How can I quickly validate if a string is alphabetic only, e.g
var str = "!";
alert(isLetter(str)); // false
var str = "a";
alert(isLetter(str)); // true
Edit : I would like to add parenthesis i.e () to an exception, so
var str = "(";
or
var str = ")";
should also return true.
Regular expression to require at least one letter, or paren, and only allow letters and paren:
function isAlphaOrParen(str) {
return /^[a-zA-Z()]+$/.test(str);
}
Modify the regexp as needed:
/^[a-zA-Z()]*$/ - also returns true for an empty string
/^[a-zA-Z()]$/ - only returns true for single characters.
/^[a-zA-Z() ]+$/ - also allows spaces
Here you go:
function isLetter(s)
{
return s.match("^[a-zA-Z\(\)]+$");
}
If memory serves this should work in javascript:
function containsOnlyLettersOrParenthesis(str)
(
return str.match(/^([a-z\(\)]+)$/i);
)
You could use Regular Expressions...
functions isLetter(str) {
return str.match("^[a-zA-Z()]+$");
}
Oops... my bad... this is wrong... it should be
functions isLetter(str) {
return "^[a-zA-Z()]+$".test(str);
}
As the other answer says... sorry
Related
As a follow up to this question (not by me), I need to replace leading numbers of an id with \\3n (where n is the number we're replacing).
Some examples:
"1foo" -> "\\31foo"
"1foo1" -> "\\31foo1"
"12foo" -> "\\31\\32foo"
"12fo3o4" -> "\\31\\32fo3o4"
"foo123" -> "foo123"
Below is a solution that replaces every instance of the number, but I don't know enough regex to make it stop once it hits a non-number.
function magic (str) {
return str.replace(/([0-9])/g, "\\3$1");
}
... Or is regex a bad way to go? I guess it would be easy enough to do it, just looping over each character of the string manually.
Here is a way to achieve what you need using a reverse string + look-ahead approach:
function revStr(str) {
return str.split('').reverse().join('');
}
var s = "12fo3o4";
document.write(revStr(revStr(s).replace(/\d(?=\d*$)/g, function (m) {
return m + "3\\\\";
}))
);
The regex is matching a number that can be followed by 0 or more numbers only until the end (which is actually start) of a reversed string (with \d(?=\d*$)). The callback allows to manipulate the match (we just add reversed \\ and 3. Then, we just reverse the result.
Just use two steps: first find the prefix, then operate on its characters:
s.replace(/^\d+/, function (m) {
return [].map.call(m, function (c) {
return '\\3' + c;
}).join('');
});
No need to emulate any features.
Here is how I would have done it:
function replace(str) {
var re = /^([\d]*)/;
var match = str.match(re)[0];
var replaced = match.replace(/([\d])/g, "\\3$1");
str = str.replace(match, replaced);
return str;
}
document.write(replace("12fo3o4"));
Don't get me wrong: the other answers are fine! My focus was more on readability.
How can I convert PascalCase string into underscore_case/snake_case string? I need to convert dots into underscores as well.
eg. convert
TypeOfData.AlphaBeta
into
type_of_data_alpha_beta
You could try the below steps.
Capture all the uppercase letters and also match the preceding optional dot character.
Then convert the captured uppercase letters to lowercase and then return back to replace function with an _ as preceding character. This will be achieved by using anonymous function in the replacement part.
This would replace the starting uppercase letter to _ + lowercase_letter.
Finally removing the starting underscore will give you the desired output.
var s = 'TypeOfData.AlphaBeta';
console.log(s.replace(/(?:^|\.?)([A-Z])/g, function (x,y){return "_" + y.toLowerCase()}).replace(/^_/, ""));
OR
var s = 'TypeOfData.AlphaBeta';
alert(s.replace(/\.?([A-Z])/g, function (x,y){return "_" + y.toLowerCase()}).replace(/^_/, ""));
any way to stop it for when a whole word is in uppercase. eg. MotorRPM into motor_rpm instead of motor_r_p_m? or BatteryAAA into battery_aaa instead of battery_a_a_a?
