JavaScript-过滤后合并2个对象

Question

我有一个对象变量的old_data数组：

let old_data = [
{
value: 'item',
text: "do_not_import",
custom: null,
updates : true, 
removes : true,        
},
{
value: 'additional_image',
text: "additional_image_mapped",
custom: null,
updates : true, 
removes : true,        
},
{
value: 'brand',
text: "brand_mapped",
custom: null,
updates : true, 
removes : true,        
},
];

现在，我正在过滤这个变量，其中text不是do_not_import

let filter_data = old_data.filter( ( item ) => {
return item.text != 'do_not_import';
});

现在，我得到了这个结果：

[ 
{ 
value: 'additional_image',
text: 'additional_image_mapped',
custom: null,
updates: true,
removes: true 
},
{ 
value: 'brand',
text: 'brand_mapped',
custom: null,
updates: true,
removes: true 
} 
]

好的。现在我有new_data可变：

let newData = [
{
value: 'item',
text: "do_not_import",
custom: null,
updates : true, 
removes : true,        
},
{
value: 'additional_image',
text: "do_not_import",
custom: null,
updates : true, 
removes : true,        
},
{
value: 'brand',
text: "do_not_import",
custom: null,
updates : true, 
removes : true,        
},
{
value: 'new',
text: "do_not_import",
custom: null,
updates : true, 
removes : true,        
},
];

现在，我想合并new_data和filter_data变量，它们应该返回来自filter_data的所有数据和来自new_data的所有数据，但从new_data它应该返回那些关键字value在filter_data变量中不相同的数据。

这意味着我想要的输出是：

[ 
{ 
value: 'additional_image',
text: 'additional_image_mapped',
custom: null,
updates: true,
removes: true
},
{ 
value: 'brand',
text: 'brand_mapped',
custom: null,
updates: true,
removes: true 
},
{
value: 'item',
text: "do_not_import",
custom: null,
updates : true, 
removes : true,        
},
{
value: 'new',
text: "do_not_import",
custom: null,
updates : true, 
removes : true,        
}
]

Answer 1

这应该完成任务-

let old_data = [
{
value: 'item',
text: "do_not_import",
custom: null,
updates : true, 
removes : true,        
},
{
value: 'additional_image',
text: "additional_image_mapped",
custom: null,
updates : true, 
removes : true,        
},
{
value: 'brand',
text: "brand_mapped",
custom: null,
updates : true, 
removes : true,        
},
];
let filter_data = old_data.filter( ( item ) => {
return item.text != 'do_not_import';
});
let newData = [
{
value: 'item',
text: "do_not_import",
custom: null,
updates : true, 
removes : true,        
},
{
value: 'additional_image',
text: "do_not_import",
custom: null,
updates : true, 
removes : true,        
},
{
value: 'brand',
text: "do_not_import",
custom: null,
updates : true, 
removes : true,        
},
{
value: 'new',
text: "do_not_import",
custom: null,
updates : true, 
removes : true,        
},
];
var values = new Set(filter_data.map(v => v.value));
var merged = [...filter_data, ...newData.filter(v => !values.has(v.value))];
console.log(merged)

Answer 2

这应该可以工作。

const combinedData=newData.reduce((acc，element)=>acc.some(e=>e.value===element.value)？acc:[…acc，element]，filteredData)；

Answer 3

const filterData = [
{
value: 'additional_image',
text: 'additional_image_mapped',
custom: null,
updates: true,
removes: true
},
{
value: 'brand',
text: 'brand_mapped',
custom: null,
updates: true,
removes: true
}
];
const newData = [
{
value: 'item',
text: "do_not_import",
custom: null,
updates: true,
removes: true,
},
{
value: 'additional_image',
text: "do_not_import",
custom: null,
updates: true,
removes: true,
},
{
value: 'brand',
text: "do_not_import",
custom: null,
updates: true,
removes: true,
},
{
value: 'new',
text: "do_not_import",
custom: null,
updates: true,
removes: true,
},
];
const consolidateArray = [];
for(const data of newData){
const getObjectFromFilterArray = filterData.find(x => x.value === data.value);
if(getObjectFromFilterArray){
consolidateArray.push(getObjectFromFilterArray);
}else{
consolidateArray.push(data);
}
}

console.log(consolidateArray);