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How to props array of regex decorators of dynamic strings? Javascript

I need to props few dynamic words regex inside child component with regex.
This is work:
In this situation my regex work but...
I need to regex some dynamic data which I don't know which is word is. All data is in my array.
What I am try?

Use the RegExp class constructor for your dynamic regexp:
const decorators = [
arrayOfWord.map((data) => {
// data is a string value:
// exp: data = Garnosny
// new RegExp(data, 'g') = /Garnosny/g
return { regex: new RegExp(data, 'g'), className: 'some css class' }
}
];

We have a couple of issues here...
1- In the second code snippet, you are missing a return statement in the arrow function passed to "map()".
2- In the third code snippet, you are calling "map()" again on the "data" array. But there is not clear what you are trying to do.
If you can explain a little bit better sure we can help.

Regex to define the number of appearances substituted [duplicate]

I'd like to know how to replace a capture group with its uppercase in JavaScript. Here's a simplified version of what I've tried so far that's not working:
> a="foobar"
'foobar'
> a.replace( /(f)/, "$1".toUpperCase() )
'foobar'
> a.replace( /(f)/, String.prototype.toUpperCase.apply("$1") )
'foobar'
Would you explain what's wrong with this code?

You can pass a function to replace.
var r = a.replace(/(f)/, function(v) { return v.toUpperCase(); });
Explanation
a.replace( /(f)/, "$1".toUpperCase())
In this example you pass a string to the replace function. Since you are using the special replace syntax ($N grabs the Nth capture) you are simply giving the same value. The toUpperCase is actually deceiving because you are only making the replace string upper case (Which is somewhat pointless because the $ and one 1 characters have no upper case so the return value will still be "$1").
a.replace( /(f)/, String.prototype.toUpperCase.apply("$1"))
Believe it or not the semantics of this expression are exactly the same.

I know I'm late to the party but here is a shorter method that is more along the lines of your initial attempts.
a.replace('f', String.call.bind(a.toUpperCase));
So where did you go wrong and what is this new voodoo?
Problem 1
As stated before, you were attempting to pass the results of a called method as the second parameter of String.prototype.replace(), when instead you ought to be passing a reference to a function
Solution 1
That's easy enough to solve. Simply removing the parameters and parentheses will give us a reference rather than executing the function.
a.replace('f', String.prototype.toUpperCase.apply)
Problem 2
If you attempt to run the code now you will get an error stating that undefined is not a function and therefore cannot be called. This is because String.prototype.toUpperCase.apply is actually a reference to Function.prototype.apply() via JavaScript's prototypical inheritance. So what we are actually doing looks more like this
a.replace('f', Function.prototype.apply)
Which is obviously not what we have intended. How does it know to run Function.prototype.apply() on String.prototype.toUpperCase()?
Solution 2
Using Function.prototype.bind() we can create a copy of Function.prototype.call with its context specifically set to String.prototype.toUpperCase. We now have the following
a.replace('f', Function.prototype.apply.bind(String.prototype.toUpperCase))
Problem 3
The last issue is that String.prototype.replace() will pass several arguments to its replacement function. However, Function.prototype.apply() expects the second parameter to be an array but instead gets either a string or number (depending on if you use capture groups or not). This would cause an invalid argument list error.
Solution 3
Luckily, we can simply substitute in Function.prototype.call() (which accepts any number of arguments, none of which have type restrictions) for Function.prototype.apply(). We have now arrived at working code!
a.replace(/f/, Function.prototype.call.bind(String.prototype.toUpperCase))
Shedding bytes!
Nobody wants to type prototype a bunch of times. Instead we'll leverage the fact that we have objects that reference the same methods via inheritance. The String constructor, being a function, inherits from Function's prototype. This means that we can substitute in String.call for Function.prototype.call (actually we can use Date.call to save even more bytes but that's less semantic).
We can also leverage our variable 'a' since it's prototype includes a reference to String.prototype.toUpperCase we can swap that out with a.toUpperCase. It is the combination of the 3 solutions above and these byte saving measures that is how we get the code at the top of this post.

