I have a string like this:
var str = "this is test
1. this is test
2. this is test
3. this is test
this is test
1. this test
2. this is test
this is test";
Also I have this regex:
/^[\s\S]*(?:^|\r?\n)\s*(\d+)(?![\s\S]*(\r?\n){2})/m
This capturing group $1 returns 2 from above string.
Now I have a position number: 65 and I want to apply that regex in this range of the string: [0 - 65]. (So I have to get 3 instead of 2). In general I want to limit that string from first to a specific position and then apply that regex on that range. How can I do that?
The simplest way is to apply it to just that substring:
var match = /^[\s\S]*(?:^|\r?\n)\s*(\d+)(?![\s\S]*(\r?\n){2})/m.exec(str.substring(0, 65));
// Note ----------------------------------------------------------------^^^^^^^^^^^^^^^^^
Example:
var str = "this is test\n1. this is test\n2. this is test\n3. this is test\nthis is test\n1. this test \n2. this is test\nthis is test";
var match = /^[\s\S]*(?:^|\r?\n)\s*(\d+)(?![\s\S]*(\r?\n){2})/m.exec(str.substring(0, 65));
// Note ----------------------------------------------------------------^^^^^^^^^^^^^^^^^
document.body.innerHTML = match ? "First capture: [" + match[1] + "]" : "(no match)";
Maybe such a build can help (source: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/exec)
var myRe = /ab*/g;
var str = 'abbcdefabh';
var myArray;
while ((myArray = myRe.exec(str)) !== null) {
var msg = 'Found ' + myArray[0] + '. ';
msg += 'Next match starts at ' + myRe.lastIndex;
console.log(msg);
}
Related
var str = "sdhdhh#gmail.com"; // true but coming false
var str1 = "sdhdhh#gmail.co.uk";
var str2 = "sdhdhh#gmail.org";
var str3 = "sdhdhh#gmail.org.uk";
var patt = new RegExp("[a-z0-9._%+-]+#[a-z0-9.-]+?[\.com]?[\.org]?[\.co.uk]?[\.org.uk]$");
console.log( str + " is " + patt.test(str));
console.log( str1 + " is " + patt.test(str1));
console.log( str2 + " is " + patt.test(str2));
console.log( str3 + " is " + patt.test(str3));
Can anyone tell me what is the mistake, my .com example is not working properly
You need
A grouping construct instead of character classes
A regex literal notation so that you do not have to double escape special chars
The ^ anchor at the start of the pattern since you need both ^ and $ to make the pattern match the entire string.
So you need to use
var patt = /^[a-z0-9._%+-]+#[a-z0-9.-]+?(?:\.com|\.org|\.co\.uk|\.org\.uk)$/;
See the regex demo.
If you need to make it case insensitive, add i flag,
var patt = /^[a-z0-9._%+-]+#[a-z0-9.-]+?(?:\.com|\.org|\.co\.uk|\.org\.uk)$/i;
I'm trying to generate a link using jQuery and need to trim the last '+' sign off the end. Is there a way to detect if there is one there, and then trim it off?
So far the code removes the word 'hotel' and replaces spaces with '+', I think I just need another replace for the '+' that shows up sometimes but not sure how to be super specific with it.
var nameSearch = name.replace("Hotel", "");
nameSearch = nameSearch.replace(/ /g, "+");
The answer to
What is the regex to remove last + sign from a string
is this
const str = "Hotel+"
const re = /\+$/; // remove the last plus if present. $ is "end of string"
console.log(str.replace(re,""))
The question is however if this is answering the actual problem at hand
If you have the string
"Ritz Hotel"
and you want to have
https://www.ritz.com
then you could trim the string:
const fullName = "Ritz Hotel",
name = fullName.replace("Hotel", "").trim().toLowerCase(),
link = `https://www.${name}.com`;
console.log(link)
// or if you want spaces to be converted in url safe format
const fullName1 = "The Ritz Hotel",
name1 = fullName1.replace("Hotel", "").trim().toLowerCase(),
link1 = new URL(`https://www.hotels.com/search?${name1}`).toString()
console.log(link1)
As an alternative to mplungjan's answer, you can use str.endsWith() for the check. If it ends on the + it will be cut out. There is no need for regex. If you can avoid regex you definitely should.
