Calculating local values when running slider function
I'm adding some variables in three different sliders.
Two of the sliders choose amount of apples and oranges.
These sliders multiply values from another slider (a customer slider).
When I pull the oranges and apples sliders, they get a value from the customer slider to multiply with, but the customer slider wont update values in the two other sliders.
I want the values to be calculated when I pull the customer slider as well.
var customers;
$('.input-1').bind('input', mainfunction);
$('.input-2').bind('input', mainfunction);
$('.customers').bind('input', customersfunction);
function mainfunction() {
var this_value = this.value;
var price = $(this).data("price");
var this_sum = this_value * price * customers;
$(this).closest("div").find(".counter").html(this_value);
$(this).closest("div").find(".sum").html(this_sum);
}
function customersfunction() {
customers = this.value;
$(this).closest("div").find(".counter").html(customers);
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="input-1" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="input-1" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>My fiddle: http://jsfiddle.net/5vnrd6zx/26/
jquery function slider
edited Nov 14 '18 at 16:15
Rory McCrossan
245k29212248
asked Nov 14 '18 at 16:01
Jørn Thomas DufeyJørn Thomas Dufey
22
add a comment |
I'm adding some variables in three different sliders.
Two of the sliders choose amount of apples and oranges.
These sliders multiply values from another slider (a customer slider).
When I pull the oranges and apples sliders, they get a value from the customer slider to multiply with, but the customer slider wont update values in the two other sliders.
I want the values to be calculated when I pull the customer slider as well.
var customers;
$('.input-1').bind('input', mainfunction);
$('.input-2').bind('input', mainfunction);
$('.customers').bind('input', customersfunction);
function mainfunction() {
var this_value = this.value;
var price = $(this).data("price");
var this_sum = this_value * price * customers;
$(this).closest("div").find(".counter").html(this_value);
$(this).closest("div").find(".sum").html(this_sum);
}
function customersfunction() {
customers = this.value;
$(this).closest("div").find(".counter").html(customers);
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="input-1" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="input-1" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>My fiddle: http://jsfiddle.net/5vnrd6zx/26/
jquery function slider
edited Nov 14 '18 at 16:15
Rory McCrossan
245k29212248
asked Nov 14 '18 at 16:01
Jørn Thomas DufeyJørn Thomas Dufey
22
add a comment |
I'm adding some variables in three different sliders.
Two of the sliders choose amount of apples and oranges.
These sliders multiply values from another slider (a customer slider).
When I pull the oranges and apples sliders, they get a value from the customer slider to multiply with, but the customer slider wont update values in the two other sliders.
I want the values to be calculated when I pull the customer slider as well.
var customers;
$('.input-1').bind('input', mainfunction);
$('.input-2').bind('input', mainfunction);
$('.customers').bind('input', customersfunction);
function mainfunction() {
var this_value = this.value;
var price = $(this).data("price");
var this_sum = this_value * price * customers;
$(this).closest("div").find(".counter").html(this_value);
$(this).closest("div").find(".sum").html(this_sum);
}
function customersfunction() {
customers = this.value;
$(this).closest("div").find(".counter").html(customers);
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="input-1" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="input-1" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>My fiddle: http://jsfiddle.net/5vnrd6zx/26/
jquery function slider
edited Nov 14 '18 at 16:15
Rory McCrossan
245k29212248
asked Nov 14 '18 at 16:01
Jørn Thomas DufeyJørn Thomas Dufey
22
I'm adding some variables in three different sliders.
Two of the sliders choose amount of apples and oranges.
These sliders multiply values from another slider (a customer slider).
When I pull the oranges and apples sliders, they get a value from the customer slider to multiply with, but the customer slider wont update values in the two other sliders.
