i tried to replace a string, provided the string regex with a value that has $ in the end.
Can anyone tell me what is happening in
Looking into mdn string replace docs, i found it is expected.
But what should one do he want to ignore this.
Means i want the replacing value should get as it is replaced, with 4 $s here.
You need to double the dollars in the replacement pattern as $$ are actually $.
.replace(/{{one}}/g, '000$$$$$$$$')
See String#replace help:
Pattern Inserts
$$ Inserts a "$".
If a user types $ in the replacement (that is, in case it is user-defined) you can just double it:
var ptrn = "{{one}}"; // regex pattern from user input
var repl = "000$$$$"; // replacement from user input
var rx = RegExp(ptrn, "g"); // building a dynamic regex
document.write("pp{{one}}pp".replace(rx, repl.replace(/\$/g, '$$$$')));
// ^--- doubling $s-----^
Related
I have a code that generates a random letter based on the word and I have tried to create a RegExp code to turn all the letters from the word to '_' except the randomly generated letter from the word.
const word = "Apple is tasty"
const randomCharacter = word[Math.floor(Math.random() * word.length)]
regex = new RegExp(/[^${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 passing a URL to a block of code in which I need to insert a new element into the regex. Pretty sure the regex is valid and the code seems right but no matter what I can't seem to execute the match for regex!
//** Incoming url's
//** url e.g. api/223344
//** api/11aa/page/2017
//** Need to match to the following
//** dir/api/12ab/page/1999
//** Hence the need to add dir at the front
var url = req.url;
//** pass in: /^\/api\/([a-zA-Z0-9-_~ %]+)(?:\/page\/([a-zA-Z0-9-_~ %]+))?$/
var re = myregex.toString();
//** Insert dir into regex: /^dir\/api\/([a-zA-Z0-9-_~ %]+)(?:\/page\/([a-zA-Z0-9-_~ %]+))?$/
var regVar = re.substr(0, 2) + 'dir' + re.substr(2);
var matchedData = url.match(regVar);
matchedData === null ? console.log('NO') : console.log('Yay');
I hope I am just missing the obvious but can anyone see why I can't match and always returns NO?
Thanks
Let's break down your regex
^\/api\/ this matches the beginning of a string, and it looks to match exactly the string "/api"
([a-zA-Z0-9-_~ %]+) this is a capturing group: this one specifically will capture anything inside those brackets, with the + indicating to capture 1 or more, so for example, this section will match abAB25-_ %
(?:\/page\/([a-zA-Z0-9-_~ %]+)) this groups multiple tokens together as well, but does not create a capturing group like above (the ?: makes it non-captuing). You are first matching a string exactly like "/page/" followed by a group exactly like mentioned in the paragraph above (that matches a-z, A-Z, 0-9, etc.
?$ is at the end, and the ? means capture 0 or more of the precending group, and the $ matches the end of the string
This regex will match this string, for example: /api/abAB25-_ %/page/abAB25-_ %
You may be able to take advantage of capturing groups, however, and use something like this instead to get similar results: ^\/api\/([a-zA-Z0-9-_~ %]+)\/page\/\1?$. Here, we are using \1 to reference that first capturing group and match exactly the same tokens it is matching. EDIT: actually, this probably won't work, since the text after /api/ and the text after /page/ will most likely be different, carrying on...
Afterwards, you are are adding "dir" to the beginning of your search, so you can now match someting like this: dir/api/abAB25-_ %/page/abAB25-_ %
You have also now converted the regex to a string, so like Crayon Violent pointed out in their comment, this will break your expected funtionality. You can fix this by using .source on your regex: var matchedData = url.match(regVar.source); https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/source
Now you can properly match a string like this: dir/api/11aa/page/2017 see this example: https://repl.it/Mj8h
As mentioned by Crayon Violent in the comments, it seems you're passing a String rather than a regular expression in the .match() function. maybe try the following:
url.match(new RegExp(regVar, "i"));
to convert the string to a regular expression. The "i" is for ignore case; don't know that's what you want. Learn more here:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp
I want to make the first character of every word in a string to uppercase.
i am referring to this article Replacement Text Case Conversion.
when i am running the regular expression ([a-zA-Z])([a-zA-Z](\s)) with the replacement text as \u$1\l$2 in my editor (sublime text) it works fine.
