Clean object from nulls and undefined
1
I've written the following function which cleans nulls and undefined from an object, but for some reason it also removes key where the value is 0.
For example:
{
key1: 'value1',
key2: 0
}
The function will remove key2, even though it should not.
Here is the function:
const cleanJson = function cleanJson (obj) {
if (Object.prototype.toString.call(obj) !== '[object Object]') return obj;
return Object.keys(obj).filter(key => obj[key] && obj[key] !== 'delete').reduce((newObj, key) => {
newObj[key] = Object.prototype.toString.call(obj[key]) !== '[object Object]' ? obj[key] : cleanJson(obj[key]);
return newObj;
}, {});
};
Please advise.
javascript node.js
edited Nov 25 '18 at 20:57
trincot
126k1688122
asked Nov 25 '18 at 20:51
David FaizDavid Faiz
92892247
add a comment |
1
I've written the following function which cleans nulls and undefined from an object, but for some reason it also removes key where the value is 0.
For example:
{
key1: 'value1',
key2: 0
}
The function will remove key2, even though it should not.
Here is the function:
const cleanJson = function cleanJson (obj) {
if (Object.prototype.toString.call(obj) !== '[object Object]') return obj;
return Object.keys(obj).filter(key => obj[key] && obj[key] !== 'delete').reduce((newObj, key) => {
newObj[key] = Object.prototype.toString.call(obj[key]) !== '[object Object]' ? obj[key] : cleanJson(obj[key]);
return newObj;
}, {});
};
Please advise.
javascript node.js
edited Nov 25 '18 at 20:57
trincot
126k1688122
asked Nov 25 '18 at 20:51
David FaizDavid Faiz
92892247
1
Please read the description of thejsontag, especially what is in capitals.
– trincot
Nov 25 '18 at 20:53
2
0is "falsy", so it gets removed. Your function also cleans up things that evaluate to false.
– Houseman
Nov 25 '18 at 20:56
add a comment |
1
1
1
I've written the following function which cleans nulls and undefined from an object, but for some reason it also removes key where the value is 0.
For example:
{
key1: 'value1',
key2: 0
}
The function will remove key2, even though it should not.
Here is the function:
const cleanJson = function cleanJson (obj) {
if (Object.prototype.toString.call(obj) !== '[object Object]') return obj;
return Object.keys(obj).filter(key => obj[key] && obj[key] !== 'delete').reduce((newObj, key) => {
newObj[key] = Object.prototype.toString.call(obj[key]) !== '[object Object]' ? obj[key] : cleanJson(obj[key]);
return newObj;
}, {});
};
Please advise.
javascript node.js
edited Nov 25 '18 at 20:57
trincot
126k1688122
asked Nov 25 '18 at 20:51
David FaizDavid Faiz
92892247
I've written the following function which cleans nulls and undefined from an object, but for some reason it also removes key where the value is 0.
For example:
{
key1: 'value1',
key2: 0
}
The function will remove key2, even though it should not.
Here is the function:
const cleanJson = function cleanJson (obj) {
if (Object.prototype.toString.call(obj) !== '[object Object]') return obj;
return Object.keys(obj).filter(key => obj[key] && obj[key] !== 'delete').reduce((newObj, key) => {
newObj[key] = Object.prototype.toString.call(obj[key]) !== '[object Object]' ? obj[key] : cleanJson(obj[key]);
return newObj;
}, {});
};
Please advise.
javascript node.js
javascript node.js
edited Nov 25 '18 at 20:57
trincot
126k1688122
asked Nov 25 '18 at 20:51
David FaizDavid Faiz
92892247
edited Nov 25 '18 at 20:57
trincot
126k1688122
asked Nov 25 '18 at 20:51
David FaizDavid Faiz
92892247
edited Nov 25 '18 at 20:57
trincot
126k1688122
edited Nov 25 '18 at 20:57
trincot
126k1688122
edited Nov 25 '18 at 20:57
trincot
126k1688122
126k1688122
asked Nov 25 '18 at 20:51
David FaizDavid Faiz
92892247
asked Nov 25 '18 at 20:51
David FaizDavid Faiz
92892247
asked Nov 25 '18 at 20:51
David FaizDavid Faiz
92892247
92892247
1
Please read the description of thejsontag, especially what is in capitals.
– trincot
Nov 25 '18 at 20:53
2
0is "falsy", so it gets removed. Your function also cleans up things that evaluate to false.
