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/

edited Nov 14 '18 at 16:15

Rory McCrossan

245k29212248

asked Nov 14 '18 at 16:01

Jørn Thomas Dufey

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/

edited Nov 14 '18 at 16:15

Rory McCrossan

245k29212248

asked Nov 14 '18 at 16:01

Jørn Thomas Dufey

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/

edited Nov 14 '18 at 16:15

Rory McCrossan

245k29212248

asked Nov 14 '18 at 16:01

Jørn Thomas Dufey

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

edited Nov 14 '18 at 16:15

Rory McCrossan

245k29212248

asked Nov 14 '18 at 16:01

Jørn Thomas Dufey

edited Nov 14 '18 at 16:15

Rory McCrossan

245k29212248

asked Nov 14 '18 at 16:01

Jørn Thomas Dufey

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

asked Nov 14 '18 at 16:01

Jørn Thomas Dufey

asked Nov 14 '18 at 16:01

Jørn Thomas Dufey

asked Nov 14 '18 at 16:01

Jørn Thomas Dufey

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 McCrossan

245k29212248

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53304261%2fcalculating-local-values-when-running-slider-function%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

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 McCrossan

245k29212248

add a comment |

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 McCrossan

245k29212248

add a comment |

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 McCrossan

245k29212248

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 McCrossan

245k29212248

answered Nov 14 '18 at 16:12

Rory McCrossan

245k29212248

answered Nov 14 '18 at 16:12

Rory McCrossan

245k29212248

answered Nov 14 '18 at 16:12

Rory McCrossan

245k29212248

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Vfrdtyky