The function finds which tv character the user compares to based on their answers to my questions. My code now is very inefficient for multiple select menus!!! Maybe an object that takes all selectmenus in html and allows me to assign array values based on the selected index of a selectmenu.
function onSelectMenuBlur() {
"use strict";
/*list of arrays that will be added to when the user selects an option in a selectmenu.*/
var rickArray = [];
var shaneArray = [];
var bobArray = [];
var carolArray = [];
var lArray = [];
var sm = document.getElementById("selectmenu");
.onchange function that determines what array will be added to depending on the option selected in the select menu. This function will add an array value of 1 once to an array. Seems like an inefficient way, especially with multiple selectmenus!
sm.onchange = function() {
if(sm.selectedIndex + 1 === 1) {
rickArray.push(1);
shaneArray.pop();
bobArray.pop();
carolArray.pop();
lArray.pop();
alert(rickArray.length);
}
else if(sm.selectedIndex + 1 === 2) {
shaneArray.push(1);
rickArray.pop();
bobArray.pop();
carolArray.pop();
lArray.pop();
alert(shaneArray.length);
}
else if(sm.selectedIndex + 1 === 3) {
bobArray.push(1);
rickArray.pop();
shaneArray.pop();
carolArray.pop();
lArray.pop();
alert(bobArray.length);
}
else if(sm.selectedIndex + 1 === 4) {
carolArray.push(1);
rickArray.pop();
shaneArray.pop();
bobArray.pop();
lArray.pop();
alert(carolArray.length);
}
else if(sm.selectedIndex + 1 === 5) {
lArray.push(1);
rickArray.pop();
shaneArray.pop();
bobArray.pop();
carolArray.pop();
alert(lArray.length);
}
else{}
};
.onblur purpose to find array with biggest length or value out of all selectmenus to determine which person associated with the array the user is like. Again seems like an inefficient way to handle!
sm.onblur = function() {
var rickL = rickArray.length;
var shaneL = shaneArray.length;
var bobL = bobArray.length;
var carolL = carolArray.length;
var lL = lArray.length;
// unfinished if else statement !!
if(rickL > shaneL && rickL > bobL && rickL > carolL && rickL > lL) {
alert("you are Rick Grimes");
}
else{
alert("you are someone else");
}
};
}
Use a 2-dimensional array instead of separate arrays for each character, and then use the selected index as an index into the array.
var characters = [[], [], [], [], []];
sm.onchange = function() {
for (var i = 0; i < characters.length; i++) {
if (i == this.selectedIndex) {
characters[i].push(1);
alert(characters[i].length);
} else {
characters[i].pop();
}
}
};
To get the character names in there, make it an array of objects.
characters = [
{ name: "Rick",
array: []
},
{ name: "Carol",
array: []
},
...
}
Then you would use characters[i].array.push(1). And then when you want to say which character they are, find the object with the longest array and then print its .name.
I have an array which looks like this:
["1,8", "4,6,8", "8,9", "6,9"]
1/ I would like to turn it in to this
[1,8,4,6,8,8,9,6,9]
2/ I would then like to find matching values, by looking for the most number:
[8]
This first has been solved with this:
var carArray = ["1,8", "4,6,8,7,7,7,7", "8,9", "6,9"];
//1) create single array
var arr = carArray.join().split(',');
//2) find most occurring
var counts = {}; //object to hold count for each occurence
var max = 0, maxOccurring;
arr.forEach(function(el){
var cnt = (counts[el] || 0); //previous count
counts[el] = ++cnt;
if(cnt > max && cnt > 1){ //only register if more than once (cnt>1)
max=cnt;
maxOccurring = el;
}
});
if(maxOccurring){
//there was an element more than once, maxOccuring contains that element
setResult('Most occuring: ' + maxOccurring + ' (' + max + ' times)');
}
else{
//3)/4) ???
setResult('sorting?');
}
//below is only for test display purposes
function setResult(res){
console.log(res);
}
3/ If the are no matching values like this
[1,8,4,6,5,7]
4/ Then I need to compare this array to another array, such as this
[6,7,4,1,2,8,9,5]
If the first number in <4> array above appears in <3> array, then get that number, ie in the above example I need to get 6. The <4> array will be static values and not change. The numbers is <3> will be dynamic.
