I am complete novice at regex and Javascript. I have the following problem: need to check into a textfield the existence of one (1) or many (n) consecutive * (asterisk) character/characters eg. * or ** or *** or infinite (n) *. Strings allowed eg. *tomato or tomato* or **tomato or tomato** or as many(n)*tomato many(n)*. So, far I had tried the following:
var str = 'a string'
var value = encodeURIComponent(str);
var reg = /([^\s]\*)|(\*[^\s])/;
if (reg.test(value) == true ) {
alert ('Watch out your asterisks!!!')
}
By your question it's hard to decipher what you're after... But let me try:
Only allow asterisks at beginning or at end
If you only allow an arbitrary number (at least one) of asterisks either at the beginning or at the end (but not on both sides) like:
*****tomato
tomato******
but not **tomato*****
Then use this regular expression:
reg = /^(?:\*+[^*]+|[^*]+\*+)$/;
Match front and back number of asterisks
If you require that the number of asterisks at the biginning matches number of asterisks at the end like
*****tomato*****
*tomato*
but not **tomato*****
then use this regular expression:
reg = /^(\*+)[^*]+\1$/;
Results?
It's unclear from your question what the results should be when each of these regular expressions match? Are strings that test positive to above regular expressions fine or wrong is on you and your requirements. As long as you have correct regular expressions you're good to go and provide the functionality you require.
I've also written my regular expressions to just exclude asterisks within the string. If you also need to reject spaces or anything else simply adjust the [^...] parts of above expressions.
Note: both regular expressions are untested but should get you started to build the one you actually need and require in your code.
If I understand correctly you're looking for a pattern like this:
var pattern = /\**[^\s*]+\**/;
this won't match strings like ***** or ** ***, but will match ***d*** *d or all of your examples that you say are valid (***tomatos etc).If I misunderstood, let me know and I'll see what I can do to help. PS: we all started out as newbies at some point, nothing to be ashamed of, let alone apologize for :)
After the edit to your question I gather the use of an asterisk is required, either at the beginning or end of the input, but the string must also contain at least 1 other character, so I propose the following solution:
var pattern = /^\*+[^\s*]+|[^\s*]+\*+$/;
'****'.match(pattern);//false
' ***tomato**'.match(pattern);//true
If, however *tomato* is not allowed, you'll have to change the regex to:
var pattern = /^\*+[^\s*]+$|^[^\s*]+\*+$/;
Here's a handy site to help you find your way in the magical world of regular expressions.
Related
How can I remove non-printable unicode characters in a multi-language input?
When users with different localizations paste strings they will sometimes unintentionally embed non-printing characters. For example:
var weird = "%E2%80%AA%E2%80%8ETest%E2%80%AC"
var displaysAs = decodeURI(weird); // Users see only "Test"
But I can't figure out how to strip the non-printing characters in a way that doesn't impact other languages like these:
encodeURI("شنط") = "%D8%B4%D9%86%D8%B7"
encodeURI("戦艦帝国") = "%E6%88%A6%E8%89%A6%E5%B8%9D%E5%9B%BD"
For example, the following attempt to repair the weird example above doesn't work:
var weird = "%E2%80%AA%E2%80%8ETest%E2%80%AC";
var displaysAs = decodeURI(weird);
var stillWeird = encodeURI(displaysAs.replace(/\s/g, ""));
// value is again "%E2%80%AA%E2%80%8ETest%E2%80%AC"
console.log('before:', weird);
console.log('after:', displaysAs);
console.log('again:', stillWeird);
.as-console-wrapper{min-height:100%}
As noted in the comments, this is largely a specification problem. I don't have an enumeration of non-printing unicode expressions. I can only observe that one can paste a unicode string into a browser Input and not be aware that it has undisplayed characters in it. I assume that some logic in the browser determines whether each unicode character will display something. This problem would be solved if I can apply that same logic to the underlying string in order to get the "display string."
Put another way: For any two unicode strings that look identical on the browser, I need a transformation that guarantees that their values are identical.
You can use the regular expression found in this other answer.
Example with array of the three strings provided in the question:
let weird = [
"%E2%80%AA%E2%80%8ETest%E2%80%AC",
"%D8%B4%D9%86%D8%B7",
"%E6%88%A6%E8%89%A6%E5%B8%9D%E5%9B%BD"
];
const expr = /[\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;
weird.map(decodeURI).forEach(el => {
let trimmed = el.replace(expr, '')
console.log(trimmed, trimmed.length);
});
If you only want to trim these non-printable characters from the beginning and/or end of the string, you'll need to assert start (^) and end ($) in the regular expression.
