X + Y gives HTMLObject (Select/option)
I'm trying to add 2 sums, that are getting fetched by an select option.
All I want is to calculate X + Y, but i get a weird error.
Code:
$( document ).ready(function() {
var id1 = $("#one").val();
$("#final_value").val(id1);
var id2 = $("#two").val();
$("#final_value").val(id2);
$("#final_value").val(this + parseInt(id1) + parseInt(id2));
});
https://jsfiddle.net/xpvt214o/957709/
javascript jquery html html5
add a comment |
I'm trying to add 2 sums, that are getting fetched by an select option.
All I want is to calculate X + Y, but i get a weird error.
Code:
$( document ).ready(function() {
var id1 = $("#one").val();
$("#final_value").val(id1);
var id2 = $("#two").val();
$("#final_value").val(id2);
$("#final_value").val(this + parseInt(id1) + parseInt(id2));
});
https://jsfiddle.net/xpvt214o/957709/
javascript jquery html html5
2
this
on your final line of code is casing the issue - remove it and you should be okay?
– Christheoreo
Nov 22 '18 at 12:13
To do: Stop trying to “add” numbers ontothis
…
– misorude
Nov 22 '18 at 12:13
Why you try to like that "this + parseInt(id1) + parseInt(id2)", why you wanna to addthis
to smth?
– Konstantin Kudelko
Nov 22 '18 at 12:14
add a comment |
I'm trying to add 2 sums, that are getting fetched by an select option.
All I want is to calculate X + Y, but i get a weird error.
Code:
$( document ).ready(function() {
var id1 = $("#one").val();
$("#final_value").val(id1);
var id2 = $("#two").val();
$("#final_value").val(id2);
$("#final_value").val(this + parseInt(id1) + parseInt(id2));
});
https://jsfiddle.net/xpvt214o/957709/
javascript jquery html html5
I'm trying to add 2 sums, that are getting fetched by an select option.
All I want is to calculate X + Y, but i get a weird error.
Code:
$( document ).ready(function() {
var id1 = $("#one").val();
$("#final_value").val(id1);
var id2 = $("#two").val();
$("#final_value").val(id2);
$("#final_value").val(this + parseInt(id1) + parseInt(id2));
});
https://jsfiddle.net/xpvt214o/957709/
javascript jquery html html5
javascript jquery html html5
edited Nov 22 '18 at 12:18
Peter B
13.3k52044
13.3k52044
asked Nov 22 '18 at 12:11
BGHBGH
132
132
2
this
on your final line of code is casing the issue - remove it and you should be okay?
– Christheoreo
Nov 22 '18 at 12:13
To do: Stop trying to “add” numbers ontothis
…
– misorude
Nov 22 '18 at 12:13
Why you try to like that "this + parseInt(id1) + parseInt(id2)", why you wanna to addthis
to smth?
– Konstantin Kudelko
Nov 22 '18 at 12:14
add a comment |
2
this
on your final line of code is casing the issue - remove it and you should be okay?
– Christheoreo
Nov 22 '18 at 12:13
To do: Stop trying to “add” numbers ontothis
…
– misorude
Nov 22 '18 at 12:13
Why you try to like that "this + parseInt(id1) + parseInt(id2)", why you wanna to addthis
to smth?
– Konstantin Kudelko
Nov 22 '18 at 12:14
2
2
this
on your final line of code is casing the issue - remove it and you should be okay?– Christheoreo
Nov 22 '18 at 12:13
this
on your final line of code is casing the issue - remove it and you should be okay?– Christheoreo
Nov 22 '18 at 12:13
To do: Stop trying to “add” numbers onto
this
…– misorude
Nov 22 '18 at 12:13
To do: Stop trying to “add” numbers onto
this
…– misorude
Nov 22 '18 at 12:13
Why you try to like that "this + parseInt(id1) + parseInt(id2)", why you wanna to add
this
to smth?– Konstantin Kudelko
Nov 22 '18 at 12:14
Why you try to like that "this + parseInt(id1) + parseInt(id2)", why you wanna to add
this
to smth?– Konstantin Kudelko
Nov 22 '18 at 12:14
add a comment |
5 Answers
5
active
oldest
votes
You have this
in your #final_value
element. I've also added a change
event to your code so it will update when one of the options change:
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = $("#one").val();
var id2 = $("#two").val();
$("#final_value").val(parseInt(id1) + parseInt(id2));
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
Thank's! That was working! Thank you so much Gary. I wish you the best :-)
– BGH
Nov 23 '18 at 8:11
add a comment |
There's a this
in the last statement, replace the last line with:
$("#final_value").val(parseInt(id1) + parseInt(id2));
Now it should work.
