Regex to not allow space between words [duplicate] - regex

I'm trying to write a regular expression to remove white spaces from just the beginning of the word, not after, and only a single space after the word.
Used RegExp:
var re = new RegExp(/^([a-zA-Z0-9]+\s?)*$/);
Test Exapmle:
1) test[space]ing - Should be allowed
2) testing - Should be allowed
3) [space]testing - Should not be allowed
4) testing[space] - Should be allowed but have to trim it
5) testing[space][space] - should be allowed but have to trim it
Only one space should be allowed. Is it possible?

To match, what you need, you can use
var re = /^([a-zA-Z0-9]+\s)*[a-zA-Z0-9]+$/;
Maybe you could shorten that a bit, but it matches _ as well
var re = /^(\w+\s)*\w+$/;

function validate(s) {
if (/^(\w+\s?)*\s*$/.test(s)) {
return s.replace(/\s+$/, '');
}
return 'NOT ALLOWED';
}
validate('test ing') // => 'test ing'
validate('testing') // => 'testing'
validate(' testing') // => 'NOT ALLOWED'
validate('testing ') // => 'testing'
validate('testing ') // => 'testing'
validate('test ing ') // => 'test ing'
BTW, new RegExp(..) is redundant if you use regular expression literal.

This one does not allow preceding and following spaces plus only one space between words. Feel free to add any special characters You want.
^([A-Za-z]+ )+[A-Za-z]+$|^[A-Za-z]+$
demo here

Working code- Inside my name.addTextChangedListener():
public void onTextChanged(CharSequence s, int start, int before, int count) {
String n = name.getText().toString();
if (n.equals(""))
name.setError("Name required");
else if (!n.matches("[\\p{Alpha}\\s]*\\b") | n.matches(".*\\s{2}.*") | n.matches("\\s.*")) {
if (n.matches("\\s.*"))
name.setError("Name cannot begin with a space");
else if (n.matches(".*\\s{2}.*"))
name.setError("Multiple spaces between texts");
else if (n.matches(".*\\s"))
name.setError("Blank space at the end of text");
else
name.setError("Non-alphabetic character entered");
}
}
You could try adapting this to your code.

var f=function(t){return Math.pow(t.split(' ').length,2)/t.trim().split(' ').length==2}
f("a a")
true
f("a a ")
false
f("a a")
false
f(" a a")
false
f("a a a")
false

Here is a solution without regular expression.
Add this script inside document.ready function it will work.
var i=0;
jQuery("input,textarea").on('keypress',function(e){
//alert();
if(jQuery(this).val().length < 1){
if(e.which == 32){
//alert(e.which);
return false;
}
}
else {
if(e.which == 32){
if(i != 0){
return false;
}
i++;
}
else{
i=0;
}
}
});

const handleChangeText = text => {
let lastLetter = text[text.length - 1];
let secondLastLetter = text[text.length - 2];
if (lastLetter === ' ' && secondLastLetter === ' ') {
return;
}
setInputText(text.trim());
};

use this
^([A-Za-z]{5,}|[\s]{1}[A-Za-z]{1,})*$
Demo:-https://regex101.com/r/3HP7hl/2

