Use toSentenceCase on function parameter Javascript - javascript

In a Codecademy assignment, I would like to add a case to the parameter of my function, so whenever a person searches on the name "JoNes", the parameter gets "translated" to "Jones". (this is just to play with Javascript, not mandatory)
According to this website, there is a case called Sentence Case (he also calls it Title Case), however, whenever I use .toSentenceCase, it returns a syntax error on Codecademy. I would like to know if the following code would normally work and that it's Codecademy that doesn't support it in this assignment (in other words: it doesn't expect it, so anything different than the expected is wrong) OR if it's not possible to add a case to the parameter of a function like this.
Bonus: If it's the latter, how would you fix the input coming in through search();, so it always corresponds to the demand of the first letter being upper-case, while the rest is lower case?
The function:
var search = function(lastName.toSentenceCase) {
var contactsLength = contacts.length;
for (var i = 0; i < contactsLength; i++) {
if (lastName === contacts[i].lastName) {
printPerson(contacts[i]);
}
}
}
search("JoNes");

Here you go:
String.prototype.toSentenceCase= function() {
return this.charAt(0).toUpperCase() + this.slice(1).toLowerCase()
}
Then:
"JoNes".toSentenceCase();
becomes: Jones

I would try this:
String.prototype.toSentenceCase = function () {
var stringArray = this.split(" ");
for (var index = 0; index < stringArray.length; index++) {
stringArray[index] = stringArray[index].substr(0,1).toUpperCase() + stringArray[index].substr(1).toLowerCase();
}
return stringArray.join(" ");
}
//test:
var testString = new String("JoNe Is A NaMe");
console.log(testString.toSentenceCase());

Related

How to find the missing next character in the array?

I have an array of characters like this:
['a','b','c','d','f']
['O','Q','R','S']
If we see that, there is one letter is missing from each of the arrays. First one has e missing and the second one has P missing. Care to be taken for the case of the character as well. So, if I have a huge Object which has all the letters in order, and check them for the next ones, and compare?
I am totally confused on what approach to follow! This is what I have got till now:
var chars = ("abcdefghijklmnopqrstuvwxyz"+"abcdefghijklmnopqrstuvwxyz".toUpperCase()).split("");
So this gives me with:
["a","b","c","d","e","f","g","h","i","j","k","l","m",
"n","o","p","q","r","s","t","u","v","w","x","y","z",
"A","B","C","D","E","F","G","H","I","J","K","L","M",
"N","O","P","Q","R","S","T","U","V","W","X","Y","Z"]
Which is awesome. Now my question is, how do I like check for the missing character in the range? Some kind of forward lookup?
I tried something like this:
Find the indexOf starting value in the source array.
Compare it with each of them.
If the comparison failed, return the one from the original array?
I think that a much better way is to check for each element in your array if the next element is the next char:
function checkMissingChar(ar) {
for (var i = 1; i < ar.length; i++) {
if (ar[i].charCodeAt(0) == ar[i-1].charCodeAt(0)+1) {
// console.log('all good');
} else {
return String.fromCharCode(ar[i-1].charCodeAt(0)+1);
}
}
return true;
}
var a = ['a','b','c','d','f']
var b = ['O','Q','R','S']
console.log(checkMissingChar(a));
console.log(checkMissingChar(b));
Not that I start to check the array with the second item because I compare it to the item before (the first in the Array).
Forward Look-Ahead or Negative Look-Ahead: Well, my solution would be some kind of that. So, if you see this, what I would do is, I'll keep track of them using the Character's Code using charCodeAt, instead of the array.
function findMissingLetter(array) {
var ords = array.map(function (v) {
return v.charCodeAt(0);
});
var prevOrd = "p";
for (var i = 0; i < ords.length; i++) {
if (prevOrd == "p") {
prevOrd = ords[i];
continue;
}
if (prevOrd + 1 != ords[i]) {
return String.fromCharCode(ords[i] - 1);
}
prevOrd = ords[i];
}
}
console.log(findMissingLetter(['a','b','c','d','f']));
console.log(findMissingLetter(['O','Q','R','S']));
Since I come from a PHP background, I use some PHP related terms like ordinal, etc. In PHP, you can get the charCode using the ord().
As Dekel's answer is better than mine, I'll try to propose somewhat more better answer:
function findMissingLetter (ar) {
for (var i = 1; i < ar.length; i++) {
if (ar[i].charCodeAt(0) != ar[i-1].charCodeAt(0)+1) {
return String.fromCharCode(ar[i-1].charCodeAt(0)+1);
}
}
return true;
}
var a = ['a','b','c','d','f']
var b = ['O','Q','R','S']
console.log(findMissingLetter(a));
console.log(findMissingLetter(b));
Shorter and Sweet.