add a comment |
Remove this
value:
From:
$("#final_value").val(this + parseInt(id1) + parseInt(id2));
To:
$("#final_value").val(parseInt(id1) + parseInt(id2));
Cause if you will use this
your function returns a string as:
[object HTMLDocument]11
add a comment |
I have altered Gary Thomas Code little bit:
Instead of parseInt(), you can use like this:
var id1 = +$("#one").val();
var id2 = +$("#two").val();
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = +$("#one").val();
var id2 = +$("#two").val();
var total = id1+id2;
$("#final_value").val(total);
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
add a comment |
Just remove this keyword from your summation as so :
$( document ).ready(function() {
var id1 = $("#one").val();
$("#final_value").val(id1);
var id2 = $("#two").val();
$("#final_value").val(id2);
$("#final_value").val(parseInt(id1) + parseInt(id2));
});
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
You have this
in your #final_value
element. I've also added a change
event to your code so it will update when one of the options change:
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = $("#one").val();
var id2 = $("#two").val();
$("#final_value").val(parseInt(id1) + parseInt(id2));
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
Thank's! That was working! Thank you so much Gary. I wish you the best :-)
– BGH
Nov 23 '18 at 8:11
add a comment |
You have this
in your #final_value
element. I've also added a change
event to your code so it will update when one of the options change:
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = $("#one").val();
var id2 = $("#two").val();
$("#final_value").val(parseInt(id1) + parseInt(id2));
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
Thank's! That was working! Thank you so much Gary. I wish you the best :-)
– BGH
Nov 23 '18 at 8:11
add a comment |
You have this
in your #final_value
element. I've also added a change
event to your code so it will update when one of the options change:
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = $("#one").val();
var id2 = $("#two").val();
$("#final_value").val(parseInt(id1) + parseInt(id2));
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
You have this
in your #final_value
element. I've also added a change
event to your code so it will update when one of the options change:
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = $("#one").val();
var id2 = $("#two").val();
$("#final_value").val(parseInt(id1) + parseInt(id2));
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = $("#one").val();
var id2 = $("#two").val();
$("#final_value").val(parseInt(id1) + parseInt(id2));
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = $("#one").val();
var id2 = $("#two").val();
$("#final_value").val(parseInt(id1) + parseInt(id2));
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
answered Nov 22 '18 at 12:16
Gary ThomasGary Thomas
1,5151415
1,5151415
Thank's! That was working! Thank you so much Gary. I wish you the best :-)
– BGH
Nov 23 '18 at 8:11
add a comment |
Thank's! That was working! Thank you so much Gary. I wish you the best :-)
– BGH
Nov 23 '18 at 8:11
Thank's! That was working! Thank you so much Gary. I wish you the best :-)
– BGH
Nov 23 '18 at 8:11
Thank's! That was working! Thank you so much Gary. I wish you the best :-)
– BGH
Nov 23 '18 at 8:11
add a comment |
There's a this
in the last statement, replace the last line with:
$("#final_value").val(parseInt(id1) + parseInt(id2));
Now it should work.
add a comment |
There's a this
in the last statement, replace the last line with:
$("#final_value").val(parseInt(id1) + parseInt(id2));
Now it should work.
add a comment |
There's a this
in the last statement, replace the last line with:
$("#final_value").val(parseInt(id1) + parseInt(id2));
Now it should work.