let str = "Hotel+";
if (str.endsWith("+")) {
str = str.substr(0, str.length - 1);
}
console.log(str);
Below you can find a function to replace all the whitespace characters with + excluding the last one:
const raw = "My Ho te l ";
function replaceSpacesWithPlus(raw) {
let rawArray = Array.from(raw);
let replArray = [];
for (let i = 0; i < rawArray.length; i++) {
const char = rawArray[i];
// handle characters 0 to n-1
if (i < rawArray.length - 1) {
if (char === ' ') {
replArray.push('+');
} else {
replArray.push(char);
}
} else {
// handle last char
if (char !== ' ' && char !== '+') {
replArray.push(char);
}
}
}
return replArray;
}
console.log(replaceSpacesWithPlus(raw));
The below snippet will remove all the existing + symbols from string.
let str = 'abcd + efg + hij';
str = str.replace(/\+/gm, '');
//output: abcd efg hij
For trim use the below snippet. It will remove the spaces from around the string.
let str = " Hello World!! "
str = str.trim();
// output: Hello World!!
If you want to replace the last + symbol only.
let str = 'abcd + efg + hij';
let lastIndex = str.lastIndexOf('+');
if (lastIndex > -1) {
let nextString = str.split('');
nextString.splice(lastIndex, 1, '');
str = nextString.join('');
}
// output: abcd + efg hij
Im trying to replace a character at a specific indexOf to uppercase.
My string is a surname plus the first letter in the last name,
looking like this: "lovisa t".
I check the position with this and it gives me the right place in the string. So the second gives me 8(in this case).
first = texten.indexOf(" ");
second = texten.indexOf(" ", first + 1);
And with this I replace the first letter to uppercase.
var name = texten.substring(0, second);
name=name.replace(/^./, name[0].toUpperCase());
But how do I replace the character at "second" to uppercase?
I tested with
name=name.replace(/.$/, name[second].toUpperCase());
But it did´t work, so any input really appreciated, thanks.
Your error is the second letter isn't in position 8, but 7.
Also this second = texten.indexOf(" ", first + 1); gives -1, not 8, because you do not have a two spaces in your string.
If you know that the string is always in the format surname space oneLetter and you want to capitalize the first letter and the last letter you can simply do this:
var name = 'something s';
name = name[0].toUpperCase() + name.substring(1, name.length - 1) + name[name.length -1].toUpperCase();
console.log(name)
Here's a version that does exactly what your question title asks for: It uppercases a specific index in a string.
function upperCaseAt(str, i) {
return str.substr(0, i) + str.charAt(i).toUpperCase() + str.substr(i + 1);
}
var str = 'lovisa t';
var i = str.indexOf(' ');
console.log(upperCaseAt(str, i + 1));
However, if you want to look for specific patterns in the string, you don't need to deal with indices.
var str = 'lovisa t';
console.log(str.replace(/.$/, function (m0) { return m0.toUpperCase(); }));
This version uses a regex to find the last character in a string and a replacement function to uppercase the match.
var str = 'lovisa t';
console.log(str.replace(/ [a-z]/, function (m0) { return m0.toUpperCase(); }));
This version is similar but instead of looking for the last character, it looks for a space followed by a lowercase letter.
var str = 'lovisa t';
console.log(str.replace(/(?:^|\s)\S/g, function (m0) { return m0.toUpperCase(); }));
Finally, here we're looking for (and uppercasing) all non-space characters that are preceded by the beginning of the string or a space character; i.e. we're uppercasing the start of each (space-separated) word.
All can be done by regex replace.
