I am currently working on a challenge to convert any amount of arbitrary numbers entered into the function below as a currency (expressed by a string where every three characters are separated by a comma). All was okay until I realized that if the length of the number entered was 4, then the comma would need to be placed as 1,234 rather than 123,4.
It seems the function has spiralled out of control a little when I wrapped the initial if statement around my for loop. (1) I keep getting thrown an 'Uncaught SyntaxError: Unexpected token else' in the console. (2) I thought it may be due trying to place an if/else around the for/if/else. Any light shed on my error would be much appreciated. Charlie
function toCurrency(price){
var newStrng = price.toString().split("");
var stringCh = [];
if(newStrng.length===4){
console.log("gotcha");
stringCh = newStrng.splice(1,0,",");
return stringCh;
} else {
for(var i = 0; i < newStrng.length; i++) {
if(i %3 === 0 && i !== 0){
stringCh.push(",");
stringCh.push(newStrng[i]);
}
} else {
stringCh.push(newStrng[i]);
}
}
var finallyDone = stringCh.join("");
return finallyDone;
}//EO function
The if statement in the following block
for(var i = 0; i < newStrng.length; i++) {
if(i %3 === 0 && i !== 0){
stringCh.push(",");
stringCh.push(newStrng[i]);
}
} else {
stringCh.push(newStrng[i]);
}
}
has an extra }, right before the else. Take it out and you should no long receive the syntax error.
You can make it easier to spot mistakes like this in the future by ensuring your code is properly indented so you can see which brackets are associated with which blocks. Though I appreciate it could be reasonably indented in the source code and may not have copied across to SO perfectly.
There is a } in the wrong place inside the for loop, that need to be moved after the else block. As it is, the if statement is located in the for loop, but the else block is located outside of the loop, which is not valid syntax.
function toCurrency(price){
var newStrng = price.toString().split("");
var stringCh = [];
if(newStrng.length===4){
console.log("gotcha");
stringCh = newStrng.splice(1,0,",");
return stringCh;
} else {
for(var i = 0; i < newStrng.length; i++) {
if(i %3 === 0 && i !== 0){
stringCh.push(",");
stringCh.push(newStrng[i]);
} else {
stringCh.push(newStrng[i]);
}
}
}
var finallyDone = stringCh.join("");
return finallyDone;
}//EO function
remove } after,
stringCh.push(newStrng[i]);
Related
I have came across an odd problem with getting JSON data like the following.
[
{
"type":"ripe",
"red":137,
"green":68,
"blue":40,
"strftime(\"%H:%M:%S\", time)":"18:46:37"
},
]
I was not able to compare this data by type using JavaScript, they both successfully went through my if statement for some odd reason. The total count for both variables is equal to 2.
let counterLoop = function() {
for (let i = 0; i < data.length; i++) {
let fruitType = JSON.stringify(data[i].type);
sortFruit(fruitType.toLowerCase());
}
}
let sortFruit = function(fruitType) {
if (fruitType.localeCompare('ripe') === 0){} {
totalRipeFruit++;
$("#totalRipeFruit").text(totalRipeFruit);
}
if (fruitType.localeCompare('unripe') === 0){} {
totalUnripeFruit++;
$("#totalUnripeFruit").text(totalUnripeFruit);
}
}
Any idea why this could be the case?
Thank you very much!
You have two problems here; First of all there is no need for the JSON.stringifyon the type, just leave it out as it will return the string containing useless quotes.
Secondly, your if statements are messed up: You have a second pair of brackets behind each one, so simply change
if (fruitType.localeCompare('ripe') === 0){} {
totalRipeFruit++;
$("#totalRipeFruit").text(totalRipeFruit);
}
if (fruitType.localeCompare('raw') === 0){} {
totalRawFruit++;
$("#totalRawFruit").text(totalRawFruit);
}
To:
if (fruitType.localeCompare('ripe') === 0) {
totalRipeFruit++;
}
if (fruitType.localeCompare('raw') === 0) {
totalRawFruit++;
}
Hello there StackOverflow people,
What I expected:
Removing the numbers of the string "23Ka5X". The loop counts the length and the if statement extracts the letters into an array letterMemory. When no letters are in the string, the message '"oh no numbers!" should be the output.
