How to assign negative value in jQuery

I have a checkbox with two options in jQuery, withdrawal and deposit. When withdrawal is selected I want to assign negative to amount.The below link shows the image.

enter image description here

edited Nov 25 '18 at 10:23

Barmar

431k36254354

asked Nov 25 '18 at 10:19

user161379

This might help you stackoverflow.com/questions/12769060/…

– Manish Thapa
Nov 25 '18 at 10:23

2

@ManishThapa What does CSS have to do with this

– Barmar
Nov 25 '18 at 10:23

1

amount = -1 * $("selector").val();

– Barmar
Nov 25 '18 at 10:24

Welcome to SO. Can you show us what you have tried so far? What happens when your user types in the amount after the transaction type has been selected?

– Peter Smith
Nov 25 '18 at 10:25

add a comment |

I have a checkbox with two options in jQuery, withdrawal and deposit. When withdrawal is selected I want to assign negative to amount.The below link shows the image.

enter image description here

edited Nov 25 '18 at 10:23

Barmar

431k36254354

asked Nov 25 '18 at 10:19

user161379

This might help you stackoverflow.com/questions/12769060/…

– Manish Thapa
Nov 25 '18 at 10:23

2

@ManishThapa What does CSS have to do with this

– Barmar
Nov 25 '18 at 10:23

1

amount = -1 * $("selector").val();

– Barmar
Nov 25 '18 at 10:24

Welcome to SO. Can you show us what you have tried so far? What happens when your user types in the amount after the transaction type has been selected?

– Peter Smith
Nov 25 '18 at 10:25

add a comment |

I have a checkbox with two options in jQuery, withdrawal and deposit. When withdrawal is selected I want to assign negative to amount.The below link shows the image.

enter image description here

edited Nov 25 '18 at 10:23

Barmar

431k36254354

asked Nov 25 '18 at 10:19

user161379

I have a checkbox with two options in jQuery, withdrawal and deposit. When withdrawal is selected I want to assign negative to amount.The below link shows the image.

enter image description here

jquery

edited Nov 25 '18 at 10:23

Barmar

431k36254354

asked Nov 25 '18 at 10:19

user161379

edited Nov 25 '18 at 10:23

Barmar

431k36254354

asked Nov 25 '18 at 10:19

user161379

edited Nov 25 '18 at 10:23

Barmar

431k36254354

edited Nov 25 '18 at 10:23

Barmar

431k36254354

edited Nov 25 '18 at 10:23

Barmar

431k36254354

asked Nov 25 '18 at 10:19

user161379

asked Nov 25 '18 at 10:19

user161379

asked Nov 25 '18 at 10:19

user161379

This might help you stackoverflow.com/questions/12769060/…

– Manish Thapa
Nov 25 '18 at 10:23

2

@ManishThapa What does CSS have to do with this

– Barmar
Nov 25 '18 at 10:23

1

amount = -1 * $("selector").val();

– Barmar
Nov 25 '18 at 10:24

Welcome to SO. Can you show us what you have tried so far? What happens when your user types in the amount after the transaction type has been selected?

– Peter Smith
Nov 25 '18 at 10:25

add a comment |

This might help you stackoverflow.com/questions/12769060/…

– Manish Thapa
Nov 25 '18 at 10:23

2

@ManishThapa What does CSS have to do with this

– Barmar
Nov 25 '18 at 10:23

1

amount = -1 * $("selector").val();

– Barmar
Nov 25 '18 at 10:24

Welcome to SO. Can you show us what you have tried so far? What happens when your user types in the amount after the transaction type has been selected?

– Peter Smith
Nov 25 '18 at 10:25

This might help you stackoverflow.com/questions/12769060/…

– Manish Thapa
Nov 25 '18 at 10:23

@ManishThapa What does CSS have to do with this

– Barmar
Nov 25 '18 at 10:23

amount = -1 * $("selector").val();

– Barmar
Nov 25 '18 at 10:24

Welcome to SO. Can you show us what you have tried so far? What happens when your user types in the amount after the transaction type has been selected?

