Develop Reference

Bubble Sort in never ending loop

I have the following data:
I basically want the 5,6,7 records to be at the very top of the list ordered by eta. The rest I am OK with as they stand. since they do not realy have an eta.
My JS function :
bubbleSortETA : function(array) {
var swapped;
do {
swapped = false;
for(var i = 0; i < array.length; i++) {
console.log('array[i]');
console.log(array[i]);
console.log('array[i+1]');
console.log(array[i+1]);
var thisRowEta;
var nextRowEta;
if (array[i] && typeof(array[i].eta) != "undefined"){
thisRowEta = array[i].eta;
}else{
thisRowEta = 1000; //so it appears at the bottom???
}
if (array[i+1] && typeof(array[i+1].eta) != "undefined"){
nextRowEta = array[i+1].eta;
}else{
nextRowEta = 1000;
}
if(array[i] && array[i + 1] && typeof(array[i].walkInDetails) != "undefined" && typeof(array[i+1].walkInDetails) != "undefined" && (thisRowEta != nextRowEta)){
this.swap(array, i, i + 1);
console.log('we swapped...');
swapped = true;
}
}
} while(swapped);
console.log(array);
return array;
}
It goes into this never ending loop. I have tried looking at output statements but am unable to understand why it is going into infinite loop.
Any tips?

Cannot read property length null error when used with regular expressions

I'm a javascript beginner doing some CodeWars.com questions. I came across this question and I'm stuck due to a "cannot read property length null" error. I've tried to look up that error and can't find what the problem is in my program.
The assignment is:
"Check to see if a string has the same amount of 'x's and 'o's. The method must return a boolean and be case insensitive. The string can contains any char."
And this is what I've written so far:
function XO(str) {
var x = "x";
var o = "o";
var numX = str.match(/x/gi).length;
var numO = str.match(/o/gi).length;
while(str.indexOf(x) > -1 || str.indexOf(o) > -1) {
if(numX == numO){
return true;
}
}
if (numX === -1 && numO === -1){
return true;
}
}
XO("xoxo");
The assignment also says that if there is neither an X or an O then the program should return true.

This will not give you that error. When there are no matches, the match function returns null and you cannot get the length of null. A few extra lines solves this issue.
function XO(str) {
var x = "x";
var o = "o";
var numX = 0;
var numO = 0;
var xMatch = str.match(/x/gi);
var oMatch = str.match(/o/gi);
if (xMatch) {
numX = xMatch.length;
}
if (oMatch) {
numO = oMatch.length;
}
while(str.indexOf(x) > -1 || str.indexOf(o) > -1) {
if(numX == numO){
return true;
} else {
return false;
}
}
if (numX === -1 && numO === -1){
return true;
} else {
return false;
}
}
console.log(XO("ddd"));

I think you are making this problem more complex than it has to be.
All you need to do is make the string lowercase(to account for case insensitive), traverse the string, and when it finds an x, add 1 to a counter, and when you find and o, decrease 1 from the counter.
If it ends at 0, you return true, else you return false. There's no need for regexes
function XO(str){
var count = 0;
str = str.toLowerCase();
for(var i = 0; i < str.length; i++){
if(str[i] === 'x') count++;
if(str[i] === 'o') count--;
}
return count === 0 ? true : false;
}

Yes you have to check the return value of match is not null before checking the length property. However
while(str.indexOf(x) > -1 || str.indexOf(o) > -1) {
if(numX == numO){
return true;
}
}
looks like an infinite loop if either string contains lower case 'x' or 'o' and there are a different number of each.
More simply:
function XO(str)
{ var matchX = str.match(/x/gi);
var matchY = str.match(/o/gi);
return (matchX && matchY) ? matchX.length == matchY.length : !matchX && !matchY;
}

Javascript condense if statement

I have an if statement in javascript that is
if(prop < 0 || trans < 0 || queue < 0 || proc < 0
|| prop > 1 || trans > 1 || queue > 1 || proc > 1
|| prop == "" || trans == "" || queue == "" || proc == ""){
Is there a way to condense this? For prop, trans, queue, and proc. I want to create an if statement if the values do not fall to be between 0 and 1 or if it has an empty string value