What I ran into:
I have been working on this for some time now but I can't find my mistake. I don't know if I missed a simple detail or made a big mess.
My feeling and console output:
var letterMemory = [];
function orderMsg(mixedMsg) {
for (var loopString = 0; loopString < mixedMsg.length; loopString++); {
if (isNaN(parseInt(mixedMsg.charAt[loopString]))); {
letterMemory.push(mixedMsg.charAt[loopString]);
return letterMemory;
} if (!isNaN(parseInt(mixedMsg.charAt[loopString]))) {
return "oh no numbers!";
}
}
}
console.log(orderMsg("23Ka5X"));
I feel like the issue is trying to push any letter into the array letterMemory via letterMemory.push(mixedMsg.charAt[loopString])
does not work how I imagine it.
I would be really grateful for your help!
I found a simple solution via .replace() but I really want to make it work with a loop and if statements since loops combined with if statements were my latest freecodecamp lessons and I want to get better at it.
The fixed code
function orderMsg(mixedMsg){
var letterMemory = []
for (var loopString = 0; loopString < mixedMsg.length; loopString++){
if (isNaN(mixedMsg[loopString])){
letterMemory.push(mixedMsg[loopString])
}
}
if (letterMemory.length){
return letterMemory
} else {
return 'oh no numbers!'
}
}
The issue was
The for loop was not executing since you terminated it with ; at the end.
charAt is a function, so you either do string.charAt(index), or you can also simply say string[index].
You are using the return statement within the for loop, so what will happen is even if the for loop ran (without the semi-colon at the end), it would run just once.
One other issue is that the variable letterMemory is declared outside the function so that means if you were to call this function twice, it would use the same letterMemory array.
-end of answer-
Additional read: you can use split, filter and ternary operator to condense the function as follows ..
function orderMsg(mixedMsg){
const letterMemory = mixedMsg.split('').filter(isNaN)
return letterMemory.length ? letterMemory : 'oh no numbers!'
}
This could be helpful,
function orderMsg(mixedMsg) {
for (var loopString = 0; loopString < mixedMsg.length; loopString++) {
if (isNaN(parseInt(mixedMsg.charAt(loopString)))) {
letterMemory.push(mixedMsg.charAt(loopString));
}
}
return letterMemory;
}
var arr = orderMsg("23s5");
if (arr.length == 0) {
console.log("oh no numbers!")
} else {
console.log(arr);
}
Use replace with regex globally, replacing all digits by an empty string:
string.replace(/[0-9]/g, "")
You have terminated for loop in the same line with ;.
charAt() is a method.
Return value after for loop ends.
var letterMemory = [];
function orderMsg(mixedMsg) {
for (var loopString = 0; loopString < mixedMsg.length; loopString++) {
var letter=parseInt(mixedMsg.charAt(loopString));
if(isNaN(letter)){
letterMemory.push(mixedMsg.charAt(loopString));
}
}
if(letterMemory.length>0){
return letterMemory;
}
else{
return "Oh no numbers!";
}
}
console.log(orderMsg("23Ka5X"));
Maybe try using .test to match the letters.
function orderMsg(str){
var result = [];
for(var letter of str){
if(/[a-zA-Z]+/g.test(letter)){
result.push(letter)
}
}
if(result.length === 0){
return 'Oh no numbers'
}
return result
}
For a more thorough explanation:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/test
Hey guys I am trying trying to right this javascript code into c++. I am doing quick sort and everything is straight forward minus the last step.