– Peter Smith
Nov 25 '18 at 10:25

add a comment |

3 Answers
3

active

oldest

votes

you can achieve that by input event on select input

$(function() {

  const selectListEle = $('select');

  selectListEle.on('input', function(e) {

    const selectVal = selectListEle.val();

    const inputVal = $('input').val();

    if (selectVal === 'withdrawal') {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    } else {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    }

  });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select>

  <option>none</option>

  <option>withdrawal</option>

</select>

<input type="number" value="">

edited Nov 25 '18 at 13:30

answered Nov 25 '18 at 10:34

Amir Fawzy

24128

jQuery is javascript by the way so no harm at all to mixing between them the opposite sometimes you have to use vanilla javascript to achieve complex code implementation such array manipulating there's noway in jQuery to do that by jQuery you have to use vanilla js

– Amir Fawzy
Nov 25 '18 at 11:10

That's right, but this can be solved by using jQuery only, it's not necessary to mix native Javascript with jQuery functionalities at this point.

– Gutelaunetyp
Nov 25 '18 at 11:12

whay if twice? selectVal === 'withdrawal' because i want to make sure if the use choose withdrawal and if it's only add negative -- inputVal constant variable i would to avoid repetitive code $('input').val() -- what i used? again i want to make sure only if the input has value add negative with that implementation i strict the behavior to avoid any error may happening

– Amir Fawzy
Nov 25 '18 at 11:18

"jQuery is javascript by the way so no harm at all to mixing between them" - generally for DOM manipulation it's much better to be consistent. eg don't use e.target.value then $(e.target).val() - stick to one or the other.

– freedomn-m
Nov 25 '18 at 11:53

add a comment |

Searching for $('input') in the whole DOM is a really bad idea.

Much cleaner:

$(function() {



  $('.selectA').on('input', function(e) {

    var selectedVal = $(e.currentTarget).val(),

      $targetInput = $('.inputA'),

      inputVal = $targetInput.val();

    if (selectedVal === 'withdrawal' && inputVal.length) {

      $targetInput.val(['-', inputVal].join(''));

    }

  });



});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select class="selectA">

  <option>none</option>

  <option>withdrawal</option>

</select>

<input class="inputA" type="number" value="">

answered Nov 25 '18 at 10:57

Gutelaunetyp

15010

add a comment |

One option is to give your select options a value of 1 for +ve and -1 for -ve, then you just multiple the two value together.

$("#inp").val() * $("#opt").val()

function calc() {

  $(this).nextAll("#result").text($("#inp").val() * $("#opt").val());

}

$("#inp").on("input", calc);

$("#opt").on("change", calc);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select id="opt">

  <option value="1">deposit</option>

  <option value="-1">withdrawal</option>

</select>

<input id="inp" type="number" value="" />

<span id='result'></span>

answered Nov 25 '18 at 11:50

freedomn-m

13k31945

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%2f53466540%2fhow-to-assign-negative-value-in-jquery%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

you can achieve that by input event on select input

$(function() {

  const selectListEle = $('select');

  selectListEle.on('input', function(e) {

    const selectVal = selectListEle.val();

    const inputVal = $('input').val();

    if (selectVal === 'withdrawal') {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    } else {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    }

  });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select>

  <option>none</option>

  <option>withdrawal</option>

</select>

<input type="number" value="">

edited Nov 25 '18 at 13:30

answered Nov 25 '18 at 10:34

Amir Fawzy

24128

jQuery is javascript by the way so no harm at all to mixing between them the opposite sometimes you have to use vanilla javascript to achieve complex code implementation such array manipulating there's noway in jQuery to do that by jQuery you have to use vanilla js

– Amir Fawzy
Nov 25 '18 at 11:10

That's right, but this can be solved by using jQuery only, it's not necessary to mix native Javascript with jQuery functionalities at this point.

– Gutelaunetyp
Nov 25 '18 at 11:12

whay if twice? selectVal === 'withdrawal' because i want to make sure if the use choose withdrawal and if it's only add negative -- inputVal constant variable i would to avoid repetitive code $('input').val() -- what i used? again i want to make sure only if the input has value add negative with that implementation i strict the behavior to avoid any error may happening

– Amir Fawzy
Nov 25 '18 at 11:18

"jQuery is javascript by the way so no harm at all to mixing between them" - generally for DOM manipulation it's much better to be consistent. eg don't use e.target.value then $(e.target).val() - stick to one or the other.