var s = 'MotorRMP';
alert(s.replace(/\.?([A-Z]+)/g, function (x,y){return "_" + y.toLowerCase()}).replace(/^_/, ""));
str.split(/\.?(?=[A-Z])/).join('_').toLowerCase();
u're welcome
var s1 = 'someTextHere';
var s2 = 'SomeTextHere';
var s3 = 'TypeOfData.AlphaBeta';
var o1 = s1.split(/\.?(?=[A-Z])/).join('_').toLowerCase();
var o2 = s2.split(/\.?(?=[A-Z])/).join('_').toLowerCase();
var o3 = s3.split(/\.?(?=[A-Z])/).join('_').toLowerCase();
console.log(o1);
console.log(o2);
console.log(o3);
Alternatively using lodash:
lodash.snakeCase(str);
Example:
_.snakeCase('TypeOfData.AlphaBeta');
// ➜ 'type_of_data_alpha_beta'
Lodash is a fine library to give shortcut to many everyday js tasks.There are many other similar string manipulation functions such as camelCase, kebabCase etc.
This solution solves the non-trailing acronym issue with the solutions above
I ported the code in 1175208 from Python to JavaScript.
Javascript Code
function camelToSnakeCase(text) {
return text.replace(/(.)([A-Z][a-z]+)/, '$1_$2').replace(/([a-z0-9])([A-Z])/, '$1_$2').toLowerCase()
}
Working Examples:
camelToSnakeCase('thisISDifficult') -> this_is_difficult
camelToSnakeCase('thisISNT') -> this_isnt
camelToSnakeCase('somethingEasyLikeThis') -> something_easy_like_this
"alphaBetaGama".replace(/([A-Z])/g, "_$1").toLowerCase() // alpha_beta_gamma
Problem - Need to convert a camel-case string ( such as a property name ) into underscore style to meet interface requirements or for meta-programming.
Explanation
This line uses a feature of regular expressions where it can return a matched result ( first pair of () is $1, second is $2, etc ).
Each match in the string is converted to have an underscore ahead of it with _$1 string provided. At that point the string looks like alpha_Beta_Gamma.
To correct the capitalization, the entire string is converted toLowerCase().
Since toLowerCase is a fairly expensive operation, its best not to put it in the looping handler for each match-case, and run it once on the entire string.
After toLowerCase it the resulting string is alpha_beta_gamma ( in this example )
This will get you pretty far: https://github.com/domchristie/humps
You will probably have to use regex replace to replace the "." with an underscore.
I found this but I edited it so suit your question.
const camelToSnakeCase = str => str.replace(/[A-Z]/g, letter => `_${letter.toLowerCase()}`).replace(/^_/,'')
Good examples for js:
Snake Case
Kebab Case
Camel Case
Pascal Case
have here
function toCamelCase(s) {
// remove all characters that should not be in a variable name
// as well underscores an numbers from the beginning of the string
s = s.replace(/([^a-zA-Z0-9_\- ])|^[_0-9]+/g, "").trim().toLowerCase();
// uppercase letters preceeded by a hyphen or a space
s = s.replace(/([ -]+)([a-zA-Z0-9])/g, function(a,b,c) {
return c.toUpperCase();
});
// uppercase letters following numbers
s = s.replace(/([0-9]+)([a-zA-Z])/g, function(a,b,c) {
return b + c.toUpperCase();
});
return s;
}
Try this function, hope it helps.
"TestString".replace(/[A-Z]/g, val => "_" + val.toLowerCase()).replace(/^_/,"")
replaces all uppercase with an underscore and lowercase, then removes the leading underscore.
A Non-Regex Answer that converts PascalCase to snake_case
Note: I understand there are tons of good answers which solve this question elegantly. I was recently working on something similar to this where I chose not to use regex. So I felt to answer a non-regex solution to this.
const toSnakeCase = (str) => {
return str.slice(0,1).toLowerCase() + str.split('').slice(1).map((char) => {
if (char == char.toUpperCase()) return '_' + char.toLowerCase();
else return char;
}).join('');
}
Eg.
inputString = "ILoveJavascript" passed onto toSnakeCase()
would become "i_love_javascript"
Can anyone tell me why does this not work for integers but works for characters? I really hate reg expressions since they are cryptic but will if I have too. Also I want to include the "-()" as well in the valid characters.