function quickSort(arr)
{
//base case if the arr is 1 or 0 then return the array
if(arr.length === 1 || arr.length === 0)
{
return arr;
}
var pivotIndex = Math.floor(arr.length/2);
var pivotValue = arr[pivotIndex];
var before = [];
var after = [];
for(var counter = 0; counter < arr.length; counter++)
{
if(counter === pivotIndex)
continue;
if(pivotValue <= arr[counter])
{
before.push(arr[counter])
}
else
{
after.push(arr[counter])
}
}
//this step I am having trouble rewriting in c++
return quickSort(after).concat(pivotValue).concat(quickSort(before));
}
I am having a hard time rewriting the recursive step in c++. I am not sure how concat 2 vector. I tried using the insert method but I keep getting an error about invalid use of void expression.
vector<int> quickSort(vector<int> arr)
{
if(arr.size() == 1 || arr.size() == 0)
{
return arr;
}
int pivotIndex = arr.size()/2;
int pivotValue = arr[pivotIndex];
vector<int> before;
vector<int> after;
//put values in before or after the piv
for(size_t counter = 0; counter < arr.size(); counter++)
{
if(counter == pivotIndex)
continue;
if(pivotValue <= arr[counter])
before.push_back( arr[counter]);
else
after.push_back( arr[counter]);
}
return //????? not sure how to do this
}
So, you realized that your core question was "how to concatenate two vectors", and you found a right answer: using insert. Now your question is about why you were getting "an error about invalid use of void expression." (That's the assumption my answer is for, at least.)
That's because you were likely trying to do something like the following:
return quickSort(after).insert( /* stuff */ );
which is wrong. In JavaScript, array.concat returns the concatenated array. It's return type is effectively Array, and so doing return arr.concat(arr2) returns an Array because arr.concat would return an Array. Further, in JavaScript, array.concat doesn't modify the array it was called on, but rather returns a new array.
In C++, however, vector.insert (#4 in the reference) returns void. That means it returns nothing. So when you try to return the result of insert, you get that error about invalid use of a void expression. Further, in C++, vector.insert does modify the vector it was called on.
So how do you use insert in this case?
vector<int> quickSort(vector<int> arr)
{
// ...
// Sort `before` and `after`
before = quickSort(before);
after = quickSort(after);
// Modify `after` and return it.
after.push_back(pivotValue);
after.insert(after.end(), before.begin(), before.end());
return after;
}
Note: My code isn't optimal and the idea of rewriting JS in C++ is also oddly specific. My answer is to simply outline the problem asked in the question, not to give a good C++ implementation of quick sort.
To concat two vector , you can use std::merge
like:std::merge(v1.begin(), v1.end(), v2.begin(), v2.end(), std::back_inserter(dst));
I wanted to use two Not and one and in booleans to test if the variable is neither upper case nor lower case.
I used this code so far but it didn't work as required:
else if ((x[i]) !== (x[i].toUpperCase()) && (x[i]!== x[i].toLowerCase()) ){
x.splice(x[i], 1);
}
This code was for a function that sorts entered strings yet uppercase are sorted first.
Here is the full code, I am also open to understanding better ways to create this function apart from boolean logic and the array methods I used.
function alpha(str){ // United States
var x = str.split(""); // [U,n,i,t,e,d,S,t,a,t,e,s]
var cap = [];
var small = [];
for (var i = 0; i<x.length; i++){
if (x[i] == x[i].toUpperCase()){
cap.push(x[i]);
}
else if ((x[i]) !== (x[i].toUpperCase()) && (x[i]!== x[i].toUpperCase()) ) {
x.splice(x[i], 1);
}
else {small.push(x[i]);}
}
var z = cap.sort();
var y = small.sort();
return z.concat(y).join("");
}
Please note the second else if statement is only useful because the code adds an empty space string at the beginning of the output, I'm not sure where it comes from, so please let me know if you have any idea how to sort this even without using the second else if.
In the ASCII table, upper case letters come first. That's why they come first when you sort alphabetically. Here's a link to a page on Wikipedia that shows the table with the upper case letters appearing first and their numerical equivalents. It's even printable.