Why don't we just look up the definition?
If we write:
a.replace(/(f)/, x => x.toUpperCase())
we might as well just say:
a.replace('f','F')
Worse, I suspect nobody realises that their examples have been working only because they were capturing the whole regex with parentheses. If you look at the definition, the first parameter passed to the replacer function is actually the whole matched pattern and not the pattern you captured with parentheses:
function replacer(match, p1, p2, p3, offset, string)
If you want to use the arrow function notation:
a.replace(/xxx(yyy)zzz/, (match, p1) => p1.toUpperCase()

Old post but it worth to extend #ChaosPandion answer for other use cases with more restricted RegEx. E.g. ensure the (f) or capturing group surround with a specific format /z(f)oo/:
> a="foobazfoobar"
'foobazfoobar'
> a.replace(/z(f)oo/, function($0,$1) {return $0.replace($1, $1.toUpperCase());})
'foobazFoobar'
// Improve the RegEx so `(f)` will only get replaced when it begins with a dot or new line, etc.
I just want to highlight the two parameters of function makes finding a specific format and replacing a capturing group within the format possible.

SOLUTION
a.replace(/(f)/,(m,g)=>g.toUpperCase())
for replace all grup occurrences use /(f)/g regexp. The problem in your code: String.prototype.toUpperCase.apply("$1") and "$1".toUpperCase() gives "$1" (try in console by yourself) - so it not change anything and in fact you call twice a.replace( /(f)/, "$1") (which also change nothing).
let a= "foobar";
let b= a.replace(/(f)/,(m,g)=>g.toUpperCase());
let c= a.replace(/(o)/g,(m,g)=>g.toUpperCase());
console.log("/(f)/ ", b);
console.log("/(o)/g", c);

Given a dictionary (object, in this case, a Map) of property, values, and using .bind() as described at answers
const regex = /([A-z0-9]+)/;
const dictionary = new Map([["hello", 123]]);
let str = "hello";
str = str.replace(regex, dictionary.get.bind(dictionary));
console.log(str);
Using a JavaScript plain object and with a function defined to get return matched property value of the object, or original string if no match is found
const regex = /([A-z0-9]+)/;
const dictionary = {
"hello": 123,
[Symbol("dictionary")](prop) {
return this[prop] || prop
}
};
let str = "hello";
str = str.replace(regex, dictionary[Object.getOwnPropertySymbols(dictionary)[0]].bind(dictionary));
console.log(str);

In the case of string conversion from CamelCase to bash_case (ie: for filenames), use a callback with ternary operator.
The captured group selected with a regexp () in the first (left) replace arg is sent to the second (right) arg that is a callback function.
x and y give the captured string (don't know why 2 times!) and index (the third one) gives the index of the beginning of the captured group in the reference string.
Therefor a ternary operator can be used not to place _ at first occurence.
let str = 'MyStringName';
str = str.replace(/([^a-z0-9])/g, (x,y,index) => {
return index != 0 ? '_' + x.toLowerCase() : x.toLowerCase();
});
console.log(str);

replacing curley braces with new text

il8n = {
'en-us':{
hello:"Hello Everyone",
goodbye:"See ya later",
welcome:"Hello how are you?",
design:"Kandi Designs"
}
};
function translate(){
var userLang = navigator.language || navigator.userLanguage;
var body = document.body || document.getElementsByTagName('body')[0];
var html = body.innerHTML;
body.innerHTML = body.innerHTML.replace(/({T_(.*?)})/g, il8n[userLang.toLowerCase()][$1.replace(/{T_|},'').toLowerCase()]);
}
translate();
Basically what I want to do is get rid of {T_Hello} to Hello Everyone but since I suck at regex my capture groups and using $1 I can't get this correctly. Can anyone assist in this a wee bit?

You have the right idea. $1, however, is not a variable where you use it. Change the second argument to replace to this:
function (match, $1) {
return il8n[userLang.toLowerCase()][$1.toLowerCase()];
}
replace can take a function as an argument. Then, shorten the capture group in the regex to this:
/{T_(.*?)}/g
This saves you some unnecessary string processing inside the replace function. On an unrelated note, regarding performance, you might want the server to send the appropriate il8n object based on the Accept-Language header, or, even better, do the insertions itself instead of relying on flaky and slow client-side code (making it cacheable).