– freedomn-m
Nov 25 '18 at 11:53

add a comment |

you can achieve that by input event on select input

$(function() {

  const selectListEle = $('select');

  selectListEle.on('input', function(e) {

    const selectVal = selectListEle.val();

    const inputVal = $('input').val();

    if (selectVal === 'withdrawal') {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    } else {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    }

  });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select>

  <option>none</option>

  <option>withdrawal</option>

</select>

<input type="number" value="">

edited Nov 25 '18 at 13:30

answered Nov 25 '18 at 10:34

Amir Fawzy

24128

jQuery is javascript by the way so no harm at all to mixing between them the opposite sometimes you have to use vanilla javascript to achieve complex code implementation such array manipulating there's noway in jQuery to do that by jQuery you have to use vanilla js

– Amir Fawzy
Nov 25 '18 at 11:10

That's right, but this can be solved by using jQuery only, it's not necessary to mix native Javascript with jQuery functionalities at this point.

– Gutelaunetyp
Nov 25 '18 at 11:12

whay if twice? selectVal === 'withdrawal' because i want to make sure if the use choose withdrawal and if it's only add negative -- inputVal constant variable i would to avoid repetitive code $('input').val() -- what i used? again i want to make sure only if the input has value add negative with that implementation i strict the behavior to avoid any error may happening

– Amir Fawzy
Nov 25 '18 at 11:18

"jQuery is javascript by the way so no harm at all to mixing between them" - generally for DOM manipulation it's much better to be consistent. eg don't use e.target.value then $(e.target).val() - stick to one or the other.

– freedomn-m
Nov 25 '18 at 11:53

add a comment |

you can achieve that by input event on select input

$(function() {

  const selectListEle = $('select');

  selectListEle.on('input', function(e) {

    const selectVal = selectListEle.val();

    const inputVal = $('input').val();

    if (selectVal === 'withdrawal') {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    } else {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    }

  });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select>

  <option>none</option>

  <option>withdrawal</option>

</select>

<input type="number" value="">

edited Nov 25 '18 at 13:30

answered Nov 25 '18 at 10:34

Amir Fawzy

24128

you can achieve that by input event on select input

$(function() {

  const selectListEle = $('select');

  selectListEle.on('input', function(e) {

    const selectVal = selectListEle.val();

    const inputVal = $('input').val();

    if (selectVal === 'withdrawal') {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    } else {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    }

  });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select>

  <option>none</option>

  <option>withdrawal</option>

</select>

<input type="number" value="">

$(function() {

  const selectListEle = $('select');

  selectListEle.on('input', function(e) {

    const selectVal = selectListEle.val();

    const inputVal = $('input').val();

    if (selectVal === 'withdrawal') {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    } else {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    }

  });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select>

  <option>none</option>

  <option>withdrawal</option>

</select>

<input type="number" value="">

$(function() {

  const selectListEle = $('select');

  selectListEle.on('input', function(e) {

    const selectVal = selectListEle.val();

    const inputVal = $('input').val();

    if (selectVal === 'withdrawal') {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    } else {

      if (inputVal) {

        $('input').val(-inputVal);

      }

    }

  });

});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select>

  <option>none</option>

  <option>withdrawal</option>

</select>

<input type="number" value="">

edited Nov 25 '18 at 13:30

answered Nov 25 '18 at 10:34

Amir Fawzy

24128

edited Nov 25 '18 at 13:30

answered Nov 25 '18 at 10:34

Amir Fawzy

24128

answered Nov 25 '18 at 10:34

Amir Fawzy

24128

answered Nov 25 '18 at 10:34

Amir Fawzy

24128

jQuery is javascript by the way so no harm at all to mixing between them the opposite sometimes you have to use vanilla javascript to achieve complex code implementation such array manipulating there's noway in jQuery to do that by jQuery you have to use vanilla js

– Amir Fawzy
Nov 25 '18 at 11:10

That's right, but this can be solved by using jQuery only, it's not necessary to mix native Javascript with jQuery functionalities at this point.

– Gutelaunetyp
Nov 25 '18 at 11:12

whay if twice? selectVal === 'withdrawal' because i want to make sure if the use choose withdrawal and if it's only add negative -- inputVal constant variable i would to avoid repetitive code $('input').val() -- what i used? again i want to make sure only if the input has value add negative with that implementation i strict the behavior to avoid any error may happening

– Amir Fawzy
Nov 25 '18 at 11:18

"jQuery is javascript by the way so no harm at all to mixing between them" - generally for DOM manipulation it's much better to be consistent. eg don't use e.target.value then $(e.target).val() - stick to one or the other.

