I am trying to make ifcondition for a large number of chars.
I can use
if (str==!||str==#||str==#||str==$||str==^||str==&)
And so on, but this seems very inefficient. I would like to get the condition to work if the char is on of those:
!##%$^&()_-+=\?/.,'][{}<>`~
Is there is any shorter and more efficient way of doing it?
for (var c0 = 1; c0 > fn.length++; c0++) {
var str = fn.charAt(c0--);
if (str ==-"!##%$^&()_-+=\?/.,'][{}<>`~") {
}
}
I want the check to accrue on every single char from the string above.
You can use a regular expression character class to check if your character matches a particular character:
/^[\!##%$\^&\(\)_\-\+=\?\/\.,'\]\[\{\}\<\>`~]$/
Here I have escape special characters so that they get treated like regular characters.
See working example below:
const regex = /^[\!##%$\^&\(\)_\-\+=\?\/\.,'\]\[\{\}\<\>`~]$/,
charA = '#', // appears in char set
charB = 'A'; // doesn't appear in char set
console.log(regex.test(charA)); // true
console.log(regex.test(charB)); // false
Alternatively, if you don't want to use regular expressions you can instead put all your characters into an array and use .includes to check if your character is in your array.
const chars = "!##%$^&()_-+=\?/.,'][{}<>`~",
charArr = [...chars],
charA = '#', // is in char set
charB = 'A'; // isn't in char set
console.log(charArr.includes(charA)); // true
console.log(charArr.includes(charB)); // false
Just use regular expressions rather than manual single character checking.
const pattern = new RegExp("!##%$^&()_-+=\?\/.,'][{}<>`~");
const exists = pattern.test(str);
if (exists) {
// code logic for special character exists in string
}
First you can use split('') to split a string into an array of characters. Next you can use .some to check if a condition is true for at least one element in the array:
"!##%$^&()_-+=\?/.,'][{}<>`~".split('').some(x => x === str)
Related
I have a textbox that a user can paste into using Ctrl+V. I would like to restrict the textbox to accept just GUIDs. I tried to write a small function that would format an input string to a GUID based on RegEx, but I can't seem to be able to do it. I tried following the below post:
Javascript string to Guid
function stringToGUID()
{
var strInput = 'b6b954d9cbac4b18b0d5a0f725695f1ca98d64e456f76';
var strOutput = strInput.replace(/([0-f]{8})([0-f]{4})([0-f]{4})([0-f]{4})([0-f]{12})/,"$1-$2-$3-$4-$5");
console.log(strOutput );
//from my understanding, the input string could be any sequence of 0-9 or a-f of any length and a valid giud patterened string would be the result in the above code. This doesn't seem to be the case;
//I would like to extract first 32 characters; how do I do that?
}
I suggest that you remove the dashes, truncate to 32 characters, and then test if the remaining characters are valid before inserting the dashes:
function stringToGUID()
{
var input = 'b6b954d9cbac4b18b0d5a0f725695f1ca98d64e456f76';
let g = input.replace("-", "");
g = g.substring(0, 32);
if (/^[0-9A-F]{32}$/i.test(g)) {
g = g.replace(/(.{8})(.{4})(.{4})(.{4})(.{12})/, "$1-$2-$3-$4-$5");
}
console.log(g);
}
stringToGUID();
(The i at the end of the regex makes it case-insensitive.)
You are already matching 32 characters with the pattern, so there is no need to get a separate operation to get 32 characters to test against.
You can replace all the hyphens with an empty string, and then match the pattern from the start of the string using ^
Then first check if there is a match, and if there is do the replacement with the 5 groups and hyphens in between. If there is not match, return the original string.
