Browse a numeric data array and add or subtract during the loop with $.each
It may be that this question has already been posted, but I have not found any clear solution to my problem.
The title of my question has already described my problem. In fact I wish to see the sum of the sum during the loop. But when you meet a negative number, it turns into a positive number. Which distorts the result.
Below an example of my loop.
Can you help me please ?
<table>
<tr id="testTable">
<td data-pos="0"></td>
<td data-pos="1"></td>
<td data-pos="2"></td>
<td data-pos="3"></td>
<td data-pos="4"></td>
<td data-pos="5"></td>
</tr>
</table>
<script>
$(document).ready(function () {
var array = [200, 400, 200, -500, 200,100];
var sum = 0;
$.each(array, function (i, val) {
val = parseFloat(val);
if ($.isNumeric(val)) {
if (val > 0) {
sum += val;
} else {
sum -= val;
}
}
$('tr#testTable td[data-pos="'+i+'"]').text(sum);
});
});
</script>
The result is 1600 instead of 600
Thanks a lot for your help
jquery
asked yesterday
Yona Smilevitch
518
add a comment |
It may be that this question has already been posted, but I have not found any clear solution to my problem.
The title of my question has already described my problem. In fact I wish to see the sum of the sum during the loop. But when you meet a negative number, it turns into a positive number. Which distorts the result.
Below an example of my loop.
Can you help me please ?
<table>
<tr id="testTable">
<td data-pos="0"></td>
<td data-pos="1"></td>
<td data-pos="2"></td>
<td data-pos="3"></td>
<td data-pos="4"></td>
<td data-pos="5"></td>
</tr>
</table>
<script>
$(document).ready(function () {
var array = [200, 400, 200, -500, 200,100];
var sum = 0;
$.each(array, function (i, val) {
val = parseFloat(val);
if ($.isNumeric(val)) {
if (val > 0) {
sum += val;
} else {
sum -= val;
}
}
$('tr#testTable td[data-pos="'+i+'"]').text(sum);
});
});
</script>
The result is 1600 instead of 600
Thanks a lot for your help
jquery
asked yesterday
Yona Smilevitch
518
Do you really need to use$.each()? Why notArray.reduce?
– Praveen Kumar Purushothaman
yesterday
Why you do not want to use reduce?
– Just code
yesterday
add a comment |
It may be that this question has already been posted, but I have not found any clear solution to my problem.
The title of my question has already described my problem. In fact I wish to see the sum of the sum during the loop. But when you meet a negative number, it turns into a positive number. Which distorts the result.
Below an example of my loop.
Can you help me please ?
<table>
<tr id="testTable">
<td data-pos="0"></td>
<td data-pos="1"></td>
<td data-pos="2"></td>
<td data-pos="3"></td>
<td data-pos="4"></td>
<td data-pos="5"></td>
</tr>
</table>
<script>
$(document).ready(function () {
var array = [200, 400, 200, -500, 200,100];
var sum = 0;
$.each(array, function (i, val) {
val = parseFloat(val);
if ($.isNumeric(val)) {
if (val > 0) {
sum += val;
} else {
sum -= val;
}
}
$('tr#testTable td[data-pos="'+i+'"]').text(sum);
});
});
</script>
The result is 1600 instead of 600
Thanks a lot for your help
jquery
asked yesterday
Yona Smilevitch
518
It may be that this question has already been posted, but I have not found any clear solution to my problem.
The title of my question has already described my problem. In fact I wish to see the sum of the sum during the loop. But when you meet a negative number, it turns into a positive number. Which distorts the result.
Below an example of my loop.
Can you help me please ?
<table>
<tr id="testTable">
<td data-pos="0"></td>
<td data-pos="1"></td>
<td data-pos="2"></td>
<td data-pos="3"></td>
<td data-pos="4"></td>
<td data-pos="5"></td>
</tr>
</table>
<script>
$(document).ready(function () {
var array = [200, 400, 200, -500, 200,100];
var sum = 0;
$.each(array, function (i, val) {
val = parseFloat(val);
if ($.isNumeric(val)) {
if (val > 0) {
sum += val;
} else {
sum -= val;
}
}
$('tr#testTable td[data-pos="'+i+'"]').text(sum);
});
});
</script>
The result is 1600 instead of 600
Thanks a lot for your help
jquery
jquery
asked yesterday
Yona Smilevitch
518
asked yesterday
Yona Smilevitch
518
asked yesterday
Yona Smilevitch
518
asked yesterday
Yona Smilevitch
518
asked yesterday
Yona Smilevitch
518
518
Do you really need to use$.each()? Why notArray.reduce?