EDIT Not the most elegant of answers, but I do have something working now. I didn't compare the original array directly with the second array, instead used simple if/else statements to do what I needed:
var carArray = ["1,5", "4", "8,2", "3,9,1,1,1"];
//1) create single array
var arr = carArray.join().split(',');
//2) find most occurring
var counts = {}; //object to hold count for each occurence
var max = 0, maxOccurring;
arr.forEach(function(el){
var cnt = (counts[el] || 0); //previous count
counts[el] = ++cnt;
if(cnt > max && cnt > 1){ //only register if more than once (cnt>1)
max=cnt;
maxOccurring = el;
}
});
if(maxOccurring){
//there was an element more than once, maxOccuring contains that element
console.log('Most occuring: ' + maxOccurring + ' (' + max + ' times)');
console.log(maxOccurring);
}
else {
// If not occuring, match from a list
if(jQuery.inArray("6", arr) !== -1) { console.log('6'); }
else if(jQuery.inArray("9", arr) !== -1) { console.log('9'); }
else if(jQuery.inArray("7", arr) !== -1) { console.log('7'); }
else if(jQuery.inArray("5", arr) !== -1) { console.log('5'); }
else if(jQuery.inArray("4", arr) !== -1) { console.log('4'); }
else if(jQuery.inArray("1", arr) !== -1) { console.log('1'); }
else { console.log('not found'); }
}
Example Fiddle
Step 1 is fairly easy by using javascript's join and split methods respectively:
var arr = carArray .join().split(',');
For step 2, several methods can be used, the most common one using an object and using the elements themselves as properties. Since you only need to get the most occurring value if there is a reoccurring value, it can be used in the same loop:
var counts = {}; //object to hold count for each occurence
var max = 0, maxOccurring;
arr.forEach(function(el){
var cnt = (counts[el] || 0); //previous count
counts[el] = ++cnt;
if(cnt > max && cnt > 1){ //only register if more than once (cnt>1)
max=cnt;
maxOccurring = el;
}
});
After the above, the variable maxOccurring will contain the reoccurring value (if any) and max will contain the times it occured
For step 4 the easiest way is to loop through the compare array and get the element that occurs in the input array:
var cmpArr = ['6','7','4','1','2','8','9','5'];
//find the first occurrence inside the cmpArr
res = function(){ for(var i= 0 ; i < cmpArr.length; i++){ if(arr.indexOf(cmpArr[i]) !== -1)return cmpArr[i];}}();
The above uses an in place function which is called immediately to be able to use return. You could also just use a loop and assign res when found, then break from the loop.
Last update, an alternate fiddle where the above is converted to a single function: http://jsfiddle.net/v9hhsdny/5/
Well first of all the following code results in four matching answers since the jQuery selectors are the same.
var questionAnswer1 = $(this).find('input[name=questionText]').val();
var questionAnswer2 = $(this).find('input[name=questionText]').val();
var questionAnswer3 = $(this).find('input[name=questionText]').val();
var questionAnswer4 = $(this).find('input[name=questionText]').val();
var carArray = [questionAnswer1, questionAnswer2, questionAnswer3, questionAnswer4];
You could use the eq(index) method of jQuery to select the appropriate element. However having 4 inputs with the same name is a bad practice.
Well lets say that the carArray has 4 different values which all consist out of comma separated numbers. You could then do the following:
var newArr = [];
carArray.forEach(function(e) {
e.split(",").forEach(function(n) {
newArr.push(n);
});
});
Well then we got to find the most occurring number. JavaScript doesn't have any functions for that so we will have to find an algorithm for that. I found the following algorithm on this stackoverflow page
var count = function(ary, classifier) {
return ary.reduce(function(counter, item) {
var p = (classifier || String)(item);
counter[p] = counter.hasOwnProperty(p) ? counter[p] + 1 : 1;
return counter;
}, {})
}
var occurances = count(newArr);
It isn't clear to me what you're trying to do in step 3 and 4, so can't answer those at the moment.
var ary = ["1,8", "4,6,8", "8,9", "6,9"];
var splitted = ary.reduce(function(acc, item) {
return acc.concat(item.split(','));
}, []);
var occurences = splitted.reduce(function(acc, item) {
if (!acc.hasOwnProperty(item)) acc[item] = 0;
acc[item] += 1;
return acc;
},{}),
biggest = Object.keys(occurences).reduce(function (acc, key) {
if (occurences[key] > acc.occurences) {
acc.name = key;
acc.occurences = occurences[key];
}
return acc;
},{'name':'none','occurences':0}).name;
var vals=["1,8", "4,6,8", "8,9", "6,9"];
// 1) turn into number array
var arrNew=[];
for(var i=0; i<vals.length; i++)
{
arrLine=vals[i].split(",");
for (var j=0;j<arrLine.length;j++) { arrNew.push (parseInt(arrLine[j])) }
}
//result:
alert(arrNew.join(";");
// 2) find most common
var found=[];
for(var i=0; i<arrNew.length; i++) {
// make an array of the number of occurrances of each value
if (found["num"+newArray[i]]) {
found["num"+newArray[i]] ++ ;
} else {
found["num"+newArray[i]]=1;
}
}
var mostCommon={count:0,val:"ROGUE"};
for (x in found) {
if (found[x] > mostCommon.count) {
mostCommon.count=found[x].count;
mostCommon.val=x;
}
}
// result :
alert(mostCommon.val);
//3) not quite sure what you meant there
// 4) unique values:
// at this point the 'found' list contains unique vals
var arrUnique=[];
for (x in found) {
arrUnique.push[x];
}
// result :
alert(arrUnique.join(";"))
//sort:
arrUnique.sort(function(a, b){return a-b});
(This won't work in most browsers) but on a side note, when ES6 becomes widely supported, your solution could look like this:
var arr1 = ["1,8", "4,6,8", "8,9", "6,9"];
var arr2 = arr1.join().split(',');
var s = Array.from(new Set(arr2)); //Array populated by unique values, ["1", "8", "4", "6", "9"]
Thought you might like to see a glimpse of the future!