I am trying to use regex in JavaScript to verify if a given password has alphabetic, numeric, and a special character. However, everything I have tried doesn't work
I have tried using
/^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])(?=.*?[!#"#%$&])[A-Za-z0-9!#"#%$&]{8,30}$/gm
and creating separate regex variables for alphabetic, numeric, and special characters:
let alpha = /^[A-Za-z]+$/i
let numer = /^[0-9]+$/i
let special = /^[!##$%^&*(),.?;":{}|<>']/i
Picture of My Code
When I log the password.match(regex) to the console I always see null
Try this:
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[^\da-zA-Z]).{8,15}$
This is from a post a long time ago. There are probably others as well. A simple search of SO will find more examples.1
Here is a visualisation of the regular expression:
Debuggex Demo
I would personally go for separate regexes, the issue with your current regexes is that you have ^ at the start and $ at the end. Which means that the password must only contain [A-Za-z] from start to finish. Then you check if the password only contains [0-9] from start to finish.
const regexes = {
alpha: /[A-Za-z]/,
number: /[0-9]/,
special: /[!##$%^&*(),.?;":{}|<>']/,
length: /^.{8,30}$/
};
["dgXHUYuDdp", "zMv4qQfZj3", "4JXyrsq!J0", "a5Z!"].forEach(password => {
let valid = Object.values(regexes).every(regex => password.match(regex));
console.log(
"password: " + JSON.stringify(password) + "\n" +
"valid: " + valid
);
});
While I suspect it may be possible to write a regex which will do position/seuqnce independent matching, my head hurts just thinking about. So even if I could work out a way of doing it, I would not implement it - code needs to be readable and parseable by human beings. Looking at what you have presented here, I think I'm a lot more familiar with regexes than you are - so even more reason not to do this.
Your alpha / numer / special regexes will only match a string containing letters, numbers or special characters, not a mixture. If you change them thus, then you can check for a match of all three (and escape the meta characters in the special regex):
let alpha = /[A-Za-z]/i;
let numer = /[0-9]/;
let special = /[!##$%\^&*(),\.?;":{}|<>\']/;
if (password.match(alpha) && password.match(numer) && password.match(speicial)) {
Hopefully a simple one!
I've been trying to get this to work for several hours now but am having no luck, as I'm fairly new to regexp I may be missing something very obvious here and was hoping someone could point me in the right direction. The pattern I want to match is as follows: -
At least 1 or more numbers + "##" + at least 1 or more numbers + "##" + at least 1 or more numbers
so a few examples of valid combinations would be: -
1##2##3
123#123#123
0##0##0
A few invalid combinations would be
a##b##c
1## ##1
I've got the following regexp like so: -
[\d+]/#/#[\d+]/#/#[\d+]
And am using it like so (note the double slashes as its inside a string): -
var patt = new RegExp("[\\d+]/#/#[\\d+]/#/#[\\d+]");
if(newFieldValue!=patt){newFieldValue=="no match"}
I also tried these but still nothing: -
if(!patt.text(newFieldValue)){newFieldValue==""}
if(patt.text(newFieldValue)){}else{newFieldValue==""}
But nothing I try is matching, where am I going wrong here?
Any pointers gratefully received, cheers!
1) I can't see any reason to use the RegExp constructor over a RegExp literal for your case. (The former is used primarily where the pattern needs to by dynamic, i.e. is contributed to by variables.)
2) You don't need a character class if there's only one type of character in it (so \d+ not [\d+]
3) You are not actually checking the pattern against the input. You don't apply RegEx by creating an instance of it and using ==; you need to use test() or match() to see if a match is made (the former if you want to check only, not capture)
4) You have == where you mean to assign (=)
if (!/\d+##\d+##\d+/.test(newFieldValue)) newFieldValue = "no match";
You put + inside the brackets, so you're matching a single character that's either a digit or +, not a sequence of digits. I also don't understand why you have / before each #, your description doesn't mention anything about this character.
Use:
var patt = /\d+##\d+##\d+/;
You should use the test method of the pat regex
if (!patt.test(newFieldValue)){ newFieldValue=="no match"; }
once you have a valid regular expression.
