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chaining sort and filter not working javascript
How do JavaScript closures work?How do I remove a property from a JavaScript object?Which equals operator (== vs ===) should be used in JavaScript comparisons?How do I sort a dictionary by value?How do I include a JavaScript file in another JavaScript file?Sort array of objects by string property valueWhat does “use strict” do in JavaScript, and what is the reasoning behind it?How to check whether a string contains a substring in JavaScript?How do I remove a particular element from an array in JavaScript?How to use foreach with array in JavaScript?
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1
I have an array. I want to sort and filter the array. I had try to chain .sort() and .filter(). The .sort() is working good, but not with the .filter(). Here is my example data and function that I had made. Whats goes wrong here?
const data = [
name: 'John',
date: '24 April 2001',
sex: 'male'
,
name: 'steve',
date: '12 August 2012',
sex: 'male'
,
name: 'natasha',
date: '13 October 1992',
sex: 'female'
,
name: 'chris',
date: '8 September 2004',
sex: 'remain unknown'
]
sortAndFilter = (arr, orderBy, order, filterBy, filterValue, dataType) =>
let result;
if (filterValue === '')
if (dataType === 'string')
switch (order)
case 'asc':
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy]));
break;
case 'dsc':
result = arr.sort((a, b) => b[orderBy].localeCompare(a[orderBy]));
break;
default:
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy]));
else if (dataType === 'date')
switch (order)
case 'asc':
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy]));
break;
case 'dsc':
result = arr.sort((a, b) => new Date(b[orderBy]) - new Date(a[orderBy]));
break;
default:
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy]));
else
if (dataType === 'string')
switch (order)
case 'asc':
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy])).filter(el => el[filterBy] === filterValue);
break;
case 'dsc':
result = arr.sort((a, b) => b[orderBy].localeCompare(a[orderBy])).filter(el => el[filterBy] === filterValue);
break;
default:
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy])).filter(el => el[filterBy] === filterValue);
else if (dataType === 'date')
switch (order)
case 'asc':
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy])).filter(el => el[filterBy] === filterValue);
break;
case 'dsc':
result = arr.sort((a, b) => new Date(b[orderBy]) - new Date(a[orderBy])).filter(el => el[filterBy] === filterValue);
break;
default:
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy])).filter(el => el[filterBy] === filterValue);
return result;
sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');
console.log('this is not filtered: ', data);
const finalData = sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');
console.log('this one is sorted and filtered: ', finalData);that is my approach. Whats wrong? or maybe, is there any better approach to achieve this? Thank you in advance.
javascript arrays sorting filter
asked Mar 21 at 23:58
AkzaAkza
6818
|
show 3 more comments
1
I have an array. I want to sort and filter the array. I had try to chain .sort() and .filter(). The .sort() is working good, but not with the .filter(). Here is my example data and function that I had made. Whats goes wrong here?
const data = [
name: 'John',
date: '24 April 2001',
sex: 'male'
,
name: 'steve',
date: '12 August 2012',
sex: 'male'
,
name: 'natasha',
date: '13 October 1992',
sex: 'female'
,
name: 'chris',
date: '8 September 2004',
sex: 'remain unknown'
]
sortAndFilter = (arr, orderBy, order, filterBy, filterValue, dataType) =>
let result;
if (filterValue === '')
if (dataType === 'string')
switch (order)
case 'asc':
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy]));
break;
case 'dsc':
result = arr.sort((a, b) => b[orderBy].localeCompare(a[orderBy]));
break;
default:
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy]));
else if (dataType === 'date')
switch (order)
case 'asc':
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy]));
break;
case 'dsc':
result = arr.sort((a, b) => new Date(b[orderBy]) - new Date(a[orderBy]));
break;
default:
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy]));
else
if (dataType === 'string')
switch (order)
case 'asc':
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy])).filter(el => el[filterBy] === filterValue);
break;
case 'dsc':
result = arr.sort((a, b) => b[orderBy].localeCompare(a[orderBy])).filter(el => el[filterBy] === filterValue);
break;
default:
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy])).filter(el => el[filterBy] === filterValue);
else if (dataType === 'date')
switch (order)
case 'asc':
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy])).filter(el => el[filterBy] === filterValue);
break;
case 'dsc':
result = arr.sort((a, b) => new Date(b[orderBy]) - new Date(a[orderBy])).filter(el => el[filterBy] === filterValue);
break;
default:
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy])).filter(el => el[filterBy] === filterValue);
return result;
sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');
console.log('this is not filtered: ', data);
const finalData = sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');
console.log('this one is sorted and filtered: ', finalData);that is my approach. Whats wrong? or maybe, is there any better approach to achieve this? Thank you in advance.
javascript arrays sorting filter
asked Mar 21 at 23:58
AkzaAkza
6818
seems working fine?