Related

split a string by commas, but commas in a token

Let's say I have this:
something,"another thing"
This can be split easily with a normal split function.
Now I want to have more complicated syntax and I do:
something,"in a string, oooh",rgba(4,2,0)
This does not work with a regular split function.
I tried using things like replacing commas inside of specific types of tokens, but that became too over-complicated and I feel there has to be a better way.
Then I tried with regular expressions, which worked, until I had to add a new feature, which wouldn't work with the regexp I had (which was pretty bad), also regexp matches can be slow, and this is supposed to be as fast as possible.
What would be a better way to solve this?
Here is the source repo for extra context https://github.com/hyprland-community/hyprparse
And the format in question is the hyprland config format
Iterate over the string keeping a context state:
None
Inside a "..."
Inside a (...)
Inside a context, comma has no separator meaning.
Limitations: This is a midnight hack!
See also Rust Playground
fn split(s: String) -> Vec<String> {
let mut context = None;
let mut i = 0;
let mut start = 0;
let mut items = Vec::new();
for c in s.chars() {
if context == Some('"') {
if c == '"' {
context = None;
}
i = i+1;
continue;
} else if context == Some('(') {
if c == ')' {
context = None;
}
i = i+1;
continue;
}
if c == '"' || c == '(' {
context = Some(c);
}
if c == ',' && context.is_none() {
items.push(s[start..i].to_string());
start = i + 1;
}
i = i+1;
}
items.push(s[start..i].to_string());
items
}
fn main() {
let s = "something,\"in a string, oooh\",rgba(4,2,0)".to_string();
println!("{:?}", split(s));
// -> ["something", "\"in a string, oooh\"", "rgba(4,2,0)"]
}
Thanks for all the help everyone, I eventually came up with my own solution with the help of people I knew IRL, here is my solution:
fn previous_rgb(index: usize, chars: Vec<char>) -> bool {
let string: String = chars[index-3..index].iter().collect();
let string = string.as_str();
if string == "gba" || string == "rgb" {
true
} else {
false
}
}
fn splitter<Str: ToString>(s: Str) -> Vec<String> {
let mut is_in_rgb = false;
let mut is_in_str = false;
let mut last = 0;
let chars: Vec<_> = s.to_string().chars().collect();
let mut final_str = vec![];
for (index, c) in chars.iter().enumerate() {
if *c == '(' && previous_rgb(index, chars.clone()) && !is_in_str {
is_in_rgb = true;
} else if *c == ')' && is_in_rgb && !is_in_str {
is_in_rgb = false;
} else if *c == '"' && is_in_str {
is_in_str = false;
} else if *c == '"' && !is_in_str {
is_in_str = true;
} else if *c == ',' && !is_in_str && !is_in_rgb {
final_str.push(chars[last..index].iter().collect());
last = index + 1
};
}
final_str.push(chars[last..].iter().collect());
final_str
}
fn main() {
let splitted = splitter(r#"test,test: rgba(5,4,3),"test2, seperated""#);
println!("{splitted:?}");
}

Capitalize First Letter Of Each Name after Hyphen "-" and Space " "

I'm currently using this String extension to Capitalize the letter of each word in a textField :
"happy sunshine" .toTitleCase() gives "Happy Sunshine"
extension StringExtension on String {
String toTitleCase() => replaceAll(RegExp(' +'), ' ')
.split(' ')
.map((str) => str.toCapitalized())
.join(' ');
String toCapitalized() =>
length > 0 ? '${this[0].toUpperCase()}${substring(1).toLowerCase()}' : '';
}
but I'd also like to Capitalize letters that come after a hyphen - with the same toTitleCase method
ex : "very-happy sunshine" .toTitleCase() would give "Very-Happy Sunshine"
Currently .toTitleCase() gives "Very-happy Sunshine" : (
I am sure a wizard with expert knowledge in regular expression can do this better but I think this solution solves your problem:
void main() {
print('happy sunshine'.toTitleCase()); // Happy Sunshine
print('very-happy sunshine'.toTitleCase()); // Very-Happy Sunshine
}
extension StringExtension on String {
String toTitleCase() => replaceAllMapped(
RegExp(r'(?<= |-|^).'), (match) => match[0]!.toUpperCase());
}
If you call the method a lot of times, you might consider having the RegExp as a cached value like:
extension StringExtension on String {
static final RegExp _toTitleCaseRegExp = RegExp(r'(?<= |-|^).');
String toTitleCase() =>
replaceAllMapped(_toTitleCaseRegExp, (match) => match[0]!.toUpperCase());
}
You can tweak your code as well. But I've used the same thing somewhere in my project so you can do something like this as well.
Working: First I'm creating an empty array looping through each character in a particular string and checking if space (" ") and hyphen ("-") are current_position - 1 then I'm making current_position to uppercase.
String capitalize(String s) {
String result = "";
for (int i = 0; i < s.length; i++) {
if (i == 0) {
result += s[i].toUpperCase();
} else if (s[i - 1] == " ") {
result += s[i].toUpperCase();
} else if (s[i - 1] == "-") {
result += s[i].toUpperCase();
} else {
result += s[i];
}
}
return result;
}