String.prototype.Contains = function (str) {
return this.indexOf(str) != -1;
};
var validChars = '0123456789';
var str = $("#textbox1").val().toString();
if (str.Contains(validChars)) {
alert("found");
} else {
alert("not found");
}
Review
String.prototype.Contains = function (str) {
return this.indexOf(str) != -1;
};
This String "method" returns true if str is contained within itself, e.g. 'hello world'.indexOf('world') != -1would returntrue`.
var validChars = '0123456789';
var str = $("#textbox1").val().toString();
The value of $('#textbox1').val() is already a string, so the .toString() isn't necessary here.
if (str.Contains(validChars)) {
alert("found");
} else {
alert("not found");
}
This is where it goes wrong; effectively, this executes '1234'.indexOf('0123456789') != -1; it will almost always return false unless you have a huge number like 10123456789.
What you could have done is test each character in str whether they're contained inside '0123456789', e.g. '0123456789'.indexOf(c) != -1 where c is a character in str. It can be done a lot easier though.
Solution
I know you don't like regular expressions, but they're pretty useful in these cases:
if ($("#textbox1").val().match(/^[0-9()]+$/)) {
alert("valid");
} else {
alert("not valid");
}
Explanation
[0-9()] is a character class, comprising the range 0-9 which is short for 0123456789 and the parentheses ().
[0-9()]+ matches at least one character that matches the above character class.
^[0-9()]+$ matches strings for which ALL characters match the character class; ^ and $ match the beginning and end of the string, respectively.
In the end, the whole expression is padded on both sides with /, which is the regular expression delimiter. It's short for new RegExp('^[0-9()]+$').
Assuming you are looking for a function to validate your input, considering a validChars parameter:
String.prototype.validate = function (validChars) {
var mychar;
for(var i=0; i < this.length; i++) {
if(validChars.indexOf(this[i]) == -1) { // Loop through all characters of your string.
return false; // Return false if the current character is not found in 'validChars' string.
}
}
return true;
};
var validChars = '0123456789';
var str = $("#textbox1").val().toString();
if (str.validate(validChars)) {
alert("Only valid characters were found! String validates!");
} else {
alert("Invalid Char found! String doesn't validate.");
}
However, This is quite a load of code for a string validation. I'd recommend looking into regexes, instead. (Jack's got a nice answer up here)
You are passing the entire list of validChars to indexOf(). You need to loop through the characters and check them one-by-one.
Demo
String.prototype.Contains = function (str) {
var mychar;
for(var i=0; i<str.length; i++)
{
mychar = this.substr(i, 1);
if(str.indexOf(mychar) == -1)
{
return false;
}
}
return this.length > 0;
};
To use this on integers, you can convert the integer to a string with String(), like this:
var myint = 33; // define integer
var strTest = String(myint); // convert to string
console.log(strTest.Contains("0123456789")); // validate against chars
I'm only guessing, but it looks like you are trying to check a phone number. One of the simple ways to change your function is to check string value with RegExp.
String.prototype.Contains = function(str) {
var reg = new RegExp("^[" + str +"]+$");
return reg.test(this);
};
But it does not check the sequence of symbols in string.
Checking phone number is more complicated, so RegExp is a good way to do this (even if you do not like it). It can look like:
String.prototype.ContainsPhone = function() {
var reg = new RegExp("^\\([0-9]{3}\\)[0-9]{3}-[0-9]{2}-[0-9]{2}$");
return reg.test(this);
};
This variant will check phones like "(123)456-78-90". It not only checks for a list of characters, but also checks their sequence in string.
Thank you all for your answers! Looks like I'll use regular expressions. I've tried all those solutions but really wanted to be able to pass in a string of validChars but instead I'll pass in a regex..