I want the values to be calculated when I pull the customer slider as well.
var customers;
$('.input-1').bind('input', mainfunction);
$('.input-2').bind('input', mainfunction);
$('.customers').bind('input', customersfunction);
function mainfunction() {
var this_value = this.value;
var price = $(this).data("price");
var this_sum = this_value * price * customers;
$(this).closest("div").find(".counter").html(this_value);
$(this).closest("div").find(".sum").html(this_sum);
}
function customersfunction() {
customers = this.value;
$(this).closest("div").find(".counter").html(customers);
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="input-1" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="input-1" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>My fiddle: http://jsfiddle.net/5vnrd6zx/26/
var customers;
$('.input-1').bind('input', mainfunction);
$('.input-2').bind('input', mainfunction);
$('.customers').bind('input', customersfunction);
function mainfunction() {
var this_value = this.value;
var price = $(this).data("price");
var this_sum = this_value * price * customers;
$(this).closest("div").find(".counter").html(this_value);
$(this).closest("div").find(".sum").html(this_sum);
}
function customersfunction() {
customers = this.value;
$(this).closest("div").find(".counter").html(customers);
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="input-1" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="input-1" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>var customers;
$('.input-1').bind('input', mainfunction);
$('.input-2').bind('input', mainfunction);
$('.customers').bind('input', customersfunction);
function mainfunction() {
var this_value = this.value;
var price = $(this).data("price");
var this_sum = this_value * price * customers;
$(this).closest("div").find(".counter").html(this_value);
$(this).closest("div").find(".sum").html(this_sum);
}
function customersfunction() {
customers = this.value;
$(this).closest("div").find(".counter").html(customers);
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="input-1" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="input-1" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>jquery function slider
jquery function slider
edited Nov 14 '18 at 16:15
Rory McCrossan
245k29212248
asked Nov 14 '18 at 16:01
Jørn Thomas DufeyJørn Thomas Dufey
22
edited Nov 14 '18 at 16:15
Rory McCrossan
245k29212248
asked Nov 14 '18 at 16:01
Jørn Thomas DufeyJørn Thomas Dufey
22
edited Nov 14 '18 at 16:15
Rory McCrossan
245k29212248
edited Nov 14 '18 at 16:15
Rory McCrossan
245k29212248
edited Nov 14 '18 at 16:15
Rory McCrossan
245k29212248
245k29212248
asked Nov 14 '18 at 16:01
Jørn Thomas DufeyJørn Thomas Dufey
22
asked Nov 14 '18 at 16:01
Jørn Thomas DufeyJørn Thomas Dufey
22
asked Nov 14 '18 at 16:01
Jørn Thomas DufeyJørn Thomas Dufey
22
22
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The issue you have is because you only update the .counter and .sum when mainfunction is called by changing a fruit slider. You also need to do this when the customers value changes as well, as the number of customers used in that calculation has changed.
As such, it makes sense to write a common function which handles the input event of all sliders, and shows the appropriate count and sum for each of them. Something like this:
var $sliders = $('.slider').on('input', function() {
$sliders.each(function() {
var $slider = $(this);
var amount = $slider.val();
var $container = $slider.closest('div');
$container.find('.counter').text(amount);
if ($slider.hasClass('fruit')) {
var price = $slider.data('price');
$container.find('.sum').text(amount * price * $customers.val());
}
});
});
var $customers = $sliders.filter('.customers');<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="slider fruit" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="slider fruit" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="slider customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>Note the use of common classes on the sliders to help genericise the logic, as well as the loop which runs through and updates all of them when any of their values has been changed.
answered Nov 14 '18 at 16:12
Rory McCrossanRory McCrossan
245k29212248
add a comment |
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The issue you have is because you only update the .counter and .sum when mainfunction is called by changing a fruit slider. You also need to do this when the customers value changes as well, as the number of customers used in that calculation has changed.
As such, it makes sense to write a common function which handles the input event of all sliders, and shows the appropriate count and sum for each of them. Something like this:
var $sliders = $('.slider').on('input', function() {
$sliders.each(function() {
var $slider = $(this);
var amount = $slider.val();
var $container = $slider.closest('div');
$container.find('.counter').text(amount);
if ($slider.hasClass('fruit')) {
var price = $slider.data('price');
$container.find('.sum').text(amount * price * $customers.val());
}
});
});
var $customers = $sliders.filter('.customers');<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="slider fruit" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="slider fruit" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="slider customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>Note the use of common classes on the sliders to help genericise the logic, as well as the loop which runs through and updates all of them when any of their values has been changed.
answered Nov 14 '18 at 16:12
Rory McCrossanRory McCrossan
245k29212248
add a comment |
The issue you have is because you only update the .counter and .sum when mainfunction is called by changing a fruit slider. You also need to do this when the customers value changes as well, as the number of customers used in that calculation has changed.