"lovisa t".replace(/(^|\s)\w/g, s=>s.toUpperCase());
Try this one (if it will be helpfull, better move constants to other place, due performance issues(yes, regexp creation is not fast)):
function normalize(str){
var LOW_DASH = /\_/g;
var NORMAL_TEXT_REGEXP = /([a-z])([A-Z])/g;
if(!str)str = '';
if(str.indexOf('_') > -1) {
str = str.replace(LOW_DASH, ' ');
}
if(str.match(NORMAL_TEXT_REGEXP)) {
str = str.replace(NORMAL_TEXT_REGEXP, '$1 $2');
}
if(str.indexOf(' ') > -1) {
var p = str.split(' ');
var out = '';
for (var i = 0; i < p.length; i++) {
if (!p[i])continue;
out += p[i].charAt(0).toUpperCase() + p[i].substring(1) + (i !== p.length - 1 ? ' ' : '');
}
return out;
} else {
return str.charAt(0).toUpperCase() + str.substring(1);
}
}
console.log(normalize('firstLast'));//First Last
console.log(normalize('first last'));//First Last
console.log(normalize('first_last'));//First Last
What is the best way to count how many spaces before the fist character of a string?
str0 = 'nospaces even with other spaces still bring back zero';
str1 = ' onespace do not care about other spaces';
str2 = ' twospaces';
Use String.prototype.search
' foo'.search(/\S/); // 4, index of first non whitespace char
EDIT:
You can search for "Non whitespace characters, OR end of input" to avoid checking for -1.
' '.search(/\S|$/)
Using the following regex:
/^\s*/
in String.prototype.match() will result in an array with a single item, the length of which will tell you how many whitespace chars there were at the start of the string.
pttrn = /^\s*/;
str0 = 'nospaces';
len0 = str0.match(pttrn)[0].length;
str1 = ' onespace do not care about other spaces';
len1 = str1.match(pttrn)[0].length;
str2 = ' twospaces';
len2 = str2.match(pttrn)[0].length;
Remember that this will also match tab chars, each of which will count as one.
You could use trimLeft() as follows
myString.length - myString.trimLeft().length
Proof it works:
let myString = ' hello there '
let spacesAtStart = myString.length - myString.trimLeft().length
console.log(spacesAtStart)
See https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/TrimLeft
str0 = 'nospaces';
str1 = ' onespace do not care about other spaces';
str2 = ' twospaces';
arr_str0 = str0.match(/^[\s]*/g);
count1 = arr_str0[0].length;
console.log(count1);
arr_str1 = str1.match(/^[\s]*/g);
count2 = arr_str1[0].length;
console.log(count2);
arr_str2 = str2.match(/^[\s]*/g);
count3 = arr_str2[0].length;
console.log(count3);
Here:
I have used regular expression to count the number of spaces before the fist character of a string.
^ : start of string.
\s : for space
[ : beginning of character group
] : end of character group
str.match(/^\s*/)[0].length
str is the string.
How can I get from this string
genre:+Drama,Comedy+cast:+Leonardo+DiCaprio,Cmelo+Hotentot+year:+1986-1990
this
genre: [Drama, Comedy],
cast: [Leonardo DiCaprio, Cmelo Hotentot],
year: [1986-1990]
with one regular expression?
This could be done using one regex and overload of replace function with replacer as a second argument. But honestly, I have to use one more replace to get rid of pluses (+) - I replaced them by a space () char:
var str = 'genre:+Drama,Comedy+cast:+Leonardo+DiCaprio,Cmelo+Hotentot+year:+1986-1990';
str = str.replace(/\+/g, ' ');
var result = str.replace(/(\w+:)(\s?)([\w,\s-]+?)(\s?)(?=\w+:|$)/g, function (m, m1, m2, m3, m4, o) {
return m1 + ' [' + m3.split(',').join(', ') + ']' + (o + m.length != str.length ? ',' : '') + '\n';
});
You could find the full example on jsfiddle.
You will not get them into arrays from the start, but it can be parsed if the order stays the same all the time.
var str = "genre:+Drama,Comedy+cast:+Leonardo+DiCaprio,Cmelo+Hotentot+year:+1986-1990";
str = str.replace(/\+/g," ");
//Get first groupings
var re = /genre:\s?(.+)\scast:\s?(.+)\syear:\s(.+)/
var parts = str.match(re)
//split to get them into an array
var genre = parts[1].split(",");
var cast = parts[2].split(",");
var years = parts[3];
console.log(genre);
You can't do this using only regular expressions cause you're trying to parse a (tiny) grammar.