– freedomn-m
Nov 25 '18 at 11:53

add a comment |

jQuery is javascript by the way so no harm at all to mixing between them the opposite sometimes you have to use vanilla javascript to achieve complex code implementation such array manipulating there's noway in jQuery to do that by jQuery you have to use vanilla js

– Amir Fawzy
Nov 25 '18 at 11:10

That's right, but this can be solved by using jQuery only, it's not necessary to mix native Javascript with jQuery functionalities at this point.

– Gutelaunetyp
Nov 25 '18 at 11:12

whay if twice? selectVal === 'withdrawal' because i want to make sure if the use choose withdrawal and if it's only add negative -- inputVal constant variable i would to avoid repetitive code $('input').val() -- what i used? again i want to make sure only if the input has value add negative with that implementation i strict the behavior to avoid any error may happening

– Amir Fawzy
Nov 25 '18 at 11:18

"jQuery is javascript by the way so no harm at all to mixing between them" - generally for DOM manipulation it's much better to be consistent. eg don't use e.target.value then $(e.target).val() - stick to one or the other.

– freedomn-m
Nov 25 '18 at 11:53

jQuery is javascript by the way so no harm at all to mixing between them the opposite sometimes you have to use vanilla javascript to achieve complex code implementation such array manipulating there's noway in jQuery to do that by jQuery you have to use vanilla js

– Amir Fawzy
Nov 25 '18 at 11:10

That's right, but this can be solved by using jQuery only, it's not necessary to mix native Javascript with jQuery functionalities at this point.

– Gutelaunetyp
Nov 25 '18 at 11:12

whay if twice? selectVal === 'withdrawal' because i want to make sure if the use choose withdrawal and if it's only add negative -- inputVal constant variable i would to avoid repetitive code $('input').val() -- what i used? again i want to make sure only if the input has value add negative with that implementation i strict the behavior to avoid any error may happening

– Amir Fawzy
Nov 25 '18 at 11:18

"jQuery is javascript by the way so no harm at all to mixing between them" - generally for DOM manipulation it's much better to be consistent. eg don't use e.target.value then $(e.target).val() - stick to one or the other.

– freedomn-m
Nov 25 '18 at 11:53

add a comment |

Searching for $('input') in the whole DOM is a really bad idea.

Much cleaner:

$(function() {



  $('.selectA').on('input', function(e) {

    var selectedVal = $(e.currentTarget).val(),

      $targetInput = $('.inputA'),

      inputVal = $targetInput.val();

    if (selectedVal === 'withdrawal' && inputVal.length) {

      $targetInput.val(['-', inputVal].join(''));

    }

  });



});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select class="selectA">

  <option>none</option>

  <option>withdrawal</option>

</select>

<input class="inputA" type="number" value="">

answered Nov 25 '18 at 10:57

Gutelaunetyp

15010

add a comment |

Searching for $('input') in the whole DOM is a really bad idea.

Much cleaner:

$(function() {



  $('.selectA').on('input', function(e) {

    var selectedVal = $(e.currentTarget).val(),

      $targetInput = $('.inputA'),

      inputVal = $targetInput.val();

    if (selectedVal === 'withdrawal' && inputVal.length) {

      $targetInput.val(['-', inputVal].join(''));

    }

  });



});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select class="selectA">

  <option>none</option>

  <option>withdrawal</option>

</select>

<input class="inputA" type="number" value="">

answered Nov 25 '18 at 10:57

Gutelaunetyp

15010

add a comment |

Searching for $('input') in the whole DOM is a really bad idea.

Much cleaner:

$(function() {



  $('.selectA').on('input', function(e) {

    var selectedVal = $(e.currentTarget).val(),

      $targetInput = $('.inputA'),

      inputVal = $targetInput.val();

    if (selectedVal === 'withdrawal' && inputVal.length) {

      $targetInput.val(['-', inputVal].join(''));

    }

  });



});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select class="selectA">

  <option>none</option>

  <option>withdrawal</option>

</select>

<input class="inputA" type="number" value="">

answered Nov 25 '18 at 10:57

Gutelaunetyp

15010

Searching for $('input') in the whole DOM is a really bad idea.

