Is there a simple way to check how many times a character appears in a String?
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4possible duplicate of Count the number of occurances of a character in a string in JavascriptMichael Mrozek– Michael Mrozek2010-10-13 05:30:04 +00:00Commented Oct 13, 2010 at 5:30
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8 Answers
You could remove any other character in the string and check the length:
str.replace(/[^a]/g, "").length
Here it is counted how many as are in str. The RegExp is described below:
[ // Start Character Group
^ // Not operator in character group
a // The character "a"
] // End character group
3 Comments
HappyHands31
Just want to point out that you can account for uppercase as well by doing
let xLength = str.replace(/[^a || A]/g, "").length;
Donald Duck is with Ukraine
@HappyHands31 You mean
let xLength = str.replace(/[^aA]/g, "").length;
Aniruddha
@HappyHands31 couldn't we just add the case insensitive flag (i) to the regex and make it
str.replace(/[^a]/gi, "").length?This counts a in below example:
str = "A man is as good as his word";
alert(str.split('a').length-1);
If you want case insensitive you'd want something like
alert(str.split( new RegExp( "a", "gi" ) ).length-1);
So that it grabs "A" and "a" ... "g" flag isn't really needed, but you do need the "i" flag
1 Comment
vdegenne
@Josh It depends on what everyone likes. For instance I like this answer because it is easy to read. But someone else will prefer another answer for it's more technical, or challenging, another person will prefer another answer because it is eating less cpu and so on.. that is why JavaScript is so nice, a taste of freedom.
Use a RegEx to count the number of "a"s in a string.
var string = 'aajlkjjskdjfAlsj;gkejflksajfjskda';
document.write(string.match(/a/gi).length);Let me explain how this works:
string.match This is a RegEx method. It searches for the specified RegEx inside the specified string (in this case, the string "string").
(/a/gi) This is the actual RegEx. It reads, "find the character a." It's very simple. It also carries two flags, the "g" and the "i". The "g" says to find ALL occurences of the character "a". Otherwise it would only find the first one, and it would never count past the number one. The second flag is "i". It makes the RegEx match all cases of that character. If that flag (i) was not there, the code above would only count 4, because it would skip the uppercase "A" in the string. Because of the "i", it will match upper and lower case. Remove the "i" if you you want to match letter case.
string.match returns an array of all of the matches, so we use the length method to retrieve the number of array entries. Simple as that!
4 Comments
Gismo Ranas
I think you should remove the i from gi, it's a source of errors. You can add it as a bonus, but in my case I was looking specifically for case sensitive match.
pbialy
the problem here is that if you don't have "
a" in string, then string.match(/a/gi) equals null and you got error Cannot read property 'length' of null.
King Friday
The 'gi' part was very helpful, Sir @Drazzah
Raphael Morgan
not gonna tag them because the comment is over 2 years old but if anyone is concerned about the problem pbialy mentioned, just put
if (string.match(/a/gi)) and that should fix the problemIn my opinion it is more convenient and safe to avoid regular expressions in this case
It's because if we want to be able to count any kind of characters then we need to consider two expressions. One for common characters and second for special characters for example like [, ], ^ and so on. It's easy to forget about it, but even if we remember it, I think we're unnecessarily expanding our code.
In this case for string str and character ch works each of these solutions:
let count = str.split(ch).length - 1
(thanks to @Sarfraz)
or
let count = str.split('').filter(x => x == ch).length
or
let count = 0
str.split('').forEach(x => x == ch ? count++ : null)
Comments
var s = "dqsskjhfds";
alert(s.length - s.replace(/a/g, "").length); // number of 'a' in the string
Comments
Very simple.
s.match(/a/g).length
const s = 'banana'
console.log(s.match(/a/g).length) // 3
Comments
var a = "acvbasbb";
var b= {};
for (let i=0;i<a.length;i++){
if((a.match(new RegExp(a[i], "g"))).length > 1){
b[a[i]]=(a.match(new RegExp(a[i], "g"))).length;
}
}
console.log(b);
2 Comments
Ahmad F
Although your code snippet might solve the issue, you should describe what’s the purpose of your code (how it solves the problem). Furthermore, you might want to check stackoverflow.com/help/how-to-answer
Nitin .
@AhmadF you can use this for > How many times a character occurs in a string. and when you run this script you ll get an object with properties (characters) which occurs more then once.
Use a RegEx to count the number of characters this way:
console.log(('#FFAA00'.match(/f+/gi) || [''])[0].length);'#FFAA00'.match(/f+/gi).length
will return the length of the array: String.prototype.match()
console.log('#FFAA00'.match(/f+/gi).length);