recheck outer if boolean true or false within it's else

My goal is to recheck variable coolFeature boolean value from it's else
if (coolFeature) { // want to run this again after value becomes true in else below
$("#Editor").data("kendoWindow").width = '600';
$("#Editor").data("kendoWindow").title("Format Feature - " + Type + ' #' + graphic.attributes.OBJECTID);
$('#ndt').hide();
} else {
$("#Editor").data("kendoWindow").title("Edit Attributes - " + Type + ' #' + graphic.attributes.OBJECTID);
$('#ndt').show();
$("#ndt").click(function() {
$(this).data('clicked', true);
$("#Editor").data("kendoWindow").hide();
coolFeature = "true"; // want to reset to True here, then run the actions under initial if
});
}
I think you need simple recursion, I can't see the entire code so I can't tell exactly.
var myFunction = function(coolFeature) {
if(coolFeature) {
console.log("Cool");
} else {
myFunction(true);
}
};
myFunction(false);

Generate all matches from a subset of regex

I need to define a bunch of vector sequences, which are all a series of L,D,R,U for left, down, right, up or x for break. There are optional parts, and either/or parts. I have been using my own invented system for noting it down, but I want to document this for other, potentially non-programmers to read.
I now want to use a subset (I don't plan on using any wildcards, or infinite repetition for example) of regex to define the vector sequence and a script to produce all possible matching strings...
/LDR/ produces ['LDR']
/LDU?R/ produces ['LDR','LDUR']
/R(LD|DR)U/ produces ['RLDU','RDRU']
/DxR[DL]U?RDRU?/ produces ['DxRDRDR','DxRDRDRU','DxRDURDR','DxRDURDRU','DxRLRDR','DxRLRDRU','DxRLURDR','DxRLURDRU']
Is there an existing library I can use to generate all matches?
EDIT
I realised I will only be needing or statements, as optional things can be specified by thing or nothing maybe a, or b, both optional could be (a|b|). Is there another language I could use to define what I am trying to do?
By translating the java code form the link provided by #Dukeling into javascript, I think I have solved my problem...
var Node = function(str){
this.bracket = false;
this.children = [];
this.s = str;
this.next = null;
this.addChild = function(child){
this.children.push(child);
}
}
var printTree = function(root,prefix){
prefix = prefix.replace(/\./g, "");
for(i in root.children){
var child = root.children[i]
printTree(child, prefix + root.s);
}
if(root.children.length < 1){
console.log(prefix + root.s);
}
}
var Stack = function(){
this.arr = []
this.push = function(item){
this.arr.push(item)
}
this.pop = function(){
return this.arr.pop()
}
this.peek = function(){
return this.arr[this.arr.length-1]
}
}
var createTree = function(s){
// this line was causing errors for `a(((b|c)d)e)f` because the `(((` was only
// replacing the forst two brackets.
// var s = s.replace(/(\(|\||\))(\(|\||\))/g, "$1.$2");
// this line fixes it
var s = s.replace(/[(|)]+/g, function(x){ return x.split('').join('.') });
var str = s.split('');
var stack = new Stack();
var root = new Node("");
stack.push(root); // start node
var justFinishedBrackets = false;
for(i in str){
var c = str[i]
if(c == '('){
stack.peek().next = new Node("Y"); // node after brackets
stack.peek().bracket = true; // node before brackets
} else if (c == '|' || c == ')'){
var last = stack.peek(); // for (ab|cd)e, remember b / d so we can add child e to it
while (!stack.peek().bracket){ // while not node before brackets
stack.pop();
}
last.addChild(stack.peek().next); // for (b|c)d, add d as child to b / c
} else {
if (justFinishedBrackets){
var next = stack.pop().next;
next.s = "" + c;
stack.push(next);
} else {
var n = new Node(""+c);
stack.peek().addChild(n);
stack.push(n);
}
}
justFinishedBrackets = (c == ')');
}
return root;
}
// Test it out
var str = "a(c|mo(r|l))e";
var root = createTree(str);
printTree(root, "");
// Prints: ace / amore / amole
I only changed one line, to allow more than two consecutive brackets to be handled, and left the original translation in the comments
I also added a function to return an array of results, instead of printing them...