– Praveen Kumar Purushothaman
yesterday
Why you do not want to use reduce?
– Just code
yesterday
add a comment |
Do you really need to use$.each()? Why notArray.reduce?
– Praveen Kumar Purushothaman
yesterday
Why you do not want to use reduce?
– Just code
yesterday
Do you really need to use
$.each()? Why not Array.reduce?– Praveen Kumar Purushothaman
yesterday
Do you really need to use
$.each()? Why not Array.reduce?– Praveen Kumar Purushothaman
yesterday
Why you do not want to use reduce?
– Just code
yesterday
Why you do not want to use reduce?
– Just code
yesterday
add a comment |
2 Answers
2
active
oldest
votes
You can actually use .reduce(). Coming to your question, the problem is in the if condition. Let's say the value is -500.
if (val > 0) {
sum += val;
} else {
// val is -500 and it comes here.
// sum will be set to sum - (-500)
// the above will be sum + 500
sum -= val;
}
Your logic here is flawed.
You just need to use:
if ($.isNumeric(val)) {
sum += val;
}
Snippet
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<table>
<tr id="testTable">
<td data-pos="0"></td>
<td data-pos="1"></td>
<td data-pos="2"></td>
<td data-pos="3"></td>
<td data-pos="4"></td>
<td data-pos="5"></td>
</tr>
</table>
<script>
$(document).ready(function() {
var array = [200, 400, 200, -500, 200, 100];
var sum = 0;
$.each(array, function(i, val) {
val = parseFloat(val);
if ($.isNumeric(val)) {
sum += val;
}
$('tr#testTable td[data-pos="' + i + '"]').text(sum);
});
});
</script>I am indeed getting 600 as the answer above.
answered yesterday
Praveen Kumar Purushothaman
131k23135181
add a comment |
Your problem sources from the if condition that you are using. You don't need to check whether the value is negative or positive, but just to add it in the sum.
When you check if the value is negative, in reality it reverses the sign of the number, which means that a negative number becomes positive.
I would recommend the use of .reduce, rather the unnecessary declaration of variables. However, the solution with $.each would be:
$.each(callback)
let array = [200, 400, 200, -500, 200, 100];
let s = 0;
$.each(array, (i,v)=> s + v);
A simpler solution with vanillaJS is:
let sum = [200, 400, 200, -500, 200,100].reduce((s,v) => s + v ,0)
answered yesterday
Marios Simou
323
The question specifically says they want to use$.each()
– Praveen Kumar Purushothaman
yesterday
Yeap. I didn't read correctly that question.
– Marios Simou
yesterday
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53944394%2fbrowse-a-numeric-data-array-and-add-or-subtract-during-the-loop-with-each%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can actually use .reduce(). Coming to your question, the problem is in the if condition. Let's say the value is -500.
if (val > 0) {
sum += val;
} else {
// val is -500 and it comes here.
// sum will be set to sum - (-500)
// the above will be sum + 500
sum -= val;
}
Your logic here is flawed.
You just need to use:
if ($.isNumeric(val)) {
sum += val;
}
Snippet
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<table>
<tr id="testTable">
<td data-pos="0"></td>
<td data-pos="1"></td>
<td data-pos="2"></td>
<td data-pos="3"></td>
<td data-pos="4"></td>
<td data-pos="5"></td>
</tr>
</table>
<script>
$(document).ready(function() {
var array = [200, 400, 200, -500, 200, 100];
var sum = 0;
$.each(array, function(i, val) {
val = parseFloat(val);
if ($.isNumeric(val)) {
sum += val;
}
$('tr#testTable td[data-pos="' + i + '"]').text(sum);
});
});
</script>I am indeed getting 600 as the answer above.
answered yesterday
Praveen Kumar Purushothaman
131k23135181
add a comment |
You can actually use .reduce(). Coming to your question, the problem is in the if condition. Let's say the value is -500.
if (val > 0) {
sum += val;
} else {
// val is -500 and it comes here.