Building off of Jordan's answer:
var checkThese = [prop, trans, queue, proc];
var result = checkTruthinessOf(checkThese);
function checkTruthinessOf(things) {
return things.every(function(el) {
return (el < 0 || el > 1 || el === "");
});
}
See Array.prototype.every()

var checkThese = [prop, trans, queue, proc];
var result = checkTruthinessOf(checkThese);
function checkTruthinessOf(things) {
var returnValue = false;
[].forEach.call(things, function(thing){
if (thing < 0 || thing > 1 || thing == "") returnValue = true;
});
return returnValue;
};

I picked up this practice from jQuery. It eliminates the extra array and just passes in as many arguments as you like. Then use rink's function to validate all it at once.
var result = checkTruthinessOf(prop, trans, queue, proc);
function checkTruthinessOf(/*unlimited arguments*/) {
return Array.prototype.every.call(arguments, function(thing) {
return (thing < 0 || thing > 1 || thing === "");
});
}

Dynamic condition in IF statement

I want to make condition of if statement dynamically in javascript,
check my code
var t = ['b','a']
if(t[0] !== 'a' && t[1] !== 'a'){console.log('remaining element')}
here t might be vary at any time say t = ['b','a','c'] then I need to write if condition like this
if(t[0] !== 'a' && t[1] !== 'a' && t[2] !== 'a'){console.log('remaining element')}
How can I rewirte this code efficiently?

You can use Array.prototype.every like this
if (t.every(function(currentElement) { return currentElement !== "a"; })) {
console.log('remaining element');
}
This works with arbitrary number of elements.
On older environments which do not support Array.prototype.every, you can use the plain for loop version
var flag = true;
for (var i = 0 ; i < t.length; i += 1) {
if (t[i] === "a") {
flag = false;
break;
}
}
if (flag) {
console.log('remaining element');
}

Javascript If statement, looking through an array

Mind has gone blank this afternoon and can't for the life of me figure out the right way to do this:
if(i!="3" && i!="4" && i!="5" && i!="6" && i!="7" && i!="8" && i!="9" && i!="2" && i!="19" && i!="18" && i!="60" && i!="61" && i!="50" && i!="49" && i!="79" && i!="78" && i!="81" && i!="82" && i!="80" && i!="70" && i!="90" && i!="91" && i!="92" && i!="93" && i!="94"){
//do stuff
}
All those numbers need to be in an array, then I can check to see if "i" is not equal to any 1 of them.

var a = [3,4,5,6,7,8,9];
if ( a.indexOf( 2 ) == -1 ) {
// do stuff
}
indexOf returns -1 if the number is not found. It returns something other than -1 if it is found. Change your logic if you want.
Wrap the numbers in quotes if you need strings ( a = ['1','2'] ). I don't know what you're dealing with so I made them numbers.
IE and other obscure/older browsers will need the indexOf method:
if (!Array.prototype.indexOf)
{
Array.prototype.indexOf = function(elt /*, from*/)
{
var len = this.length >>> 0;
var from = Number(arguments[1]) || 0;
from = (from < 0)
? Math.ceil(from)
: Math.floor(from);
if (from < 0)
from += len;
for (; from < len; from++)
{
if (from in this &&
this[from] === elt)
return from;
}
return -1;
};
}

My mind made this solution:
function not(dat, arr) { //"not" function
for(var i=0;i<arr.length;i++) {
if(arr[i] == dat){return false;}
}
return true;
}
var check = [2,3,4,5,6,7,8,9,18,19,49,50,60,61,70,78,79,80,81,82,90,91,92,93,94]; //numbers
if(not(i, check)) {
//do stuff
}

This solution is cross-browser:
var valid = true;
var cantbe = [3, 4, 5]; // Fill in all your values
for (var j in cantbe)
if (typeof cantbe[j] === "number" && i == cantbe[j]){
valid = false;
break;
}
valid will be true if i isn't a 'bad' value, false otherwise.