var getTree = function(root,prefix){
this.out = this.out || []
prefix = prefix.replace(/\./g, "");
for(i in root.children){
var child = root.children[i]
getTree(child, prefix + root.s, out);
}
if(root.children.length < 1){
this.out.push(prefix + root.s);
}
if(!prefix && !root.s){
var out = this.out;
this.out = null
return out;
}
}
// Test it
var str = "a(b|c)d";
var root = createTree(str);
console.log(getTree(root, ""));
// logs ["abd","acd"]
The last part, to allow for empty strings too, so... (ab|c|) means ab or c or nothing, and a convenience shortcut so that ab?c is translated into a(b|)c.
var getMatches = function(str){
str = str.replace(/(.)\?/g,"($1|)")
// replace all instances of `(???|)` with `(???|µ)`
// the µ will be stripped out later
str = str.replace(/\|\)/g,"|µ)")
// fix issues where last character is `)` by inserting token `µ`
// which will be stripped out later
str = str+"µ"
var root = createTree(str);
var res = getTree(root, "");
// strip out token µ
for(i in res){
res[i] = res[i].replace(/µ/g,"")
}
// return the array of results
return res
}
getMatches("a(bc|de?)?f");
// Returns: ["abcf","adef","adf","af"]
The last part is a little hack-ish as it relies on µ not being in the string (not an issue for me) and solves one bug, where a ) at the end on the input string was causing incorrect output, by inserting a µ at the end of each string, and then stripping it from the results. I would be happy for someone to suggest a better way to handle these issues, so it can work as a more general solution.
This code as it stands does everything I need. Thanks for all your help!
I'd imagine what you're trying is quite easy with a tree (as long as it's only or-statements).
Parse a(b|c)d (or any or-statement) into a tree as follows: a has children b and c, b and c have a mutual child d. b and c can both consist of 0 or more nodes (as in c could be g(e|f)h in which case (part of) the tree would be a -> g -> e/f (2 nodes) -> h -> d or c could be empty, in which case (part of) the tree would be a -> d, but an actual physical empty node may simplify things, which you should see when trying to write the code).
Generation of the tree shouldn't be too difficult with either recursion or a stack.
Once you have a tree, it's trivial to recursively iterate through the whole thing and generate all strings.
Also, here is a link to a similar question, providing a library or two.
EDIT:
"shouldn't be too difficult" - okay, maybe not
Here is a somewhat complicated example (Java) that may require some advanced knowledge about stacks.
Here is a slightly simpler version (Java) thanks to inserting a special character between each ((, )), |(, etc.
Note that neither of these are particularly efficient, the point is just to get the idea across.
Here is a JavaScript example that addresses parsing the (a|b) and (a|b|) possibilities, creates an array of possible substrings, and composes the matches based on this answer.
var regex = /\([RLUD]*\|[RLUD]*\|?\)/,
str = "R(LD|DR)U(R|L|)",
substrings = [], matches = [], str_tmp = str, find
while (find = regex.exec(str_tmp)){
var index = find.index
finds = find[0].split(/\|/)
substrings.push(str_tmp.substr(0, index))
if (find[0].match(/\|/g).length == 1)
substrings.push([finds[0].substr(1), finds[1].replace(/.$/, '')])
else if (find[0].match(/\|/g).length == 2){
substrings.push([finds[0].substr(1), ""])
substrings.push([finds[1], ""])
}
str_tmp = str_tmp.substr(index + find[0].length)
}
if (str_tmp) substrings.push([str_tmp])
console.log(substrings) //>>["R", ["LD", "DR"], "U", ["R", ""], ["L", ""]]
//compose matches
function printBin(tree, soFar, iterations) {
if (iterations == tree.length) matches.push(soFar)
else if (tree[iterations].length == 2){
printBin(tree, soFar + tree[iterations][0], iterations + 1)
printBin(tree, soFar + tree[iterations][1], iterations + 1)
}
else printBin(tree, soFar + tree[iterations], iterations + 1)
}
printBin(substrings, "", 0)
console.log(matches) //>>["RLDURL", "RLDUR", "RLDUL", "RLDU", "RDRURL", "RDRUR", "RDRUL", "RDRU"]