The function stringToGUID() by itself does not do anything except log a string that is hardcoded in the function. To extend its functionality, you can pass a parameter.
function stringToGUID(s) {
const regex = /^([0-f]{8})([0-f]{4})([0-f]{4})([0-f]{4})([0-f]{12})/;
const m = s.replace(/-+/g, '').match(regex);
return m ? `${m[1]}-${m[2]}-${m[3]}-${m[4]}-${m[5]}` : s;
}
[
'b6b954d9cbac4b18b0d5a0f725695f1ca98d64e456f76',
'b6b954d9-cbac-4b18-b0d5-a0f725695f1c',
'----54d9cbac4b18b0d5a0f725695f1ca98d64e456f76',
'!##$%'
].forEach(s => {
console.log(stringToGUID(s));
});
I have a CLI where the user can declare an alphabet and pass it to my code. My code generate a string with that alphabet
For example if the user declare these groups of alphabet abc abcAB1234 and ##1$2% I need to generate a string where every single character is at least in one group and the generated string has all the characters defined by the alphabet. No repetition are allowed (case sensitive)
So, if the alphabet is abc abcAB1234 ##1$2% the admitted output can be B#1a or #ca41% but not aA## (same character 'a' repeated) or aBcZ# ('Z' is not part of the alphabet) or aBA43 (some characters of alphabet are not presents)
I tried with this ^(?!.*([abcabcAB1234##1$2%])\1{1})(?!.*([abc])\1{1})(?!.*([abcAB1234])\1{1})(?!.*([##1$2%])\1{1})[abcabcAB1234##1$2%]{8,}$ but, obviously, doesn't work
Can someone please help me to understand where I'm wrong with my regexp?
I don't think this is possible with a RexExp. But it is easy to achieve using a Set.
const alphabet = 'abcBEL'
const wordToMatch = 'BLa'
const wordToMatch2 = 'BLaa'
const wordToMatch3 = 'zBLa'
function checkWord(alphabet, word) {
const set = new Set(alphabet.split(''))
for (const c of word){
if (!set.has(c)) return false
set.delete(c)
}
return true
}
console.log(checkWord(alphabet, wordToMatch))
console.log(checkWord(alphabet, wordToMatch2))
console.log(checkWord(alphabet, wordToMatch3))
I'm working with a string where I need to extract the first n characters up to where numbers begin. What would be the best way to do this as sometimes the string starts with a number: 7EUSA8889er898 I would need to extract 7EUSA But other string examples would be SWFX74849948, I would need to extract SWFX from that string.
Not sure how to do this with regex my limited knowledge is blocking me at this point:
^(\w{4}) that just gets me the first four characters but I don't really have a stopping point as sometimes the string could be somelongstring292894830982 which would require me to get somelongstring
Using \w will match a word character which includes characters and digits and an underscore.
You could match an optional digit [0-9]? from the start of the string ^and then match 1+ times A-Za-z
^[0-9]?[A-Za-z]+
Regex demo
const regex = /^[0-9]?[A-Za-z]+/;
[
"7EUSA8889er898",
"somelongstring292894830982",
"SWFX74849948"
].forEach(s => console.log(s.match(regex)[0]));
Can use this regex code:
(^\d+?[a-zA-Z]+)|(^\d+|[a-zA-Z]+)
I try with exmaple and good worked:
1- somelongstring292894830982 -> somelongstring
2- 7sdfsdf5456 -> 7sdfsdf
3- 875werwer54556 -> 875werwer
If you want to create function where the RegExp is parametrized by n parameter, this would be
function getStr(str,n) {
var pattern = "\\d?\\w{0,"+n+"}";
var reg = new RegExp(pattern);
var result = reg.exec(str);
if(result[0]) return result[0].substr(0,n);
}
There are answers to this but here is another way to do it.