– Anik Islam Abhi
Mar 22 at 0:11
the filter isn't
– Akza
Mar 22 at 0:12
how come? Please provide some example like what are you expecting and what is showing?
– Anik Islam Abhi
Mar 22 at 0:13
sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');, i expect the result is only john and steve, and the order is john, then steve since i order it by the name and ascending.
– Akza
Mar 22 at 0:17
1
If you run the snippet, that's what you get. However, sort mutates arrays, and filter does not... So if you check your original argument, it will have been sorted, but not filtered. The return value will be filtered, though.
– Garrett Motzner
Mar 22 at 0:18
|
show 3 more comments
1
1
1
I have an array. I want to sort and filter the array. I had try to chain .sort() and .filter(). The .sort() is working good, but not with the .filter(). Here is my example data and function that I had made. Whats goes wrong here?
const data = [
name: 'John',
date: '24 April 2001',
sex: 'male'
,
name: 'steve',
date: '12 August 2012',
sex: 'male'
,
name: 'natasha',
date: '13 October 1992',
sex: 'female'
,
name: 'chris',
date: '8 September 2004',
sex: 'remain unknown'
]
sortAndFilter = (arr, orderBy, order, filterBy, filterValue, dataType) =>
let result;
if (filterValue === '')
if (dataType === 'string')
switch (order)
case 'asc':
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy]));
break;
case 'dsc':
result = arr.sort((a, b) => b[orderBy].localeCompare(a[orderBy]));
break;
default:
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy]));
else if (dataType === 'date')
switch (order)
case 'asc':
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy]));
break;
case 'dsc':
result = arr.sort((a, b) => new Date(b[orderBy]) - new Date(a[orderBy]));
break;
default:
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy]));
else
if (dataType === 'string')
switch (order)
case 'asc':
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy])).filter(el => el[filterBy] === filterValue);
break;
case 'dsc':
result = arr.sort((a, b) => b[orderBy].localeCompare(a[orderBy])).filter(el => el[filterBy] === filterValue);
break;
default:
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy])).filter(el => el[filterBy] === filterValue);
else if (dataType === 'date')
switch (order)
case 'asc':
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy])).filter(el => el[filterBy] === filterValue);
break;
case 'dsc':
result = arr.sort((a, b) => new Date(b[orderBy]) - new Date(a[orderBy])).filter(el => el[filterBy] === filterValue);
break;
default:
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy])).filter(el => el[filterBy] === filterValue);
return result;
sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');
console.log('this is not filtered: ', data);
const finalData = sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');
console.log('this one is sorted and filtered: ', finalData);that is my approach. Whats wrong? or maybe, is there any better approach to achieve this? Thank you in advance.
javascript arrays sorting filter
asked Mar 21 at 23:58
AkzaAkza
6818
I have an array. I want to sort and filter the array. I had try to chain .sort() and .filter(). The .sort() is working good, but not with the .filter(). Here is my example data and function that I had made. Whats goes wrong here?