– Houseman
Nov 25 '18 at 20:56
add a comment |
1
Please read the description of thejsontag, especially what is in capitals.
– trincot
Nov 25 '18 at 20:53
2
0is "falsy", so it gets removed. Your function also cleans up things that evaluate to false.
– Houseman
Nov 25 '18 at 20:56
1
1
Please read the description of the
json tag, especially what is in capitals.– trincot
Nov 25 '18 at 20:53
Please read the description of the
json tag, especially what is in capitals.– trincot
Nov 25 '18 at 20:53
2
2
0 is "falsy", so it gets removed. Your function also cleans up things that evaluate to false.– Houseman
Nov 25 '18 at 20:56
0 is "falsy", so it gets removed. Your function also cleans up things that evaluate to false.– Houseman
Nov 25 '18 at 20:56
add a comment |
4 Answers
4
active
oldest
votes
2
Here is one option using Object.entries:
const myJSON = {key1:'value1', key2:0, key3:null, key4:undefined, key5:""};
const myCleanJSON = Object.entries(myJSON)
.filter(([key, value]) => (value !== null && typeof value !== 'undefined'))
.reduce((acc, b) => ((!acc.length) ? {...acc, [b[0]]: b[1] } : { [acc[0]] : acc[1], [b[0]]: b[1] }));
console.log(myCleanJSON);answered Nov 25 '18 at 21:10
zvonazvona
12.6k14059
add a comment |
2
The problem is within
obj[key] && obj[key] !== 'delete'
When obj[key] == 0, would return false.
answered Nov 25 '18 at 20:58
Guy YogevGuy Yogev
43239
add a comment |
1
Because 0 (as well as the empty string "") evaluates to false when cast to a Boolean, you need to explicitly check for undefined and null:
let a = {
key1: 'value1',
key2: 0,
key3: null
}
function cleanProps(o) {
if (Object.prototype.toString.call(o) !== '[object Object]') return o
for (key in o) {
if([undefined, null].includes(o[key])) delete o[key]
}
return o
}
console.log(cleanProps(a))answered Nov 25 '18 at 21:18
connexoconnexo
22.4k83561
add a comment |
0
I think that would solve your issue:
obj[key] !== null && obj[key] !== undefined && obj[key] !== 'delete'
answered Nov 25 '18 at 21:06
Kevin AmiranoffKevin Amiranoff
2,92651338
You can just use!=to catch bothnullandundefined
– Mark Meyer
Nov 25 '18 at 21:11
And pick one or the other?
– Kevin Amiranoff
Nov 25 '18 at 21:14
Yes, bothundefined != nullandnull != undefinedarefalse
– Mark Meyer
Nov 25 '18 at 21:16
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
Here is one option using Object.entries:
const myJSON = {key1:'value1', key2:0, key3:null, key4:undefined, key5:""};
const myCleanJSON = Object.entries(myJSON)
.filter(([key, value]) => (value !== null && typeof value !== 'undefined'))
.reduce((acc, b) => ((!acc.length) ? {...acc, [b[0]]: b[1] } : { [acc[0]] : acc[1], [b[0]]: b[1] }));
console.log(myCleanJSON);answered Nov 25 '18 at 21:10
zvonazvona
12.6k14059
add a comment |
2
Here is one option using Object.entries:
const myJSON = {key1:'value1', key2:0, key3:null, key4:undefined, key5:""};
const myCleanJSON = Object.entries(myJSON)
.filter(([key, value]) => (value !== null && typeof value !== 'undefined'))
.reduce((acc, b) => ((!acc.length) ? {...acc, [b[0]]: b[1] } : { [acc[0]] : acc[1], [b[0]]: b[1] }));
console.log(myCleanJSON);answered Nov 25 '18 at 21:10
zvonazvona
12.6k14059
add a comment |
2
2
2
Here is one option using Object.entries:
const myJSON = {key1:'value1', key2:0, key3:null, key4:undefined, key5:""};
const myCleanJSON = Object.entries(myJSON)
.filter(([key, value]) => (value !== null && typeof value !== 'undefined'))
.reduce((acc, b) => ((!acc.length) ? {...acc, [b[0]]: b[1] } : { [acc[0]] : acc[1], [b[0]]: b[1] }));
console.log(myCleanJSON);answered Nov 25 '18 at 21:10
zvonazvona
12.6k14059
Here is one option using Object.entries:
