I am trying to solve a problem, where the function accepts a string, always in lowercase, and then takes the first letter and adds it to the back of the string.
It does not seem to be working.
var fir;
function convert(str) {
str = str.split("");
fir = str[0];
str = str.join("");
switch (fir) {
case /[bcdfghjklmnpqrstvwxyz]/.test(fir):
var ind = str.substr(1);
str = ind + str[0];
break;
}
return str;
}
If I call convert("pig"); from another part of my javascript then it should return "igp"
I think this is all you need which does what you asked for - " function accepts a string, always in lowercase, and then takes the first letter and adds it to the back of the string".
function convert(str) {
str = str.substr(1) + str[0];
return str;
}
convert("pig");
If you are insistent on using what you have, I have fixed it here.
var fir;
function convert(str) {
str = str.split("");
fir = str[0];
str = str.join("");
switch (true) {
case /[bcdfghjklmnpqrstvwxyz]/.test(fir):
var ind = str.substr(1);
str = ind + str[0];
break;
}
return str;
}
convert("pig");
The only change I have done is changing switch(fir) to switch(true) which is why your switch case wasn't working.
I think you should read about the Javascript String Reference.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String
Remember:
The other way (introduced in ECMAScript 5) is to treat the string as an array-like object, where individual characters correspond to a numerical index:
Now, after reading the JS String Reference. We can do something even more smart concatenating the sliced string array without the 0 index letter with the last letter of the string array.
That way you could do something without regular expressions, switch and other stuff.
Remember to always read the references about the Global Objects, it could easy save you a lot of work and get better understand of the JS Language
Example code:
function convert(string){
return string.slice(1) + string[0];
}
Your switch statement is incorrect. Your case expression case /[bcdfghjklmnpqrstvwxyz]/.test(fir) evaluates to a boolean value while in switch expression you are passing a character in fir variable (as str[0] = 'p') which is a mismatch. So your case statement will never get executed and it will come out of switch block without doing anything. Change it to switch(true) and you should be good.
Related
I am doing a algorithm in freeCodeCamp.(https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/search-and-replace)
The task is as below:
Perform a search and replace on the sentence using the arguments provided and return the new sentence.
First argument is the sentence to perform the search and replace on.
Second argument is the word that you will be replacing (before).
Third argument is what you will be replacing the second argument with (after).
Note:
Preserve the case of the first character in the original word when you are replacing it. For example if you mean to replace the word "Book" with the word "dog", it should be replaced as "Dog"
**
myReplace("Let us get back to more Coding", "Coding", "algorithms") should return "Let us get back to more Algorithms".
myReplace("Let us go to the store", "store", "mall") should return "Let us go to the mall".
**
//if the before is uppercase, the after should be uppercase also
// str = str.replace(before, after);
var regex = /[A-Z]+/g; //check for uppercase
var newStr = "";
console.log(regex.test(before));
if (regex.test(before)) {
//if uppercase, return true, "after" convert to uppercase
after = after.toUpperCase();
newStr = after[0];
for (var i = 1; i < after.length; i++) {
//start at index=1 letter, all convert to
newStr += after[i].toLowerCase();
}
console.log(newStr);
str = str.replace(before, newStr);
} else {
str = str.replace(before, after);
}
// console.log(newStr);
console.log(str);
return str;
}
I think there should be OK for the code, but can anyone help find why the if statement can't work.
Much thanks!
The problem is that you're calling regex.test() multiple times on the same regular expression instance.
[...]
var regex = /[A-Z]+/g; //check for uppercase
var newStr = "";
console.log(regex.test(before));
if (regex.test(before)) {
//if uppercase, return true, "after" convert to uppercase
after = after.toUpperCase();
[...]
If your string is Hello_there, the first regex.test() will return true, because Hello matched. If you call regex.test() again with the same regex instance, it will have advanced in the string, and try to match starting with _there. In this case, it will fail, because _there does not begin with a capital letter between A and Z.
There are a lot of ways to fix this issue. Perhaps the simplest is to store the result of the first call to a variable, and use it everywhere you're calling regex.test():
[...]
var regex = /[A-Z]+/g; //check for uppercase
var newStr = "";
var upper_check = regex.test(before);
console.log(upper_check);
if (upper_check) {
[...]
It seems overkill to use a regex, when you really need to only check the first character. Your regex will find uppercase letters anywhere...
If the assignment is to only change one occurrence, then a regex is not really the right tool here: it does not really help to improve the code nor the efficiency. Just do:
function myReplace(str, before, after) {
if (before[0] === before[0].toUpperCase()) {
after = after[0].toUpperCase() + after.slice(1);
} else {
after = after[0].toLowerCase() + after.slice(1);
}
return str.replace(before, after);
}
function myReplace(str, before, after) {
var upperRegExp = /[A-Z]/g
var lowerRegExp = /[a-z]/g
var afterCapitalCase = after.replace(/^./, after[0].toUpperCase());
if (before[0].match(upperRegExp)) {
return str.replace(before, afterCapitalCase)
} else if (after[0].match(upperRegExp) && before[0].match(lowerRegExp)) {
return str.replace(before, after.toLowerCase());
} else {
return str.replace(before, after)
}
}
For making 1st letter of any word in Upper case, I am using below -
var HVal = "SchON";
HVal.toLowerCase().replace(/\b[a-z]/g,function(f){return f.toUpperCase();});
I get the correct output here as "Schon". However, this code does not work if my string contains umlaut vowel e.g. HVal = "SchÖN". the output here is "SchöN". How can I make this work for strings containing umlaut letters ?