// sum will be set to sum - (-500)
// the above will be sum + 500
sum -= val;
}
Your logic here is flawed.
You just need to use:
if ($.isNumeric(val)) {
sum += val;
}
Snippet
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<table>
<tr id="testTable">
<td data-pos="0"></td>
<td data-pos="1"></td>
<td data-pos="2"></td>
<td data-pos="3"></td>
<td data-pos="4"></td>
<td data-pos="5"></td>
</tr>
</table>
<script>
$(document).ready(function() {
var array = [200, 400, 200, -500, 200, 100];
var sum = 0;
$.each(array, function(i, val) {
val = parseFloat(val);
if ($.isNumeric(val)) {
sum += val;
}
$('tr#testTable td[data-pos="' + i + '"]').text(sum);
});
});
</script>I am indeed getting 600 as the answer above.
answered yesterday
Praveen Kumar Purushothaman
131k23135181
add a comment |
You can actually use .reduce(). Coming to your question, the problem is in the if condition. Let's say the value is -500.
if (val > 0) {
sum += val;
} else {
// val is -500 and it comes here.
// sum will be set to sum - (-500)
// the above will be sum + 500
sum -= val;
}
Your logic here is flawed.
You just need to use:
if ($.isNumeric(val)) {
sum += val;
}
Snippet
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<table>
<tr id="testTable">
<td data-pos="0"></td>
<td data-pos="1"></td>
<td data-pos="2"></td>
<td data-pos="3"></td>
<td data-pos="4"></td>
<td data-pos="5"></td>
</tr>
</table>
<script>
$(document).ready(function() {
var array = [200, 400, 200, -500, 200, 100];
var sum = 0;
$.each(array, function(i, val) {
val = parseFloat(val);
if ($.isNumeric(val)) {
sum += val;
}
$('tr#testTable td[data-pos="' + i + '"]').text(sum);
});
});
</script>I am indeed getting 600 as the answer above.
answered yesterday
Praveen Kumar Purushothaman
131k23135181
You can actually use .reduce(). Coming to your question, the problem is in the if condition. Let's say the value is -500.
if (val > 0) {
sum += val;
} else {
// val is -500 and it comes here.
// sum will be set to sum - (-500)
// the above will be sum + 500
sum -= val;
}
Your logic here is flawed.
You just need to use:
if ($.isNumeric(val)) {
sum += val;
}
Snippet
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<table>
<tr id="testTable">
<td data-pos="0"></td>
<td data-pos="1"></td>
<td data-pos="2"></td>
<td data-pos="3"></td>
<td data-pos="4"></td>
<td data-pos="5"></td>
</tr>
</table>
<script>
$(document).ready(function() {
var array = [200, 400, 200, -500, 200, 100];
var sum = 0;
$.each(array, function(i, val) {
val = parseFloat(val);
if ($.isNumeric(val)) {
sum += val;
}
$('tr#testTable td[data-pos="' + i + '"]').text(sum);
});
});
</script>I am indeed getting 600 as the answer above.
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<table>
<tr id="testTable">
<td data-pos="0"></td>
<td data-pos="1"></td>
<td data-pos="2"></td>
<td data-pos="3"></td>
<td data-pos="4"></td>
<td data-pos="5"></td>
</tr>
</table>
<script>
$(document).ready(function() {
var array = [200, 400, 200, -500, 200, 100];
var sum = 0;
$.each(array, function(i, val) {
val = parseFloat(val);
if ($.isNumeric(val)) {
sum += val;
}
$('tr#testTable td[data-pos="' + i + '"]').text(sum);
});
});
</script><script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<table>
<tr id="testTable">
<td data-pos="0"></td>
<td data-pos="1"></td>
<td data-pos="2"></td>
<td data-pos="3"></td>
<td data-pos="4"></td>
<td data-pos="5"></td>
</tr>
</table>
<script>
$(document).ready(function() {
var array = [200, 400, 200, -500, 200, 100];
var sum = 0;
$.each(array, function(i, val) {
val = parseFloat(val);
if ($.isNumeric(val)) {
sum += val;
}
$('tr#testTable td[data-pos="' + i + '"]').text(sum);
});
});
</script>answered yesterday
Praveen Kumar Purushothaman
131k23135181
answered yesterday
Praveen Kumar Purushothaman
131k23135181
answered yesterday
Praveen Kumar Purushothaman
131k23135181
answered yesterday
Praveen Kumar Purushothaman
131k23135181
131k23135181
add a comment |
add a comment |
Your problem sources from the if condition that you are using. You don't need to check whether the value is negative or positive, but just to add it in the sum.
