I have a regular expression that I have been using in notepad++ for search&replace to manipulate some text, and I want to incorporate it into my javascript code. This is the regular expression:
Search
(?-s)(.{150,250}\.(\[\d+\])*)\h+ and replace with \1\r\n\x20\x20\x20
In essence creating new paragraphs for every 150-250 words and indenting them.
This is what I have tried in JavaScript. For a text area <textarea name="textarea1" id="textarea1"></textarea>in the HTML. I have the following JavaScript:
function rep1() {
var re1 = new RegExp('(?-s)(.{150,250}\.(\[\d+\])*)\h+');
var re2 = new RegExp('\1\r\n\x20\x20\x20');
var s = document.getElementById("textarea1").value;
s = string.replace(re1, re2);
document.getElementById("textarea1").value = s;
}
I have also tried placing the regular expressions directly as arguments for string.replace() but that doesn't work either. Any ideas what I'm doing wrong?
Several issues:
JavaScript does not support (?-s). You would need to add modifiers separately. However, this is the default setting in JavaScript, so you can just leave it out. If it was your intention to let . also match line breaks, then use [^] instead of . in JavaScript regexes.
JavaScript does not support \h -- the horizontal white space. Instead you could use [^\S\r\n].
When passing a string literal to new RegExp be aware that backslashes are escape characters for the string literal notation, so they will not end up in the regex. So either double them, or else use JavaScript's regex literal notation
In JavaScript replace will only replace the first occurrence unless you provided the g modifier to the regular expression.
The replacement (second argument to replace) should not be a regex. It should be a string, so don't apply new RegExp to it.
The backreferences in the replacement string should be of the $1 format. JavaScript does not support \1 there.
You reference string where you really want to reference s.
This should work:
function rep1() {
var re1 = /(.{150,250}\.(\[\d+\])*)[^\S\r\n]+/g;
var re2 = '$1\r\n\x20\x20\x20';
var s = document.getElementById("textarea1").value;
s = s.replace(re1, re2);
document.getElementById("textarea1").value = s;
}
Related
I have a code that generates a random letter based on the word and I have tried to create a RegExp code to turn all the letters from the word to '_' except the randomly generated letter from the word.
const word = "Apple is tasty"
const randomCharacter = word[Math.floor(Math.random() * word.length)]
regex = new RegExp(/[^${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 looking for a specific javascript regex without the new lookahead/lookbehind features of Javascript 2018 that allows me to select text between two asterisk signs but ignores escaped characters.
In the following example only the text "test" and the included escaped characters are supposed to be selected according the rules above:
\*jdjdjdfdf*test*dfsdf\*adfasdasdasd*test**test\**sd* (Selected: "test", "test", "test\*")
During my research I found this solution Regex, everything between two characters except escaped characters /(?<!\\)(%.*?(?<!\\)%)/ but it uses negative lookbehinds which is supported in javascript 2018 but I need to support IE11 as well, so this solution doesn't work for me.
Then i found another approach which is almost getting there for me here: Javascript: negative lookbehind equivalent?. I altered the answer of Kamil Szot to fit my needs: ((?!([\\])).|^)(\*.*?((?!([\\])).|^)\*) Unfortuantely it doesn't work when two asterisks ** are in a row.
I have already invested a lot of hours and can't seem to get it right, any help is appreciated!
An example with what i have so far is here: https://www.regexpal.com/?fam=117350
I need to use the regexp in a string.replace call (str.replace(regexp|substr, newSubStr|function); so that I can wrap the found strings with a span element of a specific class.
You can use this regular expression:
(?:\\.|[^*])*\*((?:\\.|[^*])*)\*
Your code should then only take the (only) capture group of each match.
Like this:
var str = "\\*jdjdjdfdf*test*dfsdf\\*adfasdasdasd*test**test\\**sd*";
var regex = /(?:\\.|[^*])*\*((?:\\.|[^*])*)\*/g
var match;
while (match = regex.exec(str)) {
console.log(match[1]);
}
If you need to replace the matches, for instance to wrap the matches in a span tag while also dropping the asterisks, then use two capture groups:
var str = "\\*jdjdjdfdf*test*dfsdf\\*adfasdasdasd*test**test\\**sd*";
var regex = /((?:\\.|[^*])*)\*((?:\\.|[^*])*)\*/g
var result = str.replace(regex, "$1<span>$2</span>");
console.log(result);
One thing to be careful with: when you use string literals in JavaScript tests, escape the backslash (with another backslash). If you don't do that, the string actually will not have a backslash! To really get the backslash in the in-memory string, you need to escape the backslash.
const testStr = `\\*jdjdjdfdf*test*dfsdf\\*adfasdasdasd*test**test\\**sd*`;
const m = testStr.match(/\*(\\.)*t(\\.)*e(\\.)*s(\\.)*t(\\.)*\*/g).map(m => m.substr(1, m.length-2));
console.log(m);
More generic code:
const prepareRegExp = (word, delimiter = '\\*') => {
const escaped = '(\\\\.)*';
return new RegExp([
delimiter,
escaped,
[...word].join(escaped),
escaped,
delimiter
].join``, 'g');
};
const testStr = `\\*jdjdjdfdf*test*dfsdf\\*adfasdasdasd*test**test\\**sd*`;
const m = testStr
.match(prepareRegExp('test'))
.map(m => m.substr(1, m.length-2));
console.log(m);
https://instacode.dev/#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
var str='\E\\E\10.1.2.154\E\bcs\E\30877_P9999_Adult{2}_02_05_2019_0329p.pdf';
var res=str.replace('\E', '');
I am getting return like this:
\E.1.2.154EcsE877_P9999_Adult{2}_02_05_2019_0329p.pdf
I need to replace all '\E' from string and expecting output like this (\\10.1.2.154\bcs\30877_P9999_Adult{2}_02_05_2019_0329p.pdf). Some body please advise on this . I tried to do several way to fix this. No luck. When I tried with C# it's working fine.