1.
var orgArray = ['1,8', '4,6,8', '8,9', '6,9'];
var newArray = [];
for (var i in orgArray) {
var tmpArray = orgArray[i].split(',');
for (var j in tmpArray) {
newArray.push(Number(tmpArray[j]));
}
}
2.
var counts = {};
var most = null;
for (var i in newArray) {
var num = newArray[i];
if (typeof counts[num] === 'undefined') {
counts[num] = 1;
} else {
++(counts[num]);
}
if (most == null || counts[num] > counts[most]) {
most = num;
} else if (most != null && counts[num] === counts[most]) {
most = null;
}
}
I don't understand the question 3 and 4 (what "unique order" means) so I can't answer those questions.
I stored an object in one variable (Consider as datatable).
var data=[{"controlID":"A","currentValue":"10","onChange":"","onClick":""},
{"controlID":"B","currentValue":"5","onChange":"Testing(A,B)","onClick":""},
{"controlID":"C","currentValue":"-5","onChange":"Testing1(A,B)","onClick":""},
{"controlID":"D","currentValue":"","onChange":"Testing2(B,C)","onClick":""},{"controlID":"E","currentValue":"","onChange":"Testing3(C,D)","onClick":""},{"controlID":"F","currentValue":"","onChange":"","onClick":""}];
Now I know the second row key value as B. How to I Get the Third row (i.e., "C" row values)
Am new of this field. Please help us to helpful.
This function will return your index:
var FindIndexOfControlID = function(id, data){
for(var i = 0; i < data.length ; i++){
if( data[i]['controlID'] == id ){
return i;
}
}
};
Usage:
var index = FindIndexOfControlID('C', data);
Live Example
http://jsfiddle.net/urahara/medhgm7b/
NOTE
Alternatively you may also want to implement function that returns index of any specified property and value:
var FindIndexOfProperty = function(value, property, data){
for(var i = 0; i < data.length ; i++){
if( data[i][property] == value ){
return i;
}
}
};
Usage
FindIndexOfProperty('-5', 'currentValue',data); // returns 2
You can return the third row in javascript by simply executing var thirdRow = data[2]. The row will be returned as an object.
i got a group of inputs... each one has a number value.
i want to get all their values (found a method here) and then compare
then and highlight the heighest input meaning highlight the input itself
meaning i need to somehow grab its id and know which one i am comparing to...
(i hope i explained it good).
This is what i have for now taken from the link attached:
var values = [];
$("input[name='items[]']").each(function() {
values.push($(this).val());
});
try something like this
$(function(){
var higesht_val = 0;
var higesht_val_id = 0;
$("input[name='items[]']").each(function() {
var current_val = parseInt(this.value);
if(higesht_val < current_val){
higesht_val = current_val;
higesht_val_id = this.id;
}
});
alert(higesht_val); // highest value
alert(higesht_val_id);// id of highest value input
})
var highestVal = 0,
$target;
$("input[name='items[]']").each(function() {
if(parseInt($(this).val()) > highestVal){
highestVal = parseInt($(this).val());
$target = $(this);
}
});
// $target is now the input with the highest value
how about this ?
var values = [];
$("input[name='items[]']").each(function() {values.push(this);});
values.sort(function(a, b){return b.value - a.value;})
highlight(values[0]);
I have this code:
for(var i=0; i < localStorage.length; i++) {
var subjects = [];
var key, value;
key = localStorage.key(i);
value = localStorage.getItem(key);
var keysplit = key.split(".");
if(keysplit[keysplit.length] == "subj") {
subjects.push(value);
}
}
I am trying to select all the keys that have a .subj ending, but this does not seem to work. Any ideas?
The length property returns the number of items in the array, and as the index is zero based there is no item with that index.
Use length - 1 to get the last item:
if (keysplit[keysplit.length - 1] === "subj") {
Other possibilities:
if(key.substr(key.lastIndexOf('.')) == ".subj")
//or
var suffix = '.subj';
if(key.lastIndexOf(suffix) == key.length - suffix.length)
See: lastIndexOf