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
I am working on a JavaScript code that will decompose a RegExp into its base components and give a small explanation as to what it does.
My general idea is to split the input string (as a RegExp) into entries from another array.
My current code:
function interpret(regex){
var r = regex + "";
r = r.split("/");
body = r[1];
flags = r[2];
var classes = [".","\w","\d","\s","\W","\D","\S","[","]"];
var classdefs = ["any non-newline character","any word (digit or letter)","any digit (characters 0-9)","any whitespace character","any non-word (non-digit and non-letter)","any non-digit (not characters 0-9)","open matchset","close matchset"];
var quantifiers = ["*","+","?",
/{(\d+)}/g, // a{n}
/{(\d+),}/g, // a{n,}
/{(\d+),(\d+)}/g, // a{n,m}
/[+*?]\?/g // a<quant>? - lazy quantification
];
var quantDefs = ["repeated 0 or more times","repeated 1 or more times","repeated once or not at all","repeated exactly $1 time","repeated $1 or more times","repeated between $1 and $2 times"];
var escaped = ["\t","\n","\r","\.","\*","\\","\^","\?","\|"];
var escapedDefs = ["a tab","a linefeed","a carriage return","a period","an asterisk","a backslash","a carot","a question mark","a vertical bar"];
// code to split r based on entries in classes, quantifiers, and escaped.
}
Ideally, this function (lets call it splitR) will return outputs something like this:
> splitR("hello",["he","l"]);
["he", "l", "l", "o"]
> splitR("hello",["he"]);
["he", "llo"]
> splitR("hello",["he","o"]);
["he", "ll", "o"];
> splitR("5 is the square root of 25",[/\d+/g,/\w{3,}/g,"of"]);
["5", " is ", "the", " ", "square", " ", "root", " ", "of", " ", "25"]
Clearly defined, the splitR function should, in the context of the interpret function, take a RegExp and split it into its base components; e.g. \d+[0-9]\w*? should split into ["\d", "+", "[", "0-9", "]", "\w", "*", "?"]. These components are defined separately in other arrays, using a variety of RegExps (e.g. /{(\d+)}/g to find a{n}) and strings (e.g. ".").
Truly, I am stumped as to the definition of splitR. Any help is appreciated!
This will split a regex in parts, and populate a second array with description of the parts. It will skip unexpected characters, but there's no real regex syntax checking, i.e. if you start a range and don't end it, the script will not complain. I took the liberty of adding some things that were missing from your lists, like grouping brackets, start and end anchors...
function interpret(regex)
{
var body = regex.source;
var flags = (regex.global ? "g" : "") + (regex.ignoreCase ? "i" : "") + (regex.multiline ? "m" : "");
var classes = [/^\w\-\w/,/^\./,/^\\w/,/^\\d/,/^\\s/,/^\\W/,/^\\D/,/^\\S/,/^\[/,/^\]/,/^\(/,/^\)/,/^\^/,/^\$/,/^\|/];
var classDefs = ["character range","any non-newline character","any word (digit or letter)","any digit (characters 0-9)","any whitespace character","any non-word (non-digit and non-letter)","any non-digit (not characters 0-9)","any non-whitespace character","open matchset","close matchset","open group","close group","start anchor or negation","end anchor","alternative"];
var quantifiers = [/^[+*?]\?/,/^\*/,/^\+/,/^\?/,/^{(\d+)}/,/^{(\d+),}/,/^{(\d+),(\d+)}/];
var quantDefs = ["lazy quantification","repeated 0 or more times","repeated 1 or more times","repeated once or not at all","repeated exactly $1 time","repeated $1 or more times","repeated between $1 and $2 times"];
var escaped = [/^\\t/,/^\\n/,/^\\r/,/^\\\./,/^\\\*/,/^\\\+/,/^\\\-/,/^\\\\/,/^\\\^/,/^\\\$/,/^\\\?/,/^\\\|/,/^\\\[/,/^\\\]/,/^\\\(/,/^\\\)/,/^\\\{/,/^\\\}/];
var escapedDefs = ["a tab","a linefeed","a carriage return","a period","an asterisk","a plus","a minus","a backslash","a caret","a dollar sign","a question mark","a vertical bar","a square bracket","a square bracket","a bracket","a bracket","a curly bracket","a curly bracket"];
var literal = [/^[^\.\\\[\]\(\)\^\$\|\*\+\-\?\{\}]+/];
var literalDefs = ["literal text"];
var regs = classes.concat(quantifiers,escaped,literal);
var defs = classDefs.concat(quantDefs,escapedDefs,literalDefs);
var reglist = [];
var deflist = [];
while (body.length)
{
var found = false;
var chunk = null;
for (var i = 0; i < regs.length; i++)
{
chunk = body.match(regs[i]);
if (chunk)
{
reglist.push(chunk[0]);
deflist.push(defs[i]);
body = body.substr(chunk[0].length);
found = true;
break;
}
}
if (!found)
{
body = body.substr(1); // skip unexpected character
}
}
console.log(regex.source);
console.log(reglist);
console.log(deflist);
alert("see console for output");
}
var x = new RegExp("^[a-z0-9]\\^.\\.\\w\\d\\s\\W\\D\\S+(te|\\|st)*\\*\\n+\\+\\-\\}(\\W?\\?\\s{1,3})\\\\*?a{3}b{4,}c{}\\r\\t\\$$", "ig");
interpret(x);
I'm trying to make a Regex in JavaScript to match each not escaped specific characters.
Here I'm looking for all the ' characters. They can be at the beginning or the end of the string, and consecutive.
E.g.:
'abc''abc\'abc
I should get 3 matchs: the 1st, 5 and 6th character. But not 11th which escaped.
You'll have to account for cases like \\' which should match, and \\\' which shouldn't. but you don't have lookbehinds in JS, let alone variable-length lookbehinds, so you'll have to use something else.
Use the following regex:
\\.|(')
This will match both all escaped characters and the ' characters you're looking for, but the quotes will be in a capture group.
Look at this demo. The matches you're interested in are in green, the ones to ignore are in blue.
Then, in JS, ignore each match object m where !m[1].
Example:
var input = "'abc''abc\\'abc \\\\' abc";
var re = /\\.|(')/g;
var m;
var positions = [];
while (m = re.exec(input)) {
if (m[1])
positions.push(m.index);
}
var pos = [];
for (var i = 0; i < input.length; ++i) {
pos.push(positions.indexOf(i) >= 0 ? "^" : " ");
}
document.getElementById("output").innerText = input + "\n" + pos.join("");
<pre id="output"></pre>
You can use:
var s = "'abc''abc\\'abc";
var cnt=0;
s.replace(/\\?'/g, function($0) { if ($0[0] != '\\') cnt++; return $0;});
console.log(cnt);
//=> 3
What is the best way to count how many spaces before the fist character of a string?
str0 = 'nospaces even with other spaces still bring back zero';
str1 = ' onespace do not care about other spaces';
str2 = ' twospaces';
Use String.prototype.search
' foo'.search(/\S/); // 4, index of first non whitespace char
EDIT:
You can search for "Non whitespace characters, OR end of input" to avoid checking for -1.
' '.search(/\S|$/)
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.