const myJSON = {key1:'value1', key2:0, key3:null, key4:undefined, key5:""};
const myCleanJSON = Object.entries(myJSON)
.filter(([key, value]) => (value !== null && typeof value !== 'undefined'))
.reduce((acc, b) => ((!acc.length) ? {...acc, [b[0]]: b[1] } : { [acc[0]] : acc[1], [b[0]]: b[1] }));
console.log(myCleanJSON);const myJSON = {key1:'value1', key2:0, key3:null, key4:undefined, key5:""};
const myCleanJSON = Object.entries(myJSON)
.filter(([key, value]) => (value !== null && typeof value !== 'undefined'))
.reduce((acc, b) => ((!acc.length) ? {...acc, [b[0]]: b[1] } : { [acc[0]] : acc[1], [b[0]]: b[1] }));
console.log(myCleanJSON);const myJSON = {key1:'value1', key2:0, key3:null, key4:undefined, key5:""};
const myCleanJSON = Object.entries(myJSON)
.filter(([key, value]) => (value !== null && typeof value !== 'undefined'))
.reduce((acc, b) => ((!acc.length) ? {...acc, [b[0]]: b[1] } : { [acc[0]] : acc[1], [b[0]]: b[1] }));
console.log(myCleanJSON);answered Nov 25 '18 at 21:10
zvonazvona
12.6k14059
answered Nov 25 '18 at 21:10
zvonazvona
12.6k14059
answered Nov 25 '18 at 21:10
zvonazvona
12.6k14059
answered Nov 25 '18 at 21:10
zvonazvona
12.6k14059
12.6k14059
add a comment |
add a comment |
2
The problem is within
obj[key] && obj[key] !== 'delete'
When obj[key] == 0, would return false.
answered Nov 25 '18 at 20:58
Guy YogevGuy Yogev
43239
add a comment |
2
The problem is within
obj[key] && obj[key] !== 'delete'
When obj[key] == 0, would return false.
answered Nov 25 '18 at 20:58
Guy YogevGuy Yogev
43239
add a comment |
2
2
2
The problem is within
obj[key] && obj[key] !== 'delete'
When obj[key] == 0, would return false.
answered Nov 25 '18 at 20:58
Guy YogevGuy Yogev
43239
The problem is within
obj[key] && obj[key] !== 'delete'
When obj[key] == 0, would return false.
answered Nov 25 '18 at 20:58
Guy YogevGuy Yogev
43239
answered Nov 25 '18 at 20:58
Guy YogevGuy Yogev
43239
answered Nov 25 '18 at 20:58
Guy YogevGuy Yogev
43239
answered Nov 25 '18 at 20:58
Guy YogevGuy Yogev
43239
43239
add a comment |
add a comment |
1
Because 0 (as well as the empty string "") evaluates to false when cast to a Boolean, you need to explicitly check for undefined and null:
let a = {
key1: 'value1',
key2: 0,
key3: null
}
function cleanProps(o) {
if (Object.prototype.toString.call(o) !== '[object Object]') return o
for (key in o) {
if([undefined, null].includes(o[key])) delete o[key]
}
return o
}
console.log(cleanProps(a))answered Nov 25 '18 at 21:18
connexoconnexo
22.4k83561
add a comment |
1
Because 0 (as well as the empty string "") evaluates to false when cast to a Boolean, you need to explicitly check for undefined and null:
let a = {
key1: 'value1',
key2: 0,
key3: null
}
function cleanProps(o) {
if (Object.prototype.toString.call(o) !== '[object Object]') return o
for (key in o) {
if([undefined, null].includes(o[key])) delete o[key]
}
return o
}
console.log(cleanProps(a))answered Nov 25 '18 at 21:18
connexoconnexo
22.4k83561
add a comment |
1
1
1
Because 0 (as well as the empty string "") evaluates to false when cast to a Boolean, you need to explicitly check for undefined and null:
let a = {
key1: 'value1',
key2: 0,
key3: null
}
function cleanProps(o) {
if (Object.prototype.toString.call(o) !== '[object Object]') return o
for (key in o) {
if([undefined, null].includes(o[key])) delete o[key]
}
return o
}
console.log(cleanProps(a))answered Nov 25 '18 at 21:18
connexoconnexo
22.4k83561
Because 0 (as well as the empty string "") evaluates to false when cast to a Boolean, you need to explicitly check for undefined and null:
let a = {
key1: 'value1',
key2: 0,
key3: null
}
function cleanProps(o) {