However, when i am trying to do the same in javascript using replace method as below, its giving syntax errors and hence fails.
var regex = /([a-zA-Z])([a-zA-Z]*(\s)*)/gi;
var rep = "\\u$1\\l$2"; // here it is giving error
var result = input.replace(regex,rep);
How to resolve this?
I know this problem can be solved using charAt() and toUppercase() method. but I want to do it using regex with replace. :)
JS regex engine does not support lower- and uppercasing operators \u and \l in the replacement patterns. Use a callback inside String#replace:
var input = "aZ";
var regex = /([a-zA-Z])([a-zA-Z]*(\s)*)/gi;
var result = input.replace(regex, function($0,$1,$2,$3) {
return $1.toUpperCase() + $2.toLowerCase();
});
console.log(result);
Note that you can reduce your pattern to /([a-z])([a-z]*\s*)/gi.
I am holding in a field the validation format that I would need.
I need to convert different ## into a regex validation.
Is there a simple replace that can do this for me.
for example, i need to validate the account number.
sometimes it might need to be ###-###, or I'll get ####### or ##-####.
depending what is in the id="validationrule" field
I'm looking for
regex = $('#validationrule').replace("#", "[0/9]");
It also has to take into consideration that sometimes there is a dash in there.
Your question seems to be about creating regexes from a string variable (which you get from an input field that specifies the validation format).
"###-###" might turn into /^\d{3}\-\d{3}$/
"#######" might turn into /^\d{7}$/
If your validation format is built from the 2 characters # and -, this would work:
function createValidationRegEx(format){
format = format
.replace(/[^#\-]/g, '') //remove other chars
.replace(/#/g, '\\d') //convert # to \d
.replace(/\-/g, '\\-'); //convert - to \-
return new RegExp('^' + format + '$', 'g');
}
//create regexes
var format1 = createValidationRegEx('###-###');
var format2 = createValidationRegEx('#######');
//test regexes
console.log(format1.test('123-456')); // true
console.log(format2.test('123-456')); // false
console.log(format1.test('1234567')); // false
console.log(format2.test('1234567')); // true
Please note that you need to pay attention to which characters needs to be escaped when creating regexes from strings. This answer provides more details about how to solve this more generally, if you want to build more complex solutions.
If you are trying to replace the .value of an <input> element you can use .val(function), return replacement string from .replace() inside of function, chain .val() to assign result to regex. Use RegExp constructor with g flag to replace all matches of the RegExp supplied to .replace() to match characters against at string.
var regex = $("#validationrule").val(function(_, val) {
return val.replace("#", "[0/9]");
}).val();
console.log(regex);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js">
</script>
<input id="validationrule" value="#">
I am doing it wrong. I know.
I want to assign the matched text that is the result of a regex to a string var.
basically the regex is supposed to pull out anything in between two colons
so blah:xx:blahdeeblah
would result in xx
var matchedString= $(current).match('[^.:]+):(.*?):([^.:]+');
alert(matchedString);
I am looking to get this to put the xx in my matchedString variable.
I checked the jquery docs and they say that match should return an array. (string char array?)
When I run this nothing happens, No errors in the console but I tested the regex and it works outside of js. I am starting to think I am just doing the regex wrong or I am completely not getting how the match function works altogether
I checked the jquery docs and they say that match should return an array.
No such method exists for jQuery. match is a standard javascript method of a string. So using your example, this might be
var str = "blah:xx:blahdeeblah";
var matchedString = str.match(/([^.:]+):(.*?):([^.:]+)/);
alert(matchedString[2]);
// -> "xx"
However, you really don't need a regular expression for this. You can use another string method, split() to divide the string into an array of strings using a separator:
var str = "blah:xx:blahdeeblah";
var matchedString = str.split(":"); // split on the : character
alert(matchedString[1]);
// -> "xx"
String.match
String.split