This works for words, letters, but not integers. I wanted to know why it doesn't work for integers. I wanted to be able to mimic the FilteredTextBoxExtender from the ajax control toolkit in MVC by using a custom Attribute on a textBox
why does this code return false:
var text = "3b3xx";
if(text.match("/^\d?b\d+xx$/")) {
return true;
}
return false;
I can not see any problem with my regular expression.. I want to return true, if the string starts with any numbers, followed by "b", followed by any numers, followed by "xx".
That's a string, not a regex.
Remove the "".
You are passing a string where a regular expression is expected.
var text = "3b3xx";
if(text.match(/^\d?b\d+xx$/)) {
return true;
}
return false;
Why not trying this:
var text = "3b3xx";
return text.match(/^\d?b\d+xx$/);
Just lose the quotes around your regex.
Regex is an object in Javascript, not a String.
/^\d?b\d+xx$/
The replace function returns the new string with the replaces, but if there weren't any words to replace, then the original string is returned. Is there a way to know whether it actually replaced anything apart from comparing the result with the original string?
A simple option is to check for matches before you replace:
var regex = /i/g;
var newStr = str;
var replaced = str.search(regex) >= 0;
if(replaced){
newStr = newStr.replace(regex, '!');
}
If you don't want that either, you can abuse the replace callback to achieve that in a single pass:
var replaced = false;
var newStr = str.replace(/i/g, function(token){replaced = true; return '!';});
As a workaround you can implement your own callback function that will set a flag and do the replacement. The replacement argument of replace can accept functions.
Comparing the before and after strings is the easiest way to check if it did anything, there's no intrinsic support in String.replace().
[contrived example of how '==' might fail deleted because it was wrong]
Javascript replace is defected by design. Why? It has no compatibility with string replacement in callback.
For example:
"ab".replace(/(a)(b)/, "$1$2")
> "ab"
We want to verify that replace is done in single pass. I was imagine something like:
"ab".replace(/(a)(b)/, "$1$2", function replacing() { console.log('ok'); })
> "ab"
Real variant:
"ab".replace(/(a)(b)/, function replacing() {
console.log('ok');
return "$1$2";
})
> ok
> "$1$2"
But function replacing is designed to receive $0, $1, $2, offset, string and we have to fight with replacement "$1$2". The solution is:
"ab".replace(/(a)(b)/, function replacing() {
console.log('ok');
// arguments are $0, $1, ..., offset, string
return Array.from(arguments).slice(1, -2)
.reduce(function (pattern, match, index) {
// '$1' from strings like '$11 $12' shouldn't be replaced.
return pattern.replace(
new RegExp("\\$" + (index + 1) + "(?=[^\\d]|$)", "g"),
match
);
}, "$1$2");
});
> ok
> "ab"
This solution is not perfect. String replacement itself has its own WATs. For example:
"a".replace(/(a)/, "$01")
> "a"
"a".replace(/(a)/, "$001")
> "$001"
If you want to care about compatibility you have to read spec and implement all its craziness.
If your replace has a different length from the searched text, you can check the length of the string before and after. I know, this is a partial response, valid only on a subset of the problem.
OR
You can do a search. If the search is successfull you do a replace on the substring starting with the found index and then recompose the string. This could be slower because you are generating 3 strings instead of 2.
var test = "Hellllo";
var index = test.search(/ll/);
if (index >= 0) {
test = test.substr(0, index - 1) + test.substr(index).replace(/ll/g, "tt");
}
alert(test);
While this will require multiple operations, using .test() may suffice:
const regex = /foo/;
const yourString = 'foo bar';
if (regex.test(yourString)) {
console.log('yourString contains regex');
// Go ahead and do whatever else you'd like.
}
The test() method executes a search for a match between a regular expression and a specified string. Returns true or false.
With indexOf you can check wether a string contains another string.
Seems like you might want to use that.
have a look at string.match() or string.search()
After doing any RegExp method, read RegExp.lastMatch property:
/^$/.test(''); //Clear RegExp.lastMatch first, Its value will be ''
'abcd'.replace(/bc/,'12');
if(RegExp.lastMatch !== '')
console.log('has been replaced');
else
console.log('not replaced');