFFE7\uFFEF-\uFFFB\uFFFE\uFFFF]/g;
You can use ^$ to to match blank string or \s to match one line i am not sure on which language or you using i will give regualr expression in python and perl
Python
re.sub(r'^$','What ever you want to replace',variable)
re.sub(r'\s','What ever you want to replace',variable)
Perl
re =~ s/^$/What ever you want to replace/sg
re =~ s/\s/What ever you want to replace/sg
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
thanks for coming here, so i've got this little code:
while(/\[del\](.*?)\[\/del\]/i.exec(text) != null)
text = text.replace(/\[del\](.*?)\[\/del\]/i, "<s>$1</s>");
but when there are line breaks, it wont match.
Example:
[del]asdsadasdaasdadsadsadsadasdsadsa[/del] - this won't be matched
I'm really new to regex, so what I'm doing wrong?
By default in many regex flavors, the dot doesn't match the newline character. Javascript doesn't have the singleline modifier (?s) to change this behaviour. The most current trick to match all characters including newlines is to use [\s\S] that matches all that is a whitespace character and all that is not a whitespace character.
As an aside comment, you don't need to put the replace method in a while loop, since the replace will only perform a replacement if something is found. If you want to replace all occurences, just add the g command at the end of the pattern.
text = text.replace(/\[del\]([\s\S]*?)\[\/del\]/ig, "<s>$1</s>");
Note that for this specific replacement, since your del tag doesn't seem to have parameters, you can simply write:
text = text.replace(/\[(\/?)del\]/ig, "<$1s>");
(it avoids a lot of work)