const data = [
name: 'John',
date: '24 April 2001',
sex: 'male'
,
name: 'steve',
date: '12 August 2012',
sex: 'male'
,
name: 'natasha',
date: '13 October 1992',
sex: 'female'
,
name: 'chris',
date: '8 September 2004',
sex: 'remain unknown'
]
sortAndFilter = (arr, orderBy, order, filterBy, filterValue, dataType) =>
let result;
if (filterValue === '')
if (dataType === 'string')
switch (order)
case 'asc':
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy]));
break;
case 'dsc':
result = arr.sort((a, b) => b[orderBy].localeCompare(a[orderBy]));
break;
default:
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy]));
else if (dataType === 'date')
switch (order)
case 'asc':
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy]));
break;
case 'dsc':
result = arr.sort((a, b) => new Date(b[orderBy]) - new Date(a[orderBy]));
break;
default:
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy]));
else
if (dataType === 'string')
switch (order)
case 'asc':
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy])).filter(el => el[filterBy] === filterValue);
break;
case 'dsc':
result = arr.sort((a, b) => b[orderBy].localeCompare(a[orderBy])).filter(el => el[filterBy] === filterValue);
break;
default:
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy])).filter(el => el[filterBy] === filterValue);
else if (dataType === 'date')
switch (order)
case 'asc':
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy])).filter(el => el[filterBy] === filterValue);
break;
case 'dsc':
result = arr.sort((a, b) => new Date(b[orderBy]) - new Date(a[orderBy])).filter(el => el[filterBy] === filterValue);
break;
default:
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy])).filter(el => el[filterBy] === filterValue);
return result;
sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');
console.log('this is not filtered: ', data);
const finalData = sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');
console.log('this one is sorted and filtered: ', finalData);that is my approach. Whats wrong? or maybe, is there any better approach to achieve this? Thank you in advance.
const data = [
name: 'John',
date: '24 April 2001',
sex: 'male'
,
name: 'steve',
date: '12 August 2012',
sex: 'male'
,
name: 'natasha',
date: '13 October 1992',
sex: 'female'
,
name: 'chris',
date: '8 September 2004',
sex: 'remain unknown'
]
sortAndFilter = (arr, orderBy, order, filterBy, filterValue, dataType) =>
let result;
if (filterValue === '')
if (dataType === 'string')
switch (order)
case 'asc':
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy]));
break;
case 'dsc':
result = arr.sort((a, b) => b[orderBy].localeCompare(a[orderBy]));
break;
default:
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy]));
else if (dataType === 'date')
switch (order)
case 'asc':
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy]));
break;
case 'dsc':
result = arr.sort((a, b) => new Date(b[orderBy]) - new Date(a[orderBy]));
break;
default:
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy]));
else
if (dataType === 'string')
switch (order)
case 'asc':
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy])).filter(el => el[filterBy] === filterValue);
break;
case 'dsc':
result = arr.sort((a, b) => b[orderBy].localeCompare(a[orderBy])).filter(el => el[filterBy] === filterValue);
break;
default:
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy])).filter(el => el[filterBy] === filterValue);
else if (dataType === 'date')
switch (order)
case 'asc':
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy])).filter(el => el[filterBy] === filterValue);
break;
case 'dsc':
result = arr.sort((a, b) => new Date(b[orderBy]) - new Date(a[orderBy])).filter(el => el[filterBy] === filterValue);
break;
default:
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy])).filter(el => el[filterBy] === filterValue);
return result;
sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');
console.log('this is not filtered: ', data);
const finalData = sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');
console.log('this one is sorted and filtered: ', finalData);const data = [
name: 'John',
date: '24 April 2001',
sex: 'male'
,
name: 'steve',
date: '12 August 2012',
sex: 'male'
,
name: 'natasha',
date: '13 October 1992',
sex: 'female'
,
name: 'chris',
date: '8 September 2004',
sex: 'remain unknown'
]
sortAndFilter = (arr, orderBy, order, filterBy, filterValue, dataType) =>
let result;
if (filterValue === '')
if (dataType === 'string')
switch (order)
case 'asc':
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy]));
break;
case 'dsc':
result = arr.sort((a, b) => b[orderBy].localeCompare(a[orderBy]));
break;
default:
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy]));
else if (dataType === 'date')
switch (order)
case 'asc':
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy]));
break;
case 'dsc':
result = arr.sort((a, b) => new Date(b[orderBy]) - new Date(a[orderBy]));
break;
default:
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy]));
else
if (dataType === 'string')
switch (order)
case 'asc':
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy])).filter(el => el[filterBy] === filterValue);
break;
case 'dsc':
result = arr.sort((a, b) => b[orderBy].localeCompare(a[orderBy])).filter(el => el[filterBy] === filterValue);
break;
default:
result = arr.sort((a, b) => a[orderBy].localeCompare(b[orderBy])).filter(el => el[filterBy] === filterValue);
else if (dataType === 'date')
switch (order)
case 'asc':
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy])).filter(el => el[filterBy] === filterValue);
break;
case 'dsc':
result = arr.sort((a, b) => new Date(b[orderBy]) - new Date(a[orderBy])).filter(el => el[filterBy] === filterValue);
break;
default:
result = arr.sort((a, b) => new Date(a[orderBy]) - new Date(b[orderBy])).filter(el => el[filterBy] === filterValue);
return result;
sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');
console.log('this is not filtered: ', data);
const finalData = sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');
console.log('this one is sorted and filtered: ', finalData);javascript arrays sorting filter
javascript arrays sorting filter
asked Mar 21 at 23:58
AkzaAkza
6818
asked Mar 21 at 23:58
AkzaAkza
6818
edited Mar 22 at 0:31
Akza
asked Mar 21 at 23:58
AkzaAkza
6818
asked Mar 21 at 23:58
AkzaAkza
6818
asked Mar 21 at 23:58
AkzaAkza
6818
6818
seems working fine?