As such, it makes sense to write a common function which handles the input event of all sliders, and shows the appropriate count and sum for each of them. Something like this:
var $sliders = $('.slider').on('input', function() {
$sliders.each(function() {
var $slider = $(this);
var amount = $slider.val();
var $container = $slider.closest('div');
$container.find('.counter').text(amount);
if ($slider.hasClass('fruit')) {
var price = $slider.data('price');
$container.find('.sum').text(amount * price * $customers.val());
}
});
});
var $customers = $sliders.filter('.customers');<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="slider fruit" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="slider fruit" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="slider customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>Note the use of common classes on the sliders to help genericise the logic, as well as the loop which runs through and updates all of them when any of their values has been changed.
answered Nov 14 '18 at 16:12
Rory McCrossanRory McCrossan
245k29212248
add a comment |
The issue you have is because you only update the .counter and .sum when mainfunction is called by changing a fruit slider. You also need to do this when the customers value changes as well, as the number of customers used in that calculation has changed.
As such, it makes sense to write a common function which handles the input event of all sliders, and shows the appropriate count and sum for each of them. Something like this:
var $sliders = $('.slider').on('input', function() {
$sliders.each(function() {
var $slider = $(this);
var amount = $slider.val();
var $container = $slider.closest('div');
$container.find('.counter').text(amount);
if ($slider.hasClass('fruit')) {
var price = $slider.data('price');
$container.find('.sum').text(amount * price * $customers.val());
}
});
});
var $customers = $sliders.filter('.customers');<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="slider fruit" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="slider fruit" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="slider customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>Note the use of common classes on the sliders to help genericise the logic, as well as the loop which runs through and updates all of them when any of their values has been changed.
answered Nov 14 '18 at 16:12
Rory McCrossanRory McCrossan
245k29212248
The issue you have is because you only update the .counter and .sum when mainfunction is called by changing a fruit slider. You also need to do this when the customers value changes as well, as the number of customers used in that calculation has changed.
As such, it makes sense to write a common function which handles the input event of all sliders, and shows the appropriate count and sum for each of them. Something like this:
var $sliders = $('.slider').on('input', function() {
$sliders.each(function() {
var $slider = $(this);
var amount = $slider.val();
var $container = $slider.closest('div');
$container.find('.counter').text(amount);
if ($slider.hasClass('fruit')) {
var price = $slider.data('price');
$container.find('.sum').text(amount * price * $customers.val());
}
});
});
var $customers = $sliders.filter('.customers');<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="slider fruit" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="slider fruit" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="slider customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>Note the use of common classes on the sliders to help genericise the logic, as well as the loop which runs through and updates all of them when any of their values has been changed.
var $sliders = $('.slider').on('input', function() {
$sliders.each(function() {
var $slider = $(this);
var amount = $slider.val();
var $container = $slider.closest('div');
$container.find('.counter').text(amount);
if ($slider.hasClass('fruit')) {
var price = $slider.data('price');
$container.find('.sum').text(amount * price * $customers.val());
}
});
});
var $customers = $sliders.filter('.customers');<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="slider fruit" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="slider fruit" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="slider customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>var $sliders = $('.slider').on('input', function() {
$sliders.each(function() {
var $slider = $(this);
var amount = $slider.val();
var $container = $slider.closest('div');
$container.find('.counter').text(amount);
if ($slider.hasClass('fruit')) {
var price = $slider.data('price');
$container.find('.sum').text(amount * price * $customers.val());
}
});
});
var $customers = $sliders.filter('.customers');<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p>Apples: 5$</p>
<div>
<input class="slider fruit" data-price="5" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Oranges: 3$</p>
<div>
<input class="slider fruit" data-price="3" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
<span class="sum">0</span>
</div>
<p>Customers:</p>
<div>
<input class="slider customers" type="range" value="0" min="0" max="10">
<span class="counter">0</span><br>
</div>answered Nov 14 '18 at 16:12
Rory McCrossanRory McCrossan
245k29212248
answered Nov 14 '18 at 16:12
Rory McCrossanRory McCrossan
245k29212248
answered Nov 14 '18 at 16:12
Rory McCrossanRory McCrossan
245k29212248
answered Nov 14 '18 at 16:12
Rory McCrossanRory McCrossan
245k29212248
245k29212248
add a comment |
add a comment |
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