How to merge irregular object
I wanna merge 2 objects.
One object is the default form.
let data = {
weight: '',
DOB: '',
gender: '',
}
The other is irregular like below that
let temp1 = {
weight: '19.2',
}
let temp2 = {
DOB: '1992-11',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
let temp4 = undefined;
If I merge data and temp1, the result should be
let data = {
weight: '19.2',
DOB: '',
gender: ''
}
If I merge data and temp3, the result should be
let data = {
weight: '',
DOB: '1992-11',
gender: 'Male',
}
If I merge data and temp4, the result should be
let data = {
weight: '',
DOB: '',
gender: ''
}
Give me some advice to achieve it.
Thanks in advance.
javascript object
asked yesterday
zynkn
3,90621030
add a comment |
I wanna merge 2 objects.
One object is the default form.
let data = {
weight: '',
DOB: '',
gender: '',
}
The other is irregular like below that
let temp1 = {
weight: '19.2',
}
let temp2 = {
DOB: '1992-11',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
let temp4 = undefined;
If I merge data and temp1, the result should be
let data = {
weight: '19.2',
DOB: '',
gender: ''
}
If I merge data and temp3, the result should be
let data = {
weight: '',
DOB: '1992-11',
gender: 'Male',
}
If I merge data and temp4, the result should be
let data = {
weight: '',
DOB: '',
gender: ''
}
Give me some advice to achieve it.
Thanks in advance.
javascript object
asked yesterday
zynkn
3,90621030
Oh, you can't merge that way. You gotta set those values to undefined, right?
– Praveen Kumar Purushothaman
yesterday
If your default object always has the 3 keys, why not just check each different object for those properties? If they exist, take them, if not, ignore.
– tomerpacific
yesterday
@tomerpacific I can check there's existing or not, but I need to preserve the default object form.
– zynkn
yesterday
add a comment |
I wanna merge 2 objects.
One object is the default form.
let data = {
weight: '',
DOB: '',
gender: '',
}
The other is irregular like below that
let temp1 = {
weight: '19.2',
}
let temp2 = {
DOB: '1992-11',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
let temp4 = undefined;
If I merge data and temp1, the result should be
let data = {
weight: '19.2',
DOB: '',
gender: ''
}
If I merge data and temp3, the result should be
let data = {
weight: '',
DOB: '1992-11',
gender: 'Male',
}
If I merge data and temp4, the result should be
let data = {
weight: '',
DOB: '',
gender: ''
}
Give me some advice to achieve it.
Thanks in advance.
javascript object
asked yesterday
zynkn
3,90621030
I wanna merge 2 objects.
One object is the default form.
let data = {
weight: '',
DOB: '',
gender: '',
}
The other is irregular like below that
let temp1 = {
weight: '19.2',
}
let temp2 = {
DOB: '1992-11',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
let temp4 = undefined;
If I merge data and temp1, the result should be
let data = {
weight: '19.2',
DOB: '',
gender: ''
}
If I merge data and temp3, the result should be
let data = {
weight: '',
DOB: '1992-11',
gender: 'Male',
}
If I merge data and temp4, the result should be
let data = {
weight: '',
DOB: '',
gender: ''
}
Give me some advice to achieve it.
Thanks in advance.
javascript object
javascript object
asked yesterday
zynkn
3,90621030
asked yesterday
zynkn
3,90621030
asked yesterday
zynkn
3,90621030
asked yesterday
zynkn
3,90621030
asked yesterday
zynkn
3,90621030
3,90621030
Oh, you can't merge that way. You gotta set those values to undefined, right?
– Praveen Kumar Purushothaman
yesterday
If your default object always has the 3 keys, why not just check each different object for those properties? If they exist, take them, if not, ignore.
– tomerpacific
yesterday
@tomerpacific I can check there's existing or not, but I need to preserve the default object form.
– zynkn
yesterday
add a comment |
Oh, you can't merge that way. You gotta set those values to undefined, right?
– Praveen Kumar Purushothaman
yesterday
If your default object always has the 3 keys, why not just check each different object for those properties? If they exist, take them, if not, ignore.
– tomerpacific
yesterday
@tomerpacific I can check there's existing or not, but I need to preserve the default object form.
– zynkn
yesterday
Oh, you can't merge that way. You gotta set those values to undefined, right?
– Praveen Kumar Purushothaman
yesterday
Oh, you can't merge that way. You gotta set those values to undefined, right?
– Praveen Kumar Purushothaman
yesterday
If your default object always has the 3 keys, why not just check each different object for those properties? If they exist, take them, if not, ignore.
– tomerpacific
yesterday
If your default object always has the 3 keys, why not just check each different object for those properties? If they exist, take them, if not, ignore.
– tomerpacific
yesterday
@tomerpacific I can check there's existing or not, but I need to preserve the default object form.
– zynkn
yesterday
@tomerpacific I can check there's existing or not, but I need to preserve the default object form.