if (Object.prototype.toString.call(o) !== '[object Object]') return o
for (key in o) {
if([undefined, null].includes(o[key])) delete o[key]
}
return o
}
console.log(cleanProps(a))let a = {
key1: 'value1',
key2: 0,
key3: null
}
function cleanProps(o) {
if (Object.prototype.toString.call(o) !== '[object Object]') return o
for (key in o) {
if([undefined, null].includes(o[key])) delete o[key]
}
return o
}
console.log(cleanProps(a))let a = {
key1: 'value1',
key2: 0,
key3: null
}
function cleanProps(o) {
if (Object.prototype.toString.call(o) !== '[object Object]') return o
for (key in o) {
if([undefined, null].includes(o[key])) delete o[key]
}
return o
}
console.log(cleanProps(a))answered Nov 25 '18 at 21:18
connexoconnexo
22.4k83561
answered Nov 25 '18 at 21:18
connexoconnexo
22.4k83561
answered Nov 25 '18 at 21:18
connexoconnexo
22.4k83561
answered Nov 25 '18 at 21:18
connexoconnexo
22.4k83561
22.4k83561
add a comment |
add a comment |
0
I think that would solve your issue:
obj[key] !== null && obj[key] !== undefined && obj[key] !== 'delete'
answered Nov 25 '18 at 21:06
Kevin AmiranoffKevin Amiranoff
2,92651338
You can just use!=to catch bothnullandundefined
– Mark Meyer
Nov 25 '18 at 21:11
And pick one or the other?
– Kevin Amiranoff
Nov 25 '18 at 21:14
Yes, bothundefined != nullandnull != undefinedarefalse
– Mark Meyer
Nov 25 '18 at 21:16
add a comment |
0
I think that would solve your issue:
obj[key] !== null && obj[key] !== undefined && obj[key] !== 'delete'
answered Nov 25 '18 at 21:06
Kevin AmiranoffKevin Amiranoff
2,92651338
You can just use!=to catch bothnullandundefined
– Mark Meyer
Nov 25 '18 at 21:11
And pick one or the other?
– Kevin Amiranoff
Nov 25 '18 at 21:14
Yes, bothundefined != nullandnull != undefinedarefalse
– Mark Meyer
Nov 25 '18 at 21:16
add a comment |
0
0
0
I think that would solve your issue:
obj[key] !== null && obj[key] !== undefined && obj[key] !== 'delete'
answered Nov 25 '18 at 21:06
Kevin AmiranoffKevin Amiranoff
2,92651338
I think that would solve your issue:
obj[key] !== null && obj[key] !== undefined && obj[key] !== 'delete'
answered Nov 25 '18 at 21:06
Kevin AmiranoffKevin Amiranoff
2,92651338
answered Nov 25 '18 at 21:06
Kevin AmiranoffKevin Amiranoff
2,92651338
answered Nov 25 '18 at 21:06
Kevin AmiranoffKevin Amiranoff
2,92651338
answered Nov 25 '18 at 21:06
Kevin AmiranoffKevin Amiranoff
2,92651338
2,92651338
You can just use!=to catch bothnullandundefined
– Mark Meyer
Nov 25 '18 at 21:11
And pick one or the other?
– Kevin Amiranoff
Nov 25 '18 at 21:14
Yes, bothundefined != nullandnull != undefinedarefalse
– Mark Meyer
Nov 25 '18 at 21:16
add a comment |
You can just use!=to catch bothnullandundefined
– Mark Meyer
Nov 25 '18 at 21:11
And pick one or the other?
– Kevin Amiranoff
Nov 25 '18 at 21:14
Yes, bothundefined != nullandnull != undefinedarefalse
– Mark Meyer
Nov 25 '18 at 21:16
You can just use
!= to catch both null and undefined– Mark Meyer
Nov 25 '18 at 21:11
You can just use
!= to catch both null and undefined– Mark Meyer
Nov 25 '18 at 21:11
And pick one or the other?
– Kevin Amiranoff
Nov 25 '18 at 21:14
And pick one or the other?
– Kevin Amiranoff
Nov 25 '18 at 21:14
Yes, both
undefined != null and null != undefined are false– Mark Meyer
Nov 25 '18 at 21:16
Yes, both
undefined != null and null != undefined are false– Mark Meyer
Nov 25 '18 at 21:16
add a comment |
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1
Please read the description of the
jsontag, especially what is in capitals.– trincot
Nov 25 '18 at 20:53
2
0is "falsy", so it gets removed. Your function also cleans up things that evaluate to false.– Houseman
Nov 25 '18 at 20:56