static void Main(string[] args)
{
string str=#"\E\\E\10.1.2.154\E\bcs\E\30877_P9999_Adult{2}_02_05_2019_0329p.pdf";
str=str.Replace(#"\E","");
Console.WriteLine(str);
Console.Read();
}
But, I need it in JavaScript.
var str='\E\\E\10.1.2.154\E\bcs\E\30877_P9999_Adult{2}_02_05_2019_0329p.pdf';
var res=str.replace(/\\E/g, '');
console.log(str, res)
You must use RegExg to eliminate the escape inside a string in JavaScript.
In JavaScript the equivalent of the C# # prefix is String.raw followed by a template literal (notice the backtics).
And to replace all occurrences, not just one, you need to pass a regex to replace with g modifier.
var str=String.raw`\E\\E\10.1.2.154\E\bcs\E\30877_P9999_Adult{2}_02_05_2019_0329p.pdf`;
var res=str.replace(/\\E/g, '');
console.log(res);
NB: The backslash in a regular expression is an escape character, so you need \\ for one literal backslash.
If for some reason you really want to avoid the use of a regex, then there is the split/join trick, but it is a bit slower:
var str=String.raw`\E\\E\10.1.2.154\E\bcs\E\30877_P9999_Adult{2}_02_05_2019_0329p.pdf`;
var res=str.split(String.raw`\E`).join('');
console.log(res);
Older JS engines
For older JS engines that do not support String.raw you need to use standard string literals, which use the backslash as escape character. So then you need to double all of them. But this is only needed when you write the string as a literal. When you get the string via some API, then there is no need to alter the string before doing the replacement:
var str='\\E\\\\E\\10.1.2.154\\E\\bcs\\E\\30877_P9999_Adult{2}_02_05_2019_0329p.pdf';
var res=str.replace(/\\E/g, '');
console.log(res);
How to declare associative array of regex?
This is not working
var Validators = {
url : /^http(s?)://((\w+\.)?\w+\.\w+|((2[0-5]{2}|1[0-9]{2}|[0-9]{1,2})\.){3}(2[0-5]{2}|1[0-9]{2}|[0-9]{1,2}))(/)?$/gm
};
EDITED: Now working!
This will be valid in JS (like # operator in C#)
url : `/^http(s?)://((\w+\.)?\w+\.\w+|((2[0-5]{2}|1[0-9]{2}|[0-9]{1,2})\.){3}(2[0-5]{2}|1[0-9]{2}|[0-9]{1,2}))(/)?$/gm`
However, will still not work due to double escape, one in JS and other in Regex. If expression is small, perhaps naked eye can manually escape for both JS and Regex. My brain just can't :)
In order to use strings as tested on regex101.com for example, all required strings should be declared as 'row' like this:
var exp = String.raw`^(http(s?):\/\/)?(((www\.)?[a-zA-Z0-9\.\-\_]+(\.[a-zA-Z]{2,3})+)|(\b(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\b))(\/[a-zA-Z0-9\_\-\s\.\/\?\%\#\&\=]*)?$`;
var strings = [
String.raw`http://www.goo gle.com`,
String.raw`http://www.google.com`,
];
Wrap it with new RegExp() and escape slashes
var Validators = {
url : new RegExp( /^http(s?):\/\/((\w+\.)?\w+\.\w+|((2[0-5]{2}|1[0-9]{2}|[0-9]{1,2})\.){3}(2[0-5]{2}|1[0-9]{2}|[0-9]{1,2}))(\/)?$/gm )
};
Your regex has forward slashes in it. This symbol needs to be escaped because it is supposed to indicate the start and end of the expression. Try \/.
Im using this snippet to replace several characters in a string.
var badwords = eval("/foo|bar|baz/ig");
var text="foo the bar!";
document.write(text.replace(badwords, "***"));
But one of the characters I want to replace is '/'. I assume it's not working because it's a reserved character in regular expressions, but how can I get it done then?
Thanks!
You simply escape the "reserved" char in your RegExp:
var re = /abc\/def/;
You are probably having trouble with that because you are, for some reason, using a string as your RegExp and then evaling it...so odd.
var badwords = /foo|bar|baz/ig;
is all you need.
If you INISIST on using a string, then you have to escape your escape:
var badwords = eval( "/foo|ba\\/r|baz/ig" );
This gets a backslash through the JS interpreter to make it to the RegExp engine.
first of DON'T USE EVAL it's the most evil function ever and fully unnecessary here
var badwords = /foo|bar|baz/ig;
works just as well (or use the new RegExp("foo|bar|baz","ig"); constructor)
and when you want to have a / in the regex and a \ before the character you want to escape
var badwords = /\/foo|bar|baz/ig;
//or
var badwords = new RegExp("\\/foo|bar|baz","ig");//double escape to escape the backslash in the string like one has to do in java