Much cleaner:

$(function() {



  $('.selectA').on('input', function(e) {

    var selectedVal = $(e.currentTarget).val(),

      $targetInput = $('.inputA'),

      inputVal = $targetInput.val();

    if (selectedVal === 'withdrawal' && inputVal.length) {

      $targetInput.val(['-', inputVal].join(''));

    }

  });



});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select class="selectA">

  <option>none</option>

  <option>withdrawal</option>

</select>

<input class="inputA" type="number" value="">

$(function() {



  $('.selectA').on('input', function(e) {

    var selectedVal = $(e.currentTarget).val(),

      $targetInput = $('.inputA'),

      inputVal = $targetInput.val();

    if (selectedVal === 'withdrawal' && inputVal.length) {

      $targetInput.val(['-', inputVal].join(''));

    }

  });



});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select class="selectA">

  <option>none</option>

  <option>withdrawal</option>

</select>

<input class="inputA" type="number" value="">

$(function() {



  $('.selectA').on('input', function(e) {

    var selectedVal = $(e.currentTarget).val(),

      $targetInput = $('.inputA'),

      inputVal = $targetInput.val();

    if (selectedVal === 'withdrawal' && inputVal.length) {

      $targetInput.val(['-', inputVal].join(''));

    }

  });



});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select class="selectA">

  <option>none</option>

  <option>withdrawal</option>

</select>

<input class="inputA" type="number" value="">

answered Nov 25 '18 at 10:57

Gutelaunetyp

15010

answered Nov 25 '18 at 10:57

Gutelaunetyp

15010

answered Nov 25 '18 at 10:57

Gutelaunetyp

15010

answered Nov 25 '18 at 10:57

Gutelaunetyp

15010

add a comment |

One option is to give your select options a value of 1 for +ve and -1 for -ve, then you just multiple the two value together.

$("#inp").val() * $("#opt").val()

function calc() {

  $(this).nextAll("#result").text($("#inp").val() * $("#opt").val());

}

$("#inp").on("input", calc);

$("#opt").on("change", calc);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select id="opt">

  <option value="1">deposit</option>

  <option value="-1">withdrawal</option>

</select>

<input id="inp" type="number" value="" />

<span id='result'></span>

answered Nov 25 '18 at 11:50

freedomn-m

13k31945

add a comment |

One option is to give your select options a value of 1 for +ve and -1 for -ve, then you just multiple the two value together.

$("#inp").val() * $("#opt").val()

function calc() {

  $(this).nextAll("#result").text($("#inp").val() * $("#opt").val());

}

$("#inp").on("input", calc);

$("#opt").on("change", calc);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select id="opt">

  <option value="1">deposit</option>

  <option value="-1">withdrawal</option>

</select>

<input id="inp" type="number" value="" />

<span id='result'></span>

answered Nov 25 '18 at 11:50

freedomn-m

13k31945

add a comment |

One option is to give your select options a value of 1 for +ve and -1 for -ve, then you just multiple the two value together.

$("#inp").val() * $("#opt").val()

function calc() {

  $(this).nextAll("#result").text($("#inp").val() * $("#opt").val());

}

$("#inp").on("input", calc);

$("#opt").on("change", calc);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select id="opt">

  <option value="1">deposit</option>

  <option value="-1">withdrawal</option>

</select>

<input id="inp" type="number" value="" />

<span id='result'></span>

answered Nov 25 '18 at 11:50

freedomn-m

13k31945

One option is to give your select options a value of 1 for +ve and -1 for -ve, then you just multiple the two value together.

$("#inp").val() * $("#opt").val()

function calc() {

  $(this).nextAll("#result").text($("#inp").val() * $("#opt").val());

}

$("#inp").on("input", calc);

$("#opt").on("change", calc);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select id="opt">

  <option value="1">deposit</option>

  <option value="-1">withdrawal</option>

</select>

<input id="inp" type="number" value="" />

<span id='result'></span>

function calc() {

  $(this).nextAll("#result").text($("#inp").val() * $("#opt").val());

}

$("#inp").on("input", calc);

$("#opt").on("change", calc);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select id="opt">

  <option value="1">deposit</option>

  <option value="-1">withdrawal</option>

</select>

<input id="inp" type="number" value="" />

<span id='result'></span>

function calc() {

  $(this).nextAll("#result").text($("#inp").val() * $("#opt").val());

}

$("#inp").on("input", calc);

$("#opt").on("change", calc);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<select id="opt">

  <option value="1">deposit</option>

  <option value="-1">withdrawal</option>

</select>

<input id="inp" type="number" value="" />

<span id='result'></span>

answered Nov 25 '18 at 11:50

freedomn-m

13k31945

answered Nov 25 '18 at 11:50

freedomn-m

13k31945

answered Nov 25 '18 at 11:50

freedomn-m

13k31945

answered Nov 25 '18 at 11:50

freedomn-m

13k31945

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

Nsryjdtyk