There's a this
in the last statement, replace the last line with:
$("#final_value").val(parseInt(id1) + parseInt(id2));
Now it should work.
answered Nov 22 '18 at 12:14
François LanzerayFrançois Lanzeray
8318
8318
add a comment |
add a comment |
Remove this
value:
From:
$("#final_value").val(this + parseInt(id1) + parseInt(id2));
To:
$("#final_value").val(parseInt(id1) + parseInt(id2));
Cause if you will use this
your function returns a string as:
[object HTMLDocument]11
add a comment |
Remove this
value:
From:
$("#final_value").val(this + parseInt(id1) + parseInt(id2));
To:
$("#final_value").val(parseInt(id1) + parseInt(id2));
Cause if you will use this
your function returns a string as:
[object HTMLDocument]11
add a comment |
Remove this
value:
From:
$("#final_value").val(this + parseInt(id1) + parseInt(id2));
To:
$("#final_value").val(parseInt(id1) + parseInt(id2));
Cause if you will use this
your function returns a string as:
[object HTMLDocument]11
Remove this
value:
From:
$("#final_value").val(this + parseInt(id1) + parseInt(id2));
To:
$("#final_value").val(parseInt(id1) + parseInt(id2));
Cause if you will use this
your function returns a string as:
[object HTMLDocument]11
answered Nov 22 '18 at 12:15
Konstantin KudelkoKonstantin Kudelko
1278
1278
add a comment |
add a comment |
I have altered Gary Thomas Code little bit:
Instead of parseInt(), you can use like this:
var id1 = +$("#one").val();
var id2 = +$("#two").val();
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = +$("#one").val();
var id2 = +$("#two").val();
var total = id1+id2;
$("#final_value").val(total);
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
add a comment |
I have altered Gary Thomas Code little bit:
Instead of parseInt(), you can use like this:
var id1 = +$("#one").val();
var id2 = +$("#two").val();
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = +$("#one").val();
var id2 = +$("#two").val();
var total = id1+id2;
$("#final_value").val(total);
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
add a comment |
I have altered Gary Thomas Code little bit:
Instead of parseInt(), you can use like this:
var id1 = +$("#one").val();
var id2 = +$("#two").val();
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = +$("#one").val();
var id2 = +$("#two").val();
var total = id1+id2;
$("#final_value").val(total);
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
I have altered Gary Thomas Code little bit:
Instead of parseInt(), you can use like this:
var id1 = +$("#one").val();
var id2 = +$("#two").val();
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = +$("#one").val();
var id2 = +$("#two").val();
var total = id1+id2;
$("#final_value").val(total);
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = +$("#one").val();
var id2 = +$("#two").val();
var total = id1+id2;
$("#final_value").val(total);
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
$(document).ready(function() {
$("#one, #two").on("change", () => {
var id1 = +$("#one").val();
var id2 = +$("#two").val();
var total = id1+id2;
$("#final_value").val(total);
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<td>
<select class="full-width select2" name="" id="one" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<td>
<select class="full-width select2" name="" id="two" data-placeholder="" >
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</td>
<label for="final_value">Result:</label>
<input type="text" value="" id="final_value">
answered Nov 22 '18 at 12:24
Aravind Bhat KAravind Bhat K
2741214
2741214
add a comment |
add a comment |
Just remove this keyword from your summation as so :
$( document ).ready(function() {
var id1 = $("#one").val();
$("#final_value").val(id1);
var id2 = $("#two").val();
$("#final_value").val(id2);
$("#final_value").val(parseInt(id1) + parseInt(id2));
});
add a comment |
Just remove this keyword from your summation as so :
$( document ).ready(function() {
var id1 = $("#one").val();
$("#final_value").val(id1);
var id2 = $("#two").val();
$("#final_value").val(id2);
$("#final_value").val(parseInt(id1) + parseInt(id2));
});
add a comment |
Just remove this keyword from your summation as so :
$( document ).ready(function() {
var id1 = $("#one").val();
$("#final_value").val(id1);
var id2 = $("#two").val();
$("#final_value").val(id2);
$("#final_value").val(parseInt(id1) + parseInt(id2));
});
Just remove this keyword from your summation as so :
$( document ).ready(function() {
var id1 = $("#one").val();
$("#final_value").val(id1);
var id2 = $("#two").val();
$("#final_value").val(id2);
$("#final_value").val(parseInt(id1) + parseInt(id2));
});
answered Nov 22 '18 at 12:44
Voice Of The RainVoice Of The Rain
123211
123211
add a comment |
add a comment |
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2
this
on your final line of code is casing the issue - remove it and you should be okay?– Christheoreo
Nov 22 '18 at 12:13
To do: Stop trying to “add” numbers onto
this
…– misorude
Nov 22 '18 at 12:13
Why you try to like that "this + parseInt(id1) + parseInt(id2)", why you wanna to add
this
to smth?– Konstantin Kudelko
Nov 22 '18 at 12:14