– Anik Islam Abhi
Mar 22 at 0:11
the filter isn't
– Akza
Mar 22 at 0:12
how come? Please provide some example like what are you expecting and what is showing?
– Anik Islam Abhi
Mar 22 at 0:13
sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');, i expect the result is only john and steve, and the order is john, then steve since i order it by the name and ascending.
– Akza
Mar 22 at 0:17
1
If you run the snippet, that's what you get. However, sort mutates arrays, and filter does not... So if you check your original argument, it will have been sorted, but not filtered. The return value will be filtered, though.
– Garrett Motzner
Mar 22 at 0:18
|
show 3 more comments
seems working fine?
– Anik Islam Abhi
Mar 22 at 0:11
the filter isn't
– Akza
Mar 22 at 0:12
how come? Please provide some example like what are you expecting and what is showing?
– Anik Islam Abhi
Mar 22 at 0:13
sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');, i expect the result is only john and steve, and the order is john, then steve since i order it by the name and ascending.
– Akza
Mar 22 at 0:17
1
If you run the snippet, that's what you get. However, sort mutates arrays, and filter does not... So if you check your original argument, it will have been sorted, but not filtered. The return value will be filtered, though.
– Garrett Motzner
Mar 22 at 0:18
seems working fine?
– Anik Islam Abhi
Mar 22 at 0:11
seems working fine?
– Anik Islam Abhi
Mar 22 at 0:11
the filter isn't
– Akza
Mar 22 at 0:12
the filter isn't
– Akza
Mar 22 at 0:12
how come? Please provide some example like what are you expecting and what is showing?
– Anik Islam Abhi
Mar 22 at 0:13
how come? Please provide some example like what are you expecting and what is showing?
– Anik Islam Abhi
Mar 22 at 0:13
sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');, i expect the result is only john and steve, and the order is john, then steve since i order it by the name and ascending.
– Akza
Mar 22 at 0:17
sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');, i expect the result is only john and steve, and the order is john, then steve since i order it by the name and ascending.
– Akza
Mar 22 at 0:17
1
1
If you run the snippet, that's what you get. However, sort mutates arrays, and filter does not... So if you check your original argument, it will have been sorted, but not filtered. The return value will be filtered, though.
– Garrett Motzner
Mar 22 at 0:18
If you run the snippet, that's what you get. However, sort mutates arrays, and filter does not... So if you check your original argument, it will have been sorted, but not filtered. The return value will be filtered, though.
– Garrett Motzner
Mar 22 at 0:18
|
show 3 more comments
1 Answer
1
active
oldest
votes
0
Thanks to @GarretMotzner and @AnikIslamAbhi for help me in the comment section. Garret already point out whats going wrong. The function is not wrong, but I just have to store the filtered array in a new variable. Because, .filter() did not mutate the array. I've update the questions include the answer above
answered Mar 22 at 0:34
AkzaAkza
6818
1
I would say that because you are doing the sort, you probably want to do a shallow clone on the array first, so that you don't end up mutating the array. That will probably save you some bugs in the future.