Try this regex :
^(?:\d+##){2}\d+$
Demo: http://regex101.com/r/mE8aG7
With the following regex
[\d+]/#/#[\d+]/#/#[\d+]
You would only match things like:
+/#/#5/#/#+
+/#/#+/#/#+
0/#/#0/#/#0
because the regex engine sees it like on the schema below:
Something like:
((-\s)?\d+##)+\d+
I've got some string that contain invisible characters, but they are in somewhat predictable places. Typically the surround the piece of text I want to extract, and then after the 2nd occurrence I want to keep the rest of the text.
I can't seem to figure out how to both key off of the invisible characters, and exclude them from my result. To match invisibles I've been using this regex: /\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F/ which does seem to work.
Here's an example: [invisibles]Keep as match 1[invisibles]Keep as match 2
Here's what I've been using so far without success:
/([\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+)(.+)([\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+)/(.+)
I've got the capture groups in there, but it's bee a while since I've had to use regex's in this way, so I know I'm missing something important. I was hoping to just make the invisible matches non-capturing groups, but it seems that JavaScript does not support this.
Something like this seems like what you want. The second regex you have pretty much works, but the / is in totally the wrong place. Perhaps you weren't properly reading out the group data.
var s = "\x0EKeep as match 1\x0EKeep as match 2";
var r = /[\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+(.+)[\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+(.+)/;
var match = s.match(r);
var part1 = match[1];
var part2 = match[2];
Friends,
I'm new to both Javascript and Regular Expressions and hope you can help!
Within a Javascript function I need to check to see if a comma(,) appears 1 or more times. If it does then there should be one or more numbers either side of it.
e.g.
1,000.00 is ok
1,000,00 is ok
,000.00 is not ok
1,,000.00 is not ok
If these conditions are met I want the comma to be removed so 1,000.00 becomes 1000.00
What I have tried so is:
var x = '1,000.00';
var regex = new RegExp("[0-9]+,[0-9]+", "g");
var y = x.replace(regex,"");
alert(y);
When run the alert shows ".00" Which is not what I was expecting or want!
Thanks in advance for any help provided.
strong text
Edit
strong text
Thanks all for the input so far and the 3 answers given. Unfortunately I don't think I explained my question well enough.
What I am trying to achieve is:
If there is a comma in the text and there are one or more numbers either side of it then remove the comma but leave the rest of the string as is.
If there is a comma in the text and there is not at least one number either side of it then do nothing.
So using my examples from above:
1,000.00 becomes 1000.00
1,000,00 becomes 100000
,000.00 is left as ,000.00
1,,000.00 is left as 1,,000.00
Apologies for the confusion!
Your regex isn't going to be very flexible with higher orders than 1000 and it has a problem with inputs which don't have the comma. More problematically you're also matching and replacing the part of the data you're interested in!
Better to have a regex which matches the forms which are a problem and remove them.
The following matches (in order) commas at the beginning of the input, at the end of the input, preceded by a number of non digits, or followed by a number of non digits.
var y = x.replace(/^,|,$|[^0-9]+,|,[^0-9]+/g,'');
As an aside, all of this is much easier if you happen to be able to do lookbehind but almost every JS implementation doesn't.
Edit based on question update:
Ok, I won't attempt to understand why your rules are as they are, but the regex gets simpler to solve it:
var y = x.replace(/(\d),(\d)/g, '$1$2');
I would use something like the following:
^[0-9]{1,3}(,[0-9]{3})*(\.[0-9]+)$
[0-9]{1,3}: 1 to 3 digits
(,[0-9]{3})*: [Optional] More digit triplets seperated by a comma
(\.[0-9]+): [Optional] Dot + more digits
If this regex matches, you know that your number is valid. Just replace all commas with the empty string afterwards.
It seems to me you have three error conditions
",1000"
"1000,"
"1,,000"
If any one of these is true then you should reject the field, If they are all false then you can strip the commas in the normal way and move on. This can be a simple alternation:
^,|,,|,$
I would just remove anything except digits and the decimal separator ([^0-9.]) and send the output through parseFloat():
var y = parseFloat(x.replace(/[^0-9.]+/g, ""));
// invalid cases:
// - standalone comma at the beginning of the string
// - comma next to another comma
// - standalone comma at the end of the string
var i,
inputs = ['1,000.00', '1,000,00', ',000.00', '1,,000.00'],
invalid_cases = /(^,)|(,,)|(,$)/;
for (i = 0; i < inputs.length; i++) {
if (inputs[i].match(invalid_cases) === null) {
// wipe out everything but decimal and dot
inputs[i] = inputs[i].replace(/[^\d.]+/g, '');
}
}
console.log(inputs); // ["1000.00", "100000", ",000.00", "1,,000.00"]