Hey. First question here, probably extremely lame, but I totally suck in regular expressions :(
I want to extract the text from a series of strings that always have only alphabetic characters before and after a hyphen:
string = "some-text"
I need to generate separate strings that include the text before AND after the hyphen. So for the example above I would need string1 = "some" and string2 = "text"
I found this and it works for the text before the hyphen, now I only need the regex for the one after the hyphen.
Thanks.
You don't need regex for that, you can just split it instead.
var myString = "some-text";
var splitWords = myString.split("-");
splitWords[0] would then be "some", and splitWords[1] will be "text".
If you actually have to use regex for whatever reason though - the $ character marks the end of a string in regex, so -(.*)$ is a regex that will match everything after the first hyphen it finds till the end of the string. That could actually be simplified that to just -(.*) too, as the .* will match till the end of the string anyway.
Related
I have a username field in my form. I want to not allow spaces anywhere in the string. I have used this regex:
var regexp = /^\S/;
This works for me if there are spaces between the characters. That is if username is ABC DEF. It doesn't work if a space is in the beginning, e.g. <space><space>ABC. What should the regex be?
While you have specified the start anchor and the first letter, you have not done anything for the rest of the string. You seem to want repetition of that character class until the end of the string:
var regexp = /^\S*$/; // a string consisting only of non-whitespaces
Use + plus sign (Match one or more of the previous items),
var regexp = /^\S+$/
If you're using some plugin which takes string and use construct Regex to create Regex Object i:e new RegExp()
Than Below string will work
'^\\S*$'
It's same regex #Bergi mentioned just the string version for new RegExp constructor
This will help to find the spaces in the beginning, middle and ending:
var regexp = /\s/g
This one will only match the input field or string if there are no spaces. If there are any spaces, it will not match at all.
/^([A-z0-9!##$%^&*().,<>{}[\]<>?_=+\-|;:\'\"\/])*[^\s]\1*$/
Matches from the beginning of the line to the end. Accepts alphanumeric characters, numbers, and most special characters.
If you want just alphanumeric characters then change what is in the [] like so:
/^([A-z])*[^\s]\1*$/
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
Given an input text such where all spaces are replaced by n _ :
Hello_world_?. Hello_other_sentenc3___. World___________.
I want to keep the _ between words, but I want to stick each punctuation back to the last word of a sentence without any space between last word and punctuation. I want to use the the punctuation as pivot of my regex.
I wrote the following JS-Regex:
str = str.replace(/(_| )*([:punct:])*( |_)/g, "$2$3");
This fails, since it returns :
Hello_world_?. Hello_other_sentenc3_. World_._
Why it doesn't works ? How to delete all "_" between the last word and the punctuation ?
http://jsfiddle.net/9c4z5/
Try the following regex, which makes use of a positive lookahead:
str = str.replace(/_+(?=\.)/g, "");
It replaces all underscores which are immediately followed by a punctuation character with the empty string, thus removing them.
If you want to match other punctuation characters than just the period, replace the \. part with an appropriate character class.
JavaScript doesn't have :punct: in its regex implementation. I believe you'd have to list out the punctuation characters you care about, perhaps something like this:
str = str.replace(/(_| )+([.,?])/g, "$2");
That is, replace any group of _ or space that is immediately followed by punctation with just the punctuation.
Demo: http://jsfiddle.net/9c4z5/2/
the following regex
var str = "1234,john smith,jack jone";
var match = str.match(/([^,]*,[^,]*,[^ ]*)/g);
alert(match);
returns
1234,john smith,jack
But what I am trying to get is the whole string which is
1234,john smith,jack jones
Basically my script does the job only for the first whitespace between commas but I want to do it everytime there is a white space between commas.
Can anyone help me out pls.
Your pattern excludes spaces from the last section so as soon as it encounters a space in after the third comma, that's the end of the match. You might want to try this instead:
var match = str.match(/[^,]*,[^,]*,.*/g);
This will allow anything after the second comma, including spaces or more commas (since your original pattern allowed commas after the the second).
If you'd like to match pattern only on a single, use start / end anchors (^ / $) as well as the multiline flag (m), like us this:
var match = str.match(/^[^,]*,[^,]*,.*$/mg);
You can try it out with this simple demo.
Why you're not using split ?
"1234,john smith,jack jone".split(/,/)
or
"1234,john smith,jack jone".split(",")
hello I am trying what I thought would be a rather easy regex in Javascript but is giving me lots of trouble.
I want the ability to split a date via javascript splitting either by a '-','.','/' and ' '.
var date = "02-25-2010";
var myregexp2 = new RegExp("-.");
dateArray = date.split(myregexp2);
What is the correct regex for this any and all help would be great.
You need the put the characters you wish to split on in a character class, which tells the regular expression engine "any of these characters is a match". For your purposes, this would look like:
date.split(/[.,\/ -]/)
Although dashes have special meaning in character classes as a range specifier (ie [a-z] means the same as [abcdefghijklmnopqrstuvwxyz]), if you put it as the last thing in the class it is taken to mean a literal dash and does not need to be escaped.
To explain why your pattern didn't work, /-./ tells the regular expression engine to match a literal dash character followed by any character (dots are wildcard characters in regular expressions). With "02-25-2010", it would split each time "-2" is encountered, because the dash matches and the dot matches "2".
or just (anything but numbers):
date.split(/\D/);
you could just use
date.split(/-/);
or
date.split('-');
Say your string is:
let str = `word1
word2;word3,word4,word5;word7
word8,word9;word10`;
You want to split the string by the following delimiters:
Colon
Semicolon
New line
You could split the string like this:
let rawElements = str.split(new RegExp('[,;\n]', 'g'));
Finally, you may need to trim the elements in the array:
let elements = rawElements.map(element => element.trim());
Then split it on anything but numbers:
date.split(/[^0-9]/);
or just use for date strings 2015-05-20 or 2015.05.20
date.split(/\.|-/);
try this instead
date.split(/\W+/)