How to count vowels in a Javascript string with two functions?

I'm trying to write a Javascript function that counts the vowels in a string by calling another function inside that function, but when I test it in the console it returns 0.
Here is my first function that works fine and recognizes if a string is a vowel:
function isVowel(ch){
var pattern = /[aeiouAEIOU]/
return pattern.test(ch);
};
For the second function none of my ideas have worked. Here are a few examples of what I have tried so far:
This one returns me a 0:
function countVowels(str){
var count = 0;
for(var i; i <= str.length; ++i){
if(isVowel(i)){
++count;
}
}
return count;
};
I also tried the above, but removing the .length after str in the for() area.
Another example, but this one gives me an error:
function countVowels(str){
var count = 0
var pattern = /[aeiouAEIOU]/
for(var i = 1; i <= str.length(pattern); ++i){
if(isVowel(i)){
++count;
}
}
return count;
};
I've tried various other functions as well, but for the sake of keeping this post relatively short I won't continue to post them. I'm quite new to Javascript and I'm not sure what I'm doing wrong. Any help would be greatly appreciated!
Try using .match() with the g attribute on your String.
g: global
i: case insensitive
Regexp documentation
function countVowels(ch){
return ch.match(/[aeiouy]/gi).length;
}
var str = "My string";
alert(countVowels(str)); // 2
Although Robiseb answer is the way to go, I want to let you know why you code is not working (I'm referring your first attempt). Basically you made two mistakes in the loop:
As CBroe stated, you are passing i to your isVowel function. i is a integer representing the index of the loop, not the actual character inside the string. To get the character you can do str.substr(i, 1), what means "give me one character from the position i inside the string".
You are not giving a initial value to the i variable. When you create a variable, it is undefined, so you can not increment it.
alert(countVowels("hello"));
function countVowels(str) {
var count = 0;
for (var i = 0; i <= str.length; ++i) {
if (isVowel(str.substr(i, 1))) {
count++;
}
}
return count;
};
function isVowel(ch) {
var pattern = /[aeiouAEIOU]/
return pattern.test(ch);
};
UPDATE: You will see that other answers use other methods to select the character inside the string from the index. You actually have a bunch of different options. Just for reference:
str.slice(i,i+1);
str.substring(i,i+1);
str.substr(i,1));
str.charAt(i);
str[i];
i is the index, not the character. It should be:
if (isVowel(str[i])) {
count++;
}
Also, str.length(pattern) is wrong. length is a property, not a function, so it should just be str.length.
You forgot to assign the value 0 to i variable
And parameter for isVowel is the character, not the index of string
Here information about the JS language: https://stackoverflow.com/tags/javascript/info
function isVowel(ch){
var pattern = /[aeiouAEIOU]/
return pattern.test(ch);
}
function countVowels(str){
var count = 0;
// you forgot to assign the value to i variable
for(var i = 0; i < str.length; i++){
// isVowel(str[i]), not isVowel(i)
if(isVowel(str[i])){
count++;
}
}
return count;
}
console.log(countVowels('forgot'))
Obviously you should do it this way:
function isVowel(c){
var lc = c.toLowerCase();
if(lc === 'y'){
return (Math.floor(Math.random() * 2) == 0);
}
return ['a','e','i','o','u'].indexOf(lc) > -1;
}
function countVowels(s){
var i = 0;
s.split('').each(function(c){
if(isVowel(c)){
i++;
}
});
return i;
}
console.log(countVowels("the quick brown fox jumps over the lazy dog"));
Which, although less efficient and less useful than other answers, at least has the entertaining property of returning a different count 50% of the time, because sometimes Y.