– zynkn
yesterday
add a comment |
2 Answers
2
active
oldest
votes
You can use Object.assign():
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
data = Object.assign(data, temp3);
or if you don't want the data object to change:
let data2 = Object.assign({}, data, temp3);
The spread operator can also be used, which basically is the same, just that the original data object doesn't change:
data = { ...data, ...temp3 };
answered yesterday
PierreDuc
28.9k45376
It's exactly what I found, I have known thespread operatorbut I didn't know exactly how to use. Thank you a lot
– zynkn
yesterday
add a comment |
with jQuery you can use $.extend()
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp1 = {
weight: '19.2',
}
console.log($.extend(data,temp1))<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>OR use javascript spread operator ... as following
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
data={...data,...temp3}
console.log(data)If the keys are same then the value from righmost object is used
answered yesterday
Elish
1718
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use Object.assign():
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
data = Object.assign(data, temp3);
or if you don't want the data object to change:
let data2 = Object.assign({}, data, temp3);
The spread operator can also be used, which basically is the same, just that the original data object doesn't change:
data = { ...data, ...temp3 };
answered yesterday
PierreDuc
28.9k45376
It's exactly what I found, I have known thespread operatorbut I didn't know exactly how to use. Thank you a lot
– zynkn
yesterday
add a comment |
You can use Object.assign():
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
data = Object.assign(data, temp3);
or if you don't want the data object to change:
let data2 = Object.assign({}, data, temp3);
The spread operator can also be used, which basically is the same, just that the original data object doesn't change:
data = { ...data, ...temp3 };
answered yesterday
PierreDuc
28.9k45376
It's exactly what I found, I have known thespread operatorbut I didn't know exactly how to use. Thank you a lot
– zynkn
yesterday
add a comment |
You can use Object.assign():
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
data = Object.assign(data, temp3);
or if you don't want the data object to change:
let data2 = Object.assign({}, data, temp3);
The spread operator can also be used, which basically is the same, just that the original data object doesn't change:
data = { ...data, ...temp3 };
answered yesterday
PierreDuc
28.9k45376
You can use Object.assign():
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
data = Object.assign(data, temp3);
or if you don't want the data object to change:
let data2 = Object.assign({}, data, temp3);
The spread operator can also be used, which basically is the same, just that the original data object doesn't change:
data = { ...data, ...temp3 };
answered yesterday
PierreDuc
28.9k45376
edited yesterday
answered yesterday
PierreDuc
28.9k45376
answered yesterday
PierreDuc
28.9k45376
answered yesterday
PierreDuc
28.9k45376
28.9k45376
It's exactly what I found, I have known thespread operatorbut I didn't know exactly how to use. Thank you a lot
– zynkn
yesterday
add a comment |
It's exactly what I found, I have known thespread operatorbut I didn't know exactly how to use. Thank you a lot
– zynkn
yesterday
It's exactly what I found, I have known the
spread operator but I didn't know exactly how to use. Thank you a lot– zynkn
yesterday
It's exactly what I found, I have known the
spread operator but I didn't know exactly how to use. Thank you a lot– zynkn
yesterday
add a comment |
with jQuery you can use $.extend()
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp1 = {
weight: '19.2',
}
console.log($.extend(data,temp1))<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>OR use javascript spread operator ... as following
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
data={...data,...temp3}
console.log(data)If the keys are same then the value from righmost object is used
answered yesterday
Elish
1718
add a comment |
with jQuery you can use $.extend()
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp1 = {
weight: '19.2',
}
console.log($.extend(data,temp1))<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>OR use javascript spread operator ... as following
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
data={...data,...temp3}
console.log(data)If the keys are same then the value from righmost object is used
answered yesterday
Elish
1718
add a comment |
with jQuery you can use $.extend()
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp1 = {
weight: '19.2',
}
console.log($.extend(data,temp1))<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>OR use javascript spread operator ... as following
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
data={...data,...temp3}
console.log(data)If the keys are same then the value from righmost object is used
answered yesterday
Elish
1718
with jQuery you can use $.extend()
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp1 = {
weight: '19.2',
}
console.log($.extend(data,temp1))<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>OR use javascript spread operator ... as following
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
data={...data,...temp3}
console.log(data)If the keys are same then the value from righmost object is used
let data = {
weight: '',
DOB: '',
gender: '',
}
let temp1 = {
weight: '19.2',
}
console.log($.extend(data,temp1))<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>let data = {
weight: '',
DOB: '',
gender: '',
}
let temp1 = {
weight: '19.2',
}
console.log($.extend(data,temp1))<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>let data = {
weight: '',
DOB: '',
gender: '',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
data={...data,...temp3}
console.log(data)let data = {
weight: '',
DOB: '',
gender: '',
}
let temp3 = {
DOB: '1992-11',
gender: 'Male',
}
data={...data,...temp3}
console.log(data)answered yesterday
Elish
1718
edited yesterday
answered yesterday
Elish
1718
answered yesterday
Elish
1718
answered yesterday
Elish
1718
1718
add a comment |
add a comment |
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Oh, you can't merge that way. You gotta set those values to undefined, right?
– Praveen Kumar Purushothaman
yesterday
If your default object always has the 3 keys, why not just check each different object for those properties? If they exist, take them, if not, ignore.
– tomerpacific
yesterday
@tomerpacific I can check there's existing or not, but I need to preserve the default object form.
– zynkn
yesterday