Fuzzy Matches on dijit.form.ComboBox / dijit.form.FilteringSelect Subclass

I am trying to extend dijit.form.FilteringSelect with the requirement that all instances of it should match input regardless of where the characters are in the inputted text, and should also ignore whitespace and punctuation (mainly periods and dashes).
For example if an option is "J.P. Morgan" I would want to be able to select that option after typing "JP" or "P Morgan".
Now I know that the part about matching anywhere in the string can be accomplished by passing in queryExpr: "*${0}*" when creating the instance.
What I haven't figured out is how to make it ignore whitespace, periods, and dashes. I have an example of where I'm at here - http://jsfiddle.net/mNYw2/2/. Any help would be appreciated.
the thing to master in this case is the store fetch querystrings.. It will call a function in the attached store to pull out any matching items, so if you have a value entered in the autofilling inputfield, it will eventually end up similar to this in the code:
var query = { this.searchAttr: this.get("value") }; // this is not entirely accurate
this._fetchHandle = this.store.query(query, options);
this._fetchHandle.then( showResultsFunction );
So, when you define select, override the _setStoreAttr to make changes in the store query api
dojo.declare('CustomFilteringSelect', [FilteringSelect], {
constructor: function() {
//???
},
_setStoreAttr: function(store) {
this.inherited(arguments); // allow for comboboxmixin to modify it
// above line eventually calls this._set("store", store);
// so now, 'this' has 'store' set allready
// override here
this.store.query = function(query, options) {
// note that some (Memory) stores has no 'fetch' wrapper
};
}
});
EDIT: override queryEngine function as opposed to query function
Take a look at the file SimpleQueryEngine.js under dojo/store/util. This is essentially what filters the received Array items on the given String query from the FilteringSelect. Ok, it goes like this:
var MyEngine = function(query, options) {
// create our matching query function
switch(typeof query){
default:
throw new Error("Can not query with a " + typeof query);
case "object": case "undefined":
var queryObject = query;
query = function(object){
for(var key in queryObject){
var required = queryObject[key];
if(required && required.test){
if(!required.test(object[key])){
return false;
}
}else if(required != object[key]){
return false;
}
}
return true;
};
break;
case "string":
/// HERE is most likely where you can play with the reqexp matcher.
// named query
if(!this[query]){
throw new Error("No filter function " + query + " was found in store");
}
query = this[query];
// fall through
case "function":
// fall through
}
function execute(array){
// execute the whole query, first we filter
var results = arrayUtil.filter(array, query);
// next we sort
if(options && options.sort){
results.sort(function(a, b){
for(var sort, i=0; sort = options.sort[i]; i++){
var aValue = a[sort.attribute];
var bValue = b[sort.attribute];
if (aValue != bValue) {
return !!sort.descending == aValue > bValue ? -1 : 1;
}
}
return 0;
});
}
// now we paginate
if(options && (options.start || options.count)){
var total = results.length;
results = results.slice(options.start || 0, (options.start || 0) + (options.count || Infinity));
results.total = total;
}
return results;
}
execute.matches = query;
return execute;
};
new Store( { queryEngine: MyEngine });
when execute.matches is set on bottom of this function, what happens is, that the string gets called on each item. Each item has a property - Select.searchAttr - which is tested by RegExp like so: new RegExp(query).test(item[searchAttr]); or maybe a bit simpler to understand; item[searchAttr].matches(query);
I have no testing environment, but locate the inline comment above and start using console.debug..
Example:
Stpre.data = [
{ id:'WS', name: 'Will F. Smith' },
{ id:'RD', name:'Robert O. Dinero' },
{ id:'CP', name:'Cle O. Patra' }
];
Select.searchAttr = "name";
Select.value = "Robert Din"; // keyup->autocomplete->query
Select.query will become Select.queryExp.replace("${0]", Select.value), in your simple queryExp case, 'Robert Din'.. This will get fuzzy and it would be up to you to fill in the regular expression, here's something to start with
query = query.substr(1,query.length-2); // '*' be gone
var words = query.split(" ");
var exp = "";
dojo.forEach(words, function(word, idx) {
// check if last word
var nextWord = words[idx+1] ? words[idx+1] : null;
// postfix 'match-all-but-first-letter-of-nextWord'
exp += word + (nextWord ? "[^" + nextWord[0] + "]*" : "");
});
// exp should now be "Robert[^D]*Din";
// put back '*'
query = '*' + exp + '*';