Based on your il8n object keys you could do this to check for T_ followed by any uppercase letters.
/T_([A-Z]+)/
You could add lowercase or numbers to it too:
/T_([A-Za-z0-9]+)/

This should work:
body.innerHTML = body.innerHTML.replace(
/({T_(.*?)})/g,
function (s) { return il8n [userLang.toLowerCase()] [s.replace(/{T_|}/g,'').toLowerCase()]; }
);
The second argument of replace() can be a string or a function that takes so many arguments as there are ()-groups in the regular expression, which in your case is one.
Indeed, if you write
s.replace (/a/g, 'b')
then, obviously, replace() receives two strings. However, if you write
s.replace (/a/g, f(x))
then again, replace() receives two strings, the second being the string returned from f(x), so if f(x) returns 'b', then the two expressions above are equivalent. Since f(x) is executed before replace() has a chance to see it, it cannot use the special variables like $1, since at the point of call they are not defined.
If you want the code to be executed by replace(), then it should be executable code:
replace (/a/g, function (x) { return ' [' + x + '] found! '; } )
so that at the point of call, the function object is formed and this function is called inside replace with the argument $1 each time (note the global modifier g in the regex).
Note that your code will only work if placed below the text or called onload(), not above the text:
<body>
{T_hello} reader!
<script>
your code
</script>
<body>
works, but
<body>
<script>
your code
</script>
{T_hello} reader!
<body>
doesn't, because when it is executed, there is nothing still in <body>. The best place to call it is in <body onload='translate()'>.
I also corrected a typo in your code: s.replace(/{T_|} should be s.replace(/{T_|}/.
And another error in your code is: s.replace(/{T_|}/ should be s.replace(/{T_|}/g, because it is supposed to make two replacements, not one.

How to add variable inside a Character Class using javascript?

I need to add a variable inside a Character Class when using javascript regular expression. What I tried did not work. What the correct syntax for accomplishing this.
For example,
if(/[variable]/.test(Log))
{
}

You can use the following syntax
var re = new RegExp(variable);
and then use
re.test(Log)

Try this instead:
var regex = new RegExp(variable);
if(regex.test(Log)){
// your logic
}
Hope this will help !!

a='.*'
var patt=new RegExp(a);
patt.test('asdsd')

Actually,
if(/[variable]/.test(Log)) {
}
Is equivalent to:
if(new RegExp("["+variable+"]").test(Log)) {
}
Explanation:
Whenever you use a regex with the // notation, such as:
/[\d*]"+/
You can use the RegExp object and pass the regex as a escaped string. This way, the above regex is equivalent to:
new RegExp("[\\d*]\"+")
The modifiers can be passed as second argument. So that /anything/gm is the same as new RegExp("anything", "gm").

Should I regexp.test before I string.replace?

When I want to replace some parts of a string, should I call replace directly like this?
var r1 = /"\+((:?[\w\.]+)(:?(:?\()(:?.*?)(:?\))|$){0,1})\+"/g;
arg = arg.replace(r1, function(outer, inner){
return eval(inner);
});
Or test for a match first, and then replace if it's a hit, like this?
var r1 = /"\+((:?[\w\.]+)(:?(:?\()(:?.*?)(:?\))|$){0,1})\+"/g;
if (r1.test(arg)) {
arg = arg.replace(r1, function(outer, inner){
return eval(inner);
});
}
I guess this boils down to how the string.replace(regex, string) function works. Will it go into my callback even if there is no match, or will it then simply return arg? In that case I assume the calling replace directly is the right way to go to avoid having the regex engine match the string twice?

You don have to use test. The function in replace only executed when a match occurs.
No matches: No function (eval) call
1 match: 1 call
2 matches: 2 calls
etc.
Also, why are you using eval? eval executes the parameter, as if it's a JavaScript expression. Since you know the input format, it's likely that you're able to achieve the same behaviour without eval.