The solution is
var HVal = "SchÖN dasdNdsad";
HVal.toLowerCase().replace(/(^[a-z])|(\s[a-z])/g,function(f){return f.toUpperCase();});
Use can something like this
var HVal = "SchÖN";
HVal=capitalize(HVal);
alert(HVal);
function capitalize(string) {
return string.charAt(0).toUpperCase() + string.slice(1).toLowerCase();
}
I want to after type Title of post automatically take value and create slug. My code works fine with English Latin characters but problem is when I type characters 'čćšđž'. Code replace first type of this characters in string but if character is repeated than is a problem. So, for testing purpose this title 'šžđčćžđš čćšđžčćšđž čćšđžčć ćčšđžšžčćšđ ćčšžčć' is converted to this slug 'szdcc'.
This is my jquery code:
$('input[name=title]').on('blur', function() {
var slugElement = $('input[name=slug]');
if(slugElement.val()) {
return;
}
slugElement.val(this.value.toLowerCase().replace('ž', 'z').replace('č','c').replace('š', 's').replace('ć', 'c').replace('đ', 'd').replace(/[^a-z0-9-]+/g, '-').replace(/^-+|-+$/g, ''));
});
How to solve this problems? Also is it possible to this few characters put in same replace() function?
Try this:
function clearText(inp) {
var wrong = 'čćšđž';
var right = 'ccsdz';
var re = new RegExp('[' + wrong + ']', 'ig');
return inp.replace(re, function (m) { return right.charAt(wrong.indexOf(m)); });
}
replace() only replaces the first occurrence unless regex is used with global modifier. You will need to change them all to regular expression.
replace(/ž/g, "z")
As far as I know, it will not be possible to use a single replace() in your case.
If you are concerned with chaining a bunch of .replace() together, you might be better off writing some custom code to replace these characters.
var newStr = "";
for (var i = 0; i < str.length; i++) {
var c = str.charAt(i);
switch (c) {
case "ž": newStr += "z"; break;
default: newStr += c; break;
}
}
I understand the .split() function quite well. But what I can seem to figure out is how to split in certain places but not in others. Sounds confusing? Well I mean for example, lets say I use .split(",") on the following string:
div:(li,div),div
Is it possible to split it so that only the commas ouside of the parentheses get split.
So the string above with the split method should return:
['div:(li,div)', 'div']
Of course at the moment it is also splitting the first comma inside of the parentheses, returning:
['div:(li', 'div)', 'div']
Is there some way to make this work like I desire?
If your expected strings are not going to become more complicated than this, you don't have to worry about writing code to parse them. Regex will work just fine.
http://jsfiddle.net/dC5HN/1/
var str = "div:(li,div),div:(li,div),div";
var parts = str.split(/,(?=(?:[^\)]|\([^\)]*\))*$)/g);
console.log(parts);
outputs:
["div:(li,div)", "div:(li,div)", "div"]
REGEX is not built for this sort of thing, which is essentially parsing.
When faced with this sort of situation previously I've first temporarily replaced the parenthesised parts with a placeholder, then split, then replaced the placeholders with the original parenthised parts.
A bit hacky, but it works:
var str = 'div:(li,div),div',
repls = [];
//first strip out parenthesised parts and store in array
str = str.replace(/\([^\)]*\)/g, function($0) {
repls.push($0);
return '*repl'+(repls.length - 1)+'*';
});
//with the parenthisised parts removed, split the string then iteratively
//reinstate the removed parenthisised parts
var pieces = str.split(',').map(function(val, index) {
return val.replace(/\*repl(\d+)\*/, function($0, $1) {
return repls[$1];
});
});
//test
console.log(pieces); //["div:(li,div)","div"]
This function will split whatever you specify in splitChar, but ignore that value if inside parenthesis:
function customSplit(stringToSplit, splitChar){
var arr = new Array();
var isParenOpen = 0
var curChar;
var curString = "";
for (var i = 0; i < stringToSplit.length; i++) {
curChar = stringToSplit.substr(i, 1);
switch(curChar) {
case "(":
isParenOpen++;
break;
case ")":
if(isParenOpen > 0) isParenOpen--;
break;
case splitChar:
if (isParenOpen < 1) {
arr.push(curString);
curString = "";
continue;
}
}
curString += curChar;
}
if (curString.length > 0) {
arr.push(curString);
}
return arr;
}
How can I quickly validate if a string is alphabetic only, e.g
var str = "!";
alert(isLetter(str)); // false
var str = "a";
alert(isLetter(str)); // true
Edit : I would like to add parenthesis i.e () to an exception, so
var str = "(";
or
var str = ")";
should also return true.
Regular expression to require at least one letter, or paren, and only allow letters and paren:
function isAlphaOrParen(str) {
return /^[a-zA-Z()]+$/.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