– Garrett Motzner
Mar 22 at 16:56
@GarrettMotzner noted. Thank you for your advice. I really appreciate it
– Akza
Mar 24 at 16:51
add a comment |
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Thanks to @GarretMotzner and @AnikIslamAbhi for help me in the comment section. Garret already point out whats going wrong. The function is not wrong, but I just have to store the filtered array in a new variable. Because, .filter() did not mutate the array. I've update the questions include the answer above
answered Mar 22 at 0:34
AkzaAkza
6818
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I would say that because you are doing the sort, you probably want to do a shallow clone on the array first, so that you don't end up mutating the array. That will probably save you some bugs in the future.
– Garrett Motzner
Mar 22 at 16:56
@GarrettMotzner noted. Thank you for your advice. I really appreciate it
– Akza
Mar 24 at 16:51
add a comment |
0
Thanks to @GarretMotzner and @AnikIslamAbhi for help me in the comment section. Garret already point out whats going wrong. The function is not wrong, but I just have to store the filtered array in a new variable. Because, .filter() did not mutate the array. I've update the questions include the answer above
answered Mar 22 at 0:34
AkzaAkza
6818
1
I would say that because you are doing the sort, you probably want to do a shallow clone on the array first, so that you don't end up mutating the array. That will probably save you some bugs in the future.
– Garrett Motzner
Mar 22 at 16:56
@GarrettMotzner noted. Thank you for your advice. I really appreciate it
– Akza
Mar 24 at 16:51
add a comment |
0
0
0
Thanks to @GarretMotzner and @AnikIslamAbhi for help me in the comment section. Garret already point out whats going wrong. The function is not wrong, but I just have to store the filtered array in a new variable. Because, .filter() did not mutate the array. I've update the questions include the answer above
answered Mar 22 at 0:34
AkzaAkza
6818
Thanks to @GarretMotzner and @AnikIslamAbhi for help me in the comment section. Garret already point out whats going wrong. The function is not wrong, but I just have to store the filtered array in a new variable. Because, .filter() did not mutate the array. I've update the questions include the answer above
answered Mar 22 at 0:34
AkzaAkza
6818
answered Mar 22 at 0:34
AkzaAkza
6818
answered Mar 22 at 0:34
AkzaAkza
6818
answered Mar 22 at 0:34
AkzaAkza
6818
6818
1
I would say that because you are doing the sort, you probably want to do a shallow clone on the array first, so that you don't end up mutating the array. That will probably save you some bugs in the future.
– Garrett Motzner
Mar 22 at 16:56
@GarrettMotzner noted. Thank you for your advice. I really appreciate it
– Akza
Mar 24 at 16:51
add a comment |
1
I would say that because you are doing the sort, you probably want to do a shallow clone on the array first, so that you don't end up mutating the array. That will probably save you some bugs in the future.
– Garrett Motzner
Mar 22 at 16:56
@GarrettMotzner noted. Thank you for your advice. I really appreciate it
– Akza
Mar 24 at 16:51
1
1
I would say that because you are doing the sort, you probably want to do a shallow clone on the array first, so that you don't end up mutating the array. That will probably save you some bugs in the future.
– Garrett Motzner
Mar 22 at 16:56
I would say that because you are doing the sort, you probably want to do a shallow clone on the array first, so that you don't end up mutating the array. That will probably save you some bugs in the future.
– Garrett Motzner
Mar 22 at 16:56
@GarrettMotzner noted. Thank you for your advice. I really appreciate it
– Akza
Mar 24 at 16:51
@GarrettMotzner noted. Thank you for your advice. I really appreciate it
– Akza
Mar 24 at 16:51
add a comment |
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seems working fine?
– Anik Islam Abhi
Mar 22 at 0:11
the filter isn't
– Akza
Mar 22 at 0:12
how come? Please provide some example like what are you expecting and what is showing?
– Anik Islam Abhi
Mar 22 at 0:13
sortAndFilter(data, 'name', 'asc', 'sex', 'male', 'string');, i expect the result is only john and steve, and the order is john, then steve since i order it by the name and ascending.
– Akza
Mar 22 at 0:17
1
If you run the snippet, that's what you get. However, sort mutates arrays, and filter does not... So if you check your original argument, it will have been sorted, but not filtered. The return value will be filtered, though.
– Garrett Motzner
Mar 22 at 0:18