Best way to create string filter comparation?

My goal's create a filter search function, in particular actually I'm using .indexOf method that allow me to check if two string are equal. The problem's that if I've the compare string with space break like this: Hair Cut.
Example:
String to search: Hair
String contained in the object: Hair Cut
var cerca = $('#filter_service').val();
for(var i = 0; i < GlobalVariables.availableServices.length; i++) {
if (cerca.toLowerCase().contains(GlobalVariables.availableServices[i].name.toLowerCase()) != -1) {
console.log(GlobalVariables.availableServices[i].name)
}
}
How you can see I valorize the variable cerca that contains the string Hair in the example. I compare this with an object variable, how I said, the problem is if I insert the string Hair I get no response in console, also if I insert the string with break space like the compare string Hair Cut I get the console response.
How I can print a result also when the variable cerca is equal to the first character of the compair string? In particular Hai?
I don't know if I was clear, hope yes.
.contains() is for checking DOM element children. You said above that you are using .indexOf to check, but it doesn't look like you use it in your code?
var cerca = $('#filter_service').val();
var searchIn;
for(var i = 0; i < GlobalVariables.availableServices.length; i++) {
searchIn = GlobalVariables.availableServices[i].name.toLowerCase().split(' ');
for (j = 0; j < searchIn.length; j++) {
if (cerca.toLowerCase().split(' ').indexOf(searchIn[j].toLowerCase()) >= 0) {
console.log(GlobalVariables.availableServices[i].name);
}
}
}
$('#filter_service').on('input', function() {
var inputStr = $('#filter_service').val();
var similar = [];
for (i = 0; i < GlobalVariables.availableServices.length; i++) {
if (GlobalVariables.availableServices[i].name.toLowerCase().indexOf(inputStr.toLowerCase) >= 0) {
similar[similar.length] = GlobalVariables.availableServices[i].name;
}
}
// At this point, you can do whatever you want with the similar service
// names (all of the possible result names are included in the array, similar[].)
});
I can't test that code right now, but in theory, it should work.
Here is a JSFiddle demo:
http://jsfiddle.net/MrGarretto/vrp5pghr/
EDIT: Updated and fixed my errors
EDIT 2: Added the 'possible results' solution
EDIT 3: Added a JSFiddle

Using for loop to find specific characters in string

I am attempting the bean counting example in the functions chapter of the book Eloquent Javascript. My function is returning a blank.
Without giving me the full answer (I'm working through this example to learn), can someone tell me why my code is not printing any text?"
var string = "donkey puke on me boot thar be thar be!";
for (var i = 0; i <= string.length; i++);
function getB(){
if (string.charAt(i) == "b")
return i;
else return "";
}
console.log(getB());
There is something wrong about how you're trying to implement this feature.
First of all I think it's better if you have a function that accepts the string and the char as parameters in order to call it whenever you want.
Example of calling :
getChar('this is my custom string', 'c') -> it should search character `c` in `this is my custom string`
getChar('this is another custom string', 'b') -> it should search character `b` in `this is another custom string`
Example of implementation :
var getChar = function(string, char){
for(var i=0;i<string.length;i++)
{
if(string.charAt(i)==char) console.log(i);
}
}
Now, try to make it not case-sensitive and instead of console.log the output try to return a sorted array with character positions
Use This,
var string = "donkey puke on me boot thar be thar be!";
for (var i = 0; i <= string.length; i++) {
if (string.charAt(i) == "b") {
console.log(i);
}
}
if you want to print every position your value is at, you can program something like this:
var string = "donkey puke on me boot thar be thar be!";
for (var i = 0; i <= string.length; i++)
{
getChar(i, "b");
}
function getChar(i, input)
{
if (string.charAt(i) == input)
console.log(i);
}
another example: collect all b positions:
var string = "donkey puke on me boot thar be thar be!";
function getB(string){
var placesOfB = [];
for (var i = 0; i < string.length; i++) {
if (string.charAt(i) == "b") {
placesOfB.push(i);
}
}
return placesOfB;
}
console.log(getB(string));
tip: your for has no body (putting ; after it just loops without doing anything)...
and it does not make sense to define a function inside that for.
Without giving you the full answer, I'll just give you pointers:
1. Your for loop is not complete - it doesn't do anything.
2. Your getB() function needs to accept string parameter in order to perform some action on it.
3. The if..else statement doesn't have opening and closing brackets {}

