For stop work when my object is copied without referency
I need to remove one attribute from my object (indexvariacaoatributo), but i need that the attribute stay in the root object.
This is my algorithm that receive from the variable variacoes my object and after i remove the attribute indexvariacaoatributo from my object, this way works but too remove from my root object:
var variacoes = categoriaForm.variacoes
if(variacoes[0].estoque_variacao == null || variacoes[0].estoque_variacao == 0){
variacoes = ;
}
//Retira do objeto o indexvariacaoatributo
for(let i=0;i<variacoes.length;i++){
for (let j=0;j<variacoes[i].atributo.length;j++){
console.log(variacoes[i].atributo[j].indexvariacaoatributo);
delete variacoes[i].atributo[j].indexvariacaoatributo
}
}
console.log(variacoes);
I try to copy without referency in this ways:
var variacoes = Object.assign({}, categoriaForm.variacoes);
Also tried:
var variacoes = { ...categoriaForm.variacoes };
But when i print my variable variacao i see that my attribute still stay in my object. I put a console.log and don't print the element inside the for.
There's something more that i need to do when i copy the object in this way?
Thanks
angular typescript
asked Nov 20 '18 at 18:46
Renato VeroneseRenato Veronese
4011
add a comment |
I need to remove one attribute from my object (indexvariacaoatributo), but i need that the attribute stay in the root object.
This is my algorithm that receive from the variable variacoes my object and after i remove the attribute indexvariacaoatributo from my object, this way works but too remove from my root object:
var variacoes = categoriaForm.variacoes
if(variacoes[0].estoque_variacao == null || variacoes[0].estoque_variacao == 0){
variacoes = ;
}
//Retira do objeto o indexvariacaoatributo
for(let i=0;i<variacoes.length;i++){
for (let j=0;j<variacoes[i].atributo.length;j++){
console.log(variacoes[i].atributo[j].indexvariacaoatributo);
delete variacoes[i].atributo[j].indexvariacaoatributo
}
}
console.log(variacoes);
I try to copy without referency in this ways:
var variacoes = Object.assign({}, categoriaForm.variacoes);
Also tried:
var variacoes = { ...categoriaForm.variacoes };
But when i print my variable variacao i see that my attribute still stay in my object. I put a console.log and don't print the element inside the for.
There's something more that i need to do when i copy the object in this way?
Thanks
angular typescript
asked Nov 20 '18 at 18:46
Renato VeroneseRenato Veronese
4011
Try var variacoes = JSON.parse(JSON.stringify(categoriaForm.variacoes)). This makes sure that changes to one does not affect the other and so the issue of copying can be avoided
– Minu
Nov 20 '18 at 18:57
add a comment |
I need to remove one attribute from my object (indexvariacaoatributo), but i need that the attribute stay in the root object.
This is my algorithm that receive from the variable variacoes my object and after i remove the attribute indexvariacaoatributo from my object, this way works but too remove from my root object:
var variacoes = categoriaForm.variacoes
if(variacoes[0].estoque_variacao == null || variacoes[0].estoque_variacao == 0){
variacoes = ;
}
//Retira do objeto o indexvariacaoatributo
for(let i=0;i<variacoes.length;i++){
for (let j=0;j<variacoes[i].atributo.length;j++){
console.log(variacoes[i].atributo[j].indexvariacaoatributo);
delete variacoes[i].atributo[j].indexvariacaoatributo
}
}
console.log(variacoes);
I try to copy without referency in this ways:
var variacoes = Object.assign({}, categoriaForm.variacoes);
Also tried:
var variacoes = { ...categoriaForm.variacoes };
But when i print my variable variacao i see that my attribute still stay in my object. I put a console.log and don't print the element inside the for.
There's something more that i need to do when i copy the object in this way?
Thanks
angular typescript
asked Nov 20 '18 at 18:46
Renato VeroneseRenato Veronese
4011
I need to remove one attribute from my object (indexvariacaoatributo), but i need that the attribute stay in the root object.
This is my algorithm that receive from the variable variacoes my object and after i remove the attribute indexvariacaoatributo from my object, this way works but too remove from my root object:
var variacoes = categoriaForm.variacoes
if(variacoes[0].estoque_variacao == null || variacoes[0].estoque_variacao == 0){
variacoes = ;
}
//Retira do objeto o indexvariacaoatributo
for(let i=0;i<variacoes.length;i++){
for (let j=0;j<variacoes[i].atributo.length;j++){
console.log(variacoes[i].atributo[j].indexvariacaoatributo);
delete variacoes[i].atributo[j].indexvariacaoatributo
}
}
console.log(variacoes);
I try to copy without referency in this ways:
var variacoes = Object.assign({}, categoriaForm.variacoes);
Also tried:
var variacoes = { ...categoriaForm.variacoes };
But when i print my variable variacao i see that my attribute still stay in my object. I put a console.log and don't print the element inside the for.
There's something more that i need to do when i copy the object in this way?
Thanks
angular typescript
angular typescript
asked Nov 20 '18 at 18:46
Renato VeroneseRenato Veronese
4011
asked Nov 20 '18 at 18:46
Renato VeroneseRenato Veronese
4011
asked Nov 20 '18 at 18:46
Renato VeroneseRenato Veronese
4011
asked Nov 20 '18 at 18:46
Renato VeroneseRenato Veronese
4011
asked Nov 20 '18 at 18:46
Renato VeroneseRenato Veronese
4011
4011
Try var variacoes = JSON.parse(JSON.stringify(categoriaForm.variacoes)). This makes sure that changes to one does not affect the other and so the issue of copying can be avoided
– Minu
Nov 20 '18 at 18:57
add a comment |
Try var variacoes = JSON.parse(JSON.stringify(categoriaForm.variacoes)). This makes sure that changes to one does not affect the other and so the issue of copying can be avoided
– Minu
Nov 20 '18 at 18:57
Try var variacoes = JSON.parse(JSON.stringify(categoriaForm.variacoes)). This makes sure that changes to one does not affect the other and so the issue of copying can be avoided
– Minu
Nov 20 '18 at 18:57
Try var variacoes = JSON.parse(JSON.stringify(categoriaForm.variacoes)). This makes sure that changes to one does not affect the other and so the issue of copying can be avoided
– Minu
Nov 20 '18 at 18:57
add a comment |
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Try var variacoes = JSON.parse(JSON.stringify(categoriaForm.variacoes)). This makes sure that changes to one does not affect the other and so the issue of copying can be avoided
– Minu
Nov 20 '18 at 18:57