This is my string converted into javascript object.
{"text" : "Must consist of alphabetical characters and spaces only", regexp:"/^[a-z\\s]+$/i"}
I need regexp to use it for validation but it won’t work because of the double quotes and \s escape sequence.
To make it work the value of regexp must be {"text" : "Must consist of alphabetical characters and spaces only", regexp : /^[a-z\s]+$/i}.
I also used this new RegExp(object.regexp) and any other way I can possibly think but with no luck at all.
Any help is appreciated!
Try split-ing out the part that you want, before putting it into the new RegExp constructor:
var regexVariable = new RegExp(object.regexp.split("/")[1]);
That will trim off the string representation of the regex "boundaries", as well as the "i" flag, and leave you with just the "guts" of the regex.
Pushing the result of that to the console results in the following regex: /^[a-z\s]+$/
Edit:
Not sure if you want to "read" the case insensitivity from the value in the object or not, but, if you do, you can expand the use of the split a little more to get any flags included automatically:
var aRegexParts = object.regexp.split("/");
var regexVariable = new RegExp(aRegexParts[1], aRegexParts[2]);
Logging that in the console results in the first regex that I posted, but with the addition of the "i" flag: /^[a-z\s]+$/i
Borrowing the example #RoryMcCrossan made, you can use a regular expression to parse your regular expression.
var object = {
"text": "Must consist of alphabetical characters and spaces only",
"regexp": "/^[a-z\\s]+$/i"
}
// parse out the main regex and any additional flags.
var extracted_regex = object.regexp.match(/\/(.*?)\/([ig]+)?/);
var re = new RegExp(extracted_regex[1], extracted_regex[2]);
// don't use document.write in production! this is just so that it's
// easier to see the values in stackoverflow's editor.
document.write('<b>regular expression:</b> ' + re + '<br>');
document.write('<b>string:</b> ' + object.text + '<br>');
document.write('<b>evaluation:</b> ' + re.test(object.text));
not used regex in Java but the regular expression itself should look something like :
"^([aA-zZ] | \s)*$"
If Java uses regular expression as I am used to them [a-z] will only capture lowercase characters
Hope this helps even if it's just a little (would add this as a comment instead of answer but need 50 rep)
Related
I have a code that generates a random letter based on the word and I have tried to create a RegExp code to turn all the letters from the word to '_' except the randomly generated letter from the word.
const word = "Apple is tasty"
const randomCharacter = word[Math.floor(Math.random() * word.length)]
regex = new RegExp(/[^${randomCharacter}&\/\\#,+()$~%.'":;*?<>{}\s]/gi)
hint = word.replace(regex,'_')
I want to change all the letters to '_' except the randomly generated word. The above code for some reason does not work and shows the result: A___e __ ta_t_ and I'm not able to figure out what to do.
The final result I want is something like this: A____ __ _a___
Is there a way with regex to change all the alphabets and numbers '/[^a-zA-Z0-9]/g' to '_' except the randomly generated letter?
I'm listing all the expressions I want to include on my above code because I'm not able to figure out a way to do include and exclude at the same time using the variable with regex.
You can't do string interpolation inside of a RegExp literal (/.../). Meaning your placeholder ${randomCharacter} will not evaluate to its value in the template, but is instead interpreted literally as the string "${randomCharacter}".
If you want to use template literals, initialize your regex variable with a RegExp constructor instead, like:
const regex = new RegExp(`[^${randomCharacter}&\\/\\\#,+()$~%.'":;*?<>{}\\s]`, "gi");
See the MDN RegExp documentation for an explanation on the differences between the literal notation and constructor function, most notably:
The constructor of the regular expression object [...] results in runtime compilation of the regular expression. Use the constructor function when [...] you don't know the pattern and obtain it from another source, such as user input.
/(?:[^A\s])/
test it on regex101
just replace A in [^A\s] with you character that you want to ommit from replacement
demo:
const word = "Apple is tasty";
const randomCharacter = 'a';//word[Math.floor(Math.random() * word.length)];
regex = new RegExp('(?:[^' + randomCharacter + '\\s])', 'gi');
hint = word.replaceAll(regex, '_');
console.log(hint)
I am developing a small character encoder generator where the user input their text and on the click of a button, it outputs the encoded version.
I've defined an object of the characters that need to be encoded like so:
map = {
'©' : '©',
'&' : '&'
},
And here is the loop that gets the values from the map and replaces them:
Object.keys(map).forEach(function (ico) {
var icoE = ico.replace(/([.?*+^$[\]\\(){}|-])/g, "\\$1");
raw = raw.replace( new RegExp(icoE, 'g'), map[ico] );
});
I am them simply outputting the result to a textarea. This all works fine, however the problem I'm facing is this.