javascript return all combination of a number

I am trying to get all combination of a number. For example, input "123" should return ["123", "231", "213", "312", "321", "132"].
Here is my function:
function swapDigits(input) {
for (var i = 0; i++; i < input.length - 1) {
var output = [];
var inter = input.slice(i, i + 1);
var left = (input.slice(0, i) + input.slice(i + 1, input)).split("");
for (var j = 0; j++; j <= left.length) {
var result = left.splice(j, 0, inter).join("");
output.push(result);
}
}
console.log(output);
return output;
}
However this function returns undefined, could anyone tell me what's going wrong?
The errors with the for loop and scope have already been mentioned. Besides that, the splice method will change the string that it operates on. This means that the inner loop will never terminate because left keeps on growing, so j never reaches left.length.
If you are new to a language, I would suggest starting with an implementation that is close to the algorithm that you want to implement. Then, once you are comfortable with it, use more advanced language constructs.
See this fiddle for an example. This is the algorithm code:
function getPermutations(input)
{
if(input.length <= 1)
{
return [input];
}
var character = input[0];
var returnArray = [];
var subPermutes = getPermutations(input.slice(1));
debugOutput('Returned array: ' + subPermutes);
for(var subPermuteIndex = 0; subPermuteIndex < subPermutes.length; subPermuteIndex++ )
{
var subPermute = subPermutes[subPermuteIndex];
for(var charIndex = 0; charIndex <= subPermute.length; charIndex++)
{
var pre = subPermute.slice( 0, charIndex );
var post = subPermute.slice( charIndex );
returnArray.push(pre+character+post);
debugOutput(pre + '_' + character + '_' + post );
}
}
return returnArray;
}
Basically, this will walk to the end of the string and work its way back constructing all permutations of sub-strings. It is easiest to see this from the debug output for 1234. Note that 'Returned array' refers to the array that was created by the permutations of the sub-string. Also note that the current character is placed in every position in that array. The current character is shown between _ such as the 1 in 432_1_.
Returned array: 4
_3_4
4_3_
Returned array: 34,43
_2_34
3_2_4
34_2_
_2_43
4_2_3
43_2_
Returned array: 234,324,342,243,423,432
_1_234
2_1_34
23_1_4
234_1_
_1_324
3_1_24
32_1_4
324_1_
_1_342
3_1_42
34_1_2
342_1_
_1_243
2_1_43
24_1_3
243_1_
_1_423
4_1_23
42_1_3
423_1_
_1_432
4_1_32
43_1_2
432_1_
This algorithm doesn't enforce uniqueness. So, if you have a string 22 then you will get two results - 22,22. Also, this algorithm uses recursion which I think is quite intuitive in this case, however there are pure iterative implementations if you look for them.
There are several errors in that code.
You have the order of the parts of the for statement incorrect. The order is initialization, test, increment. So for (/* init */ ; /* test */ ; /* increment */)
You're creating a new array for each iteration of your outer loop.
I'm making this a CW because I haven't checked for further errors than the above.

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