var string1 = '7EUSA8889er898';
var string2 = 'SWFX74849948';
var Extract = function (args) {
var C = args.split(''); // Split string in array
var NI = []; // Store indexes of all numbers
// Loop through list -> if char is a number add its index
C.map(function (I) { return /^\d+$/.test(I) === true ? NI.push(C.indexOf(I)) : ''; });
// Get the items between the first and second occurence of a number
return C.slice(NI[0] === 0 ? NI[0] + 1 : 0, NI[1]).join('');
};
console.log(Extract(string1));
console.log(Extract(string2));
Output
EUSA
SWFX7
Since it's hard to tell what you are trying to match, I'd go with a general regex
^\d?\D+(?=\d)
I need a way to match a word against a string and not get false positives. Let me give an example of what I mean:
"/thing" should match the string "/a/thing"
"/thing" should match the string "/a/thing/that/is/here"
"/thing" should NOT match the string "/a/thing_foo"
Basically, it should match if the exact characters are there in the first string and the second, but not if there are run-ons in the second (such as an underscore like in thing_foo).
Right now, I'm doing this, which is not working.
let found = b.includes(a); // true
Hopefully my question is clear enough. Thanks for the help!
Boy did this turn in to a classic XY Problem.
If I had to guess, you want to know if a path contains a particular segment.
In that case, split the string on a positive lookahead for '/' and use Array.prototype.includes()
const paths = ["/a/thing", "/a/thing/that/is/here", "/a/thing_foo"]
const search = '/thing'
paths.forEach(path => {
const segments = path.split(/(?=\/)/)
console.log('segments', segments)
console.info(path, ':', segments.includes(search))
})
Using the positive lookahead expression /(?=\/)/ allows us to split the string on / whilst maintaining the / prefix in each segment.
Alternatively, if you're still super keen in using a straight regex solution, you'll want something like this
const paths = ["/a/thing", "/a/thing/that/is/here", "/a/thing_foo", "/a/thing-that/is/here"]
const search = '/thing'
const rx = new RegExp(search + '\\b') // note the escaped backslash
paths.forEach(path => {
console.info(path, ':', rx.test(path))
})
Note that this will return false positives if the search string is followed by a hyphen or tilde as those are considered to be word boundaries. You would need a more complex pattern and I think the first solution handles these cases better.
I'd recommend using regular expressions...
e.g. The following regular expression /\/thing$/ - matches anything that ends with /thing.
console.log(/\/thing$/.test('/a/thing')) // true
console.log(/\/thing$/.test('/a/thing_foo')) // false
Update: To use a variable...
var search = '/thing'
console.log(new RegExp(search + '$').test('/a/thing')) // true
console.log(new RegExp(search + '$').test('/a/thing_foo')) // false
Simply with following regex you can do it
var a = "/a/thing";
var b = "/a/thing/that/is/here";
var c = "/a/thing_foo";
var pattern = new RegExp(/(:?(thing)(([^_])|$))/);
pattern.test(a) // true
pattern.test(b) // true
pattern.test(c) // false
What I'd like to achieve is splitting a string like this, i.e. the delimiters are the indexes where the character before that index is different from the character after that index:
"AAABBCCCCDEEE" -> ["AAA", "BB", "CCCC", "D", "EEE"]
I've been trying to make up a concise solution, but I ended up with this rather verbose code: http://jsfiddle.net/b39aM/1/.
var arr = [], // output
text = "AAABBCCCCDEEE", // input
current;
for(var i = 0; i < text.length; i++) {
var char = text[i];
if(char !== current) { // new letter
arr.push(char); // create new array element
current = char; // update current
} else { // current letter continued
arr[arr.length - 1] += char; // append letter to last element
}
}
It's naive and I don't like it:
I'm manually iterating over each character, and I'm appending to the array character by character
It's a little too long for the simple thing I want to achieve
I was thinking of using a regexp but I'm not sure what the regexp should be. Is it possible to define a regexp that means "one character and a different character following"?
Or more generally, is there a more elegant solution for achieving this splitting method?
Yes, you can use a regular expression:
"AAABBCCCCDEEE".match(/(.)\1*/g)
Here . will match any character and \1* will match any following characters that are the same as the formerly matched one. And with a global match you’ll get all matching sequences.