© is replaced with © however the & symbol at the beginning of this is then converted to & so it ends up being ©.
I see why this is happening however I'm not sure how to go about ensuring that & is not replaced within character encoded strings.
Here is a JSFiddle for a live preview of what I mean:
http://jsfiddle.net/4m3nw/1/
Any help would be much appreciated
Prelude: Apart from regex, an idea worth considering is something like this JS function that already handles html entities. Now, on to the regex question.
HTML Special Characters, Negative Lookahead
In HTML, special characters can look not only like © but also like —, and they can have upper-case characters.
To replace ampersands that are not immediately followed by a hash or word characters and a semicolon, you can use something like this:
&(?!(?:#[0-9]+|[a-z]+);)
See the demo.
Make sure to use the i flag to activate case-insensitive mode
& matches the literal ampersand
The negative lookahead (?!(?:#[0-9]+|[a-z]+);) asserts that it is not followed by...
(?:#[0-9]+|[a-z]+) a hash and digits, | OR letters...
then a semicolon.
Reference
Lookahead and Lookbehind Zero-Length Assertions
Mastering Lookahead and Lookbehind
The problem is that since you process the same string you replace the &in ©. If you re-order your map then that seemingly solves the problem. However according to the ECMAScript specifications, this is not a given, so you would be relying on implementation details of the ECMAScript engine used.
What you can do to make sure it will always work is to swap the keys so that & is always processed first:
map = {
'©' : '©',
'&' : '&'
};
var keys = Object.keys(map);
keys[keys.indexOf('&')] = keys[0];
keys[0] = '&';
keys.forEach(function (ico) {
var icoE = ico.replace(/([.?*+^$[\]\\(){}|-])/g, "\\$1");
raw = raw.replace( new RegExp(icoE, 'g'), map[ico] );
});
Obviously you need to add checks for the &'s existence if it isn't always there.
jsFiddle Demo.
Probably the simplest code change is to reorder your map by putting the ampersand on top.
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+
My objective is to replace all characters which are not dash (-) or not number or not letters in any language in a string.All of the #!()[], and all other signs to be replaced with empty string. All occurences of - should not be replaced also.
I have used for this the XRegExp plugin but it seems I cannot find the magic solution :)
I have tryed like this :
var txt = "Ad СТИНГ (ALI) - Englishmen In New York";
var regex = new XRegExp('\\p{^N}\\p{^L}',"g");
var b = XRegExp.replace(txt, regex, "")
but the result is : AСТИН(AL EnglishmeINeYork ... which is kind of weird
If I try to add also the condition for not removing the '-' character leads to make the RegEx invalid.
\\p{^N}\\p{^L} means a non-number followed by a non-letter.
Try [^\\p{N}\\p{L}-] that means a non-number, non-letter, non-dash.
A jsfiddle where to do some tests... The third XRegExp is the one you asked.
\p{^N}\p{^L}
is a non-number followed by a non-letter. You probably meant to say a character that is neither a letter nor a number:
[^\p{N}\p{L}]
// all non letters/numbers in a string => /[^a-zA-z0-9]/g
I dont know XRegExp.
but in js Regexp you can replace it by
b.replace(/[^a-zA-z0-9]/g,'')
My problem start with like-
var str='0|31|2|03|.....|4|2007'
str=str.replace(/[^|]\d*[^|]/,'5');
so the output becomes like:"0|5|2|03|....|4|2007" so it replaces 31->5
But this doesn't work for replacing other segments when i change code like this:
str=str.replace(/[^|]{2}\d*[^|]/,'6');
doesn't change 2->6.
What actually i am missing here.Any help?
I think a regular expression is a bad solution for that problem. I'd rather do something like this:
var str = '0|31|2|03|4|2007';
var segments = str.split("|");
segments[1] = "35";
segments[2] = "123";
Can't think of a good way to solve this with a regexp.
Here is a specific regex solution which replaces the number following the first | pipe symbol with the number 5:
var re = /^((?:\d+\|){1})\d+/;
return text.replace(re, '$15');
If you want to replace the digits following the third |, simply change the {1} portion of the regex to {3}
Here is a generalized function that will replace any given number slot (zero-based index), with a specified new number:
function replaceNthNumber(text, n, newnum) {
var re = new RegExp("^((?:\\d+\\|){"+ n +'})\\d+');
return text.replace(re, '$1'+ newnum);
}
Firstly, you don't have to escape | in the character set, because it doesn't have any special meaning in character sets.
Secondly, you don't put quantifiers in character sets.
And finally, to create a global matching expression, you have to use the g flag.
[^\|] means anything but a '|', so in your case it only matches a digit. So it will only match anything with 2 or more digits.
Second you should put the {2} outside of the []-brackets
I'm not sure what you want to achieve here.