Also, I took the liberty of simplifying your code a little. Seems like .splice() was not necessary.
function alpha( str ) {
var x = str.split(""); // [U,n,i,t,e,d,S,t,a,t,e,s]
var cap = [];
var small = [];
var length = x.length;
for (var i = 0; i < length; i++) {
if (x[i] === x[i].toUpperCase()) {
cap.push(x[i]);
} else if (x[i] === x[i].toLowerCase()) {
small.push(x[i]);
}
}
return cap.sort().concat(small.sort()).join("");
}
Maybe explain what you're trying to do? It most likely has been done before in some form and you definitely came to the right place to find an answer.
Is this what you want to do?
var str = "United States";
function alpha(str) {
return str.split('').sort().join('');
}
alert(alpha(str));
In all programming languages (as far as i know), boolean expressions are always evaluated from the left to the right with brackets of course.
So in the following example my_func() is called first, and then if there is the chance that the complete expression becomes true my_other_func() is called
if (my_func() && my_other_func()) {
// I only get here if my_func() AND my_other_func() return true
// If my_func() returns false, my_other_func() is never called
}
The same is true for the "or" operator in the following example
if (my_func() || my_other_func()) {
// I only get here if my_func() OR my_other_func() return true
// If my_func() returns true, my_other_func() is not called
}
So back to your code, in details this part (I reformated it a bit for better readability):
if (x[i] == x[i].toUpperCase()){
// only uppercase here
cap.push(x[i]);
} else if (x[i] !== x[i].toUpperCase() && x[i] !== x[i].toUpperCase()) {
// tested twice the same thing, so Im really sure that its not uppercase :D
// only lowercase here
x.splice(x[i], 1);
} else {
// I will never reach this
small.push(x[i]);
}
Im not sure what you want to do, but I hope the comments help to understand your code.
I've been writing a javascript function which returns true if the value matches one of about 4 values (just 3 in the example below). The problem is, when I have just two values the function works correctly, but adding a third breaks the code.
I'm pretty new to javascript and I'm guessing there's a much better way of doing this? I've tried searching but found nothing as of yet.
Any help is much appreciated.
function isValid(elem, helperMsg){
var sn6 = /[sS][nN]6/;
var sn5 = /[sS][nN]5/;
var sn38 = /[sS][nN]38/;
if(elem.value.match(sn6 || sn5 || sn38)){
//do stuff
return true;
}else{
return false;
}
}
Edit:
Here's my second attempt with an array:
function isLocal(elem, helperMsg){
var validPostcodes=new Array();
validPostcodes[0]= /[wW][rR]12/;
validPostcodes[1]= /[cC][vV]35/;
validPostcodes[2]= /[sS][nN]99/;
validPostcodes[3]= /[sS][nN]6/;
validPostcodes[4]= /[sS][nN]5/;
validPostcodes[5]= /[sS][nN]38/;
validPostcodes[6]= /[oO][xX]29/;
validPostcodes[7]= /[oO][xX]28/;
var i = 0;
for (i = 0; i < validPostcodes.length; ++i) {
if(elem.value.match(validPostcodes[i])){
// do stuff
return true;
}else{
alert(helperMsg);
elem.focus();
return false;
}
}
}
a || b || c
is an expression that evaluates to a boolean. That means that you're running either match(true) or match(false). You must write it as:
match(a) || match(b) || match(c)
Another option would be to store them in an array and loop over it. That would mean if the number of patterns grew you wouldn't have to change code other than the list of patterns. Another approach, though limited to this situation, might be to change the pattern to one that is equivalent to or-ing the three options together (untested, and I'm a bit rusty on regex):
elem.value.match(/[sSnN][6|5|38]/)
Array based example:
var patterns = [/../, /.../];
for (var i = 0; i < patterns.length; ++i) {
if (elem.value.match(patterns[i])) { return true; }
}
In real code, I would probably format it like this:
function isValid(elem, helperMsg){
var patterns = [/../, /.../],
i = 0;
for (i = 0; i < patterns.length; ++i) {
if (elem.value.match(patterns[i])) {
return true;
}
}
}
That's just a habit though since JavaScript hoists variables to the top of their scope. It's by no means required to declare the variables like that.