When you check if the value is negative, in reality it reverses the sign of the number, which means that a negative number becomes positive.
I would recommend the use of .reduce, rather the unnecessary declaration of variables. However, the solution with $.each would be:
$.each(callback)
let array = [200, 400, 200, -500, 200, 100];
let s = 0;
$.each(array, (i,v)=> s + v);
A simpler solution with vanillaJS is:
let sum = [200, 400, 200, -500, 200,100].reduce((s,v) => s + v ,0)
answered yesterday
Marios Simou
323
The question specifically says they want to use$.each()
– Praveen Kumar Purushothaman
yesterday
Yeap. I didn't read correctly that question.
– Marios Simou
yesterday
add a comment |
Your problem sources from the if condition that you are using. You don't need to check whether the value is negative or positive, but just to add it in the sum.
When you check if the value is negative, in reality it reverses the sign of the number, which means that a negative number becomes positive.
I would recommend the use of .reduce, rather the unnecessary declaration of variables. However, the solution with $.each would be:
$.each(callback)
let array = [200, 400, 200, -500, 200, 100];
let s = 0;
$.each(array, (i,v)=> s + v);
A simpler solution with vanillaJS is:
let sum = [200, 400, 200, -500, 200,100].reduce((s,v) => s + v ,0)
answered yesterday
Marios Simou
323
The question specifically says they want to use$.each()
– Praveen Kumar Purushothaman
yesterday
Yeap. I didn't read correctly that question.
– Marios Simou
yesterday
add a comment |
Your problem sources from the if condition that you are using. You don't need to check whether the value is negative or positive, but just to add it in the sum.
When you check if the value is negative, in reality it reverses the sign of the number, which means that a negative number becomes positive.
I would recommend the use of .reduce, rather the unnecessary declaration of variables. However, the solution with $.each would be:
$.each(callback)
let array = [200, 400, 200, -500, 200, 100];
let s = 0;
$.each(array, (i,v)=> s + v);
A simpler solution with vanillaJS is:
let sum = [200, 400, 200, -500, 200,100].reduce((s,v) => s + v ,0)
answered yesterday
Marios Simou
323
Your problem sources from the if condition that you are using. You don't need to check whether the value is negative or positive, but just to add it in the sum.
When you check if the value is negative, in reality it reverses the sign of the number, which means that a negative number becomes positive.
I would recommend the use of .reduce, rather the unnecessary declaration of variables. However, the solution with $.each would be:
$.each(callback)
let array = [200, 400, 200, -500, 200, 100];
let s = 0;
$.each(array, (i,v)=> s + v);
A simpler solution with vanillaJS is:
let sum = [200, 400, 200, -500, 200,100].reduce((s,v) => s + v ,0)
answered yesterday
Marios Simou
323
edited yesterday
answered yesterday
Marios Simou
323
answered yesterday
Marios Simou
323
answered yesterday
Marios Simou
323
323
The question specifically says they want to use$.each()
– Praveen Kumar Purushothaman
yesterday
Yeap. I didn't read correctly that question.
– Marios Simou
yesterday
add a comment |
The question specifically says they want to use$.each()
– Praveen Kumar Purushothaman
yesterday
Yeap. I didn't read correctly that question.
– Marios Simou
yesterday
The question specifically says they want to use
$.each()– Praveen Kumar Purushothaman
yesterday
The question specifically says they want to use
$.each()– Praveen Kumar Purushothaman
yesterday
Yeap. I didn't read correctly that question.
– Marios Simou
yesterday
Yeap. I didn't read correctly that question.
– Marios Simou
yesterday
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53944394%2fbrowse-a-numeric-data-array-and-add-or-subtract-during-the-loop-with-each%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Do you really need to use
$.each()? Why notArray.reduce?– Praveen Kumar Purushothaman
yesterday
Why you do not want to use reduce?
– Just code
yesterday