难以理解为什么原始对象在示例中的 .map 之后受到影响
Having trouble understanding why original object is affected after .map within example
我一直无法理解为什么这里的温度会受到影响。当查看控制台时运行这段代码:
const objectsArray = [
{'id': 1, 'name': 'Erik', 'yearsCompleted': 2, 'status': 'student'},
{'id': 2, 'name': 'Carol', 'yearsCompleted': 1, 'status': 'student'},
{'id': 3, 'name': 'Sarah', 'yearsCompleted': 4, 'status': 'student'}
];
temp = {
actions: objectsArray.map(student => {
if (student.yearsCompleted === 4) student.status = 'wow';
return student
})
}
console.log(temp)
temp2 = {
actions: objectsArray.map(student => {
if (student.yearsCompleted === 4) student.status = 'foo';
return student
})
}
console.log(temp)
console.log(temp2)
我最终得到以下结果:
{ actions:
[ { id: 1, name: 'Erik', yearsCompleted: 2, status: 'student' },
{ id: 2, name: 'Carol', yearsCompleted: 1, status: 'student' },
{ id: 3, name: 'Sarah', yearsCompleted: 4, status: 'wow' } ] }
{ actions:
[ { id: 1, name: 'Erik', yearsCompleted: 2, status: 'student' },
{ id: 2, name: 'Carol', yearsCompleted: 1, status: 'student' },
{ id: 3, name: 'Sarah', yearsCompleted: 4, status: 'foo' } ] }
{ actions:
[ { id: 1, name: 'Erik', yearsCompleted: 2, status: 'student' },
{ id: 2, name: 'Carol', yearsCompleted: 1, status: 'student' },
{ id: 3, name: 'Sarah', yearsCompleted: 4, status: 'foo' } ] }
为什么在这种情况下温度会受到影响?将 temp 的操作 属性 设置为 objectsArray.map 函数时,我们是否获得了对 objectsArray 的引用?我认为在这种情况下温度不会受到影响。谢谢。
这是因为 student.status = 'foo'
更改了 temp
数组中的实际项目。您需要使用 spread operator(...
) 制作副本并 return ,如下所示。
const objectsArray = [{
'id': 1,
'name': 'Erik',
'yearsCompleted': 2,
'status': 'student'
},
{
'id': 2,
'name': 'Carol',
'yearsCompleted': 1,
'status': 'student'
},
{
'id': 3,
'name': 'Sarah',
'yearsCompleted': 4,
'status': 'student'
}
];
temp = {
actions: objectsArray.map(student => {
if (student.yearsCompleted === 4) {
return { ...student,
status: 'wow'
}
};
return student
})
}
console.log(temp)
temp2 = {
actions: objectsArray.map(student => {
if (student.yearsCompleted === 4) {
return { ...student,
status: 'foo'
}
};
return student
})
}
console.log(temp)
console.log(temp2)
我一直无法理解为什么这里的温度会受到影响。当查看控制台时运行这段代码:
const objectsArray = [
{'id': 1, 'name': 'Erik', 'yearsCompleted': 2, 'status': 'student'},
{'id': 2, 'name': 'Carol', 'yearsCompleted': 1, 'status': 'student'},
{'id': 3, 'name': 'Sarah', 'yearsCompleted': 4, 'status': 'student'}
];
temp = {
actions: objectsArray.map(student => {
if (student.yearsCompleted === 4) student.status = 'wow';
return student
})
}
console.log(temp)
temp2 = {
actions: objectsArray.map(student => {
if (student.yearsCompleted === 4) student.status = 'foo';
return student
})
}
console.log(temp)
console.log(temp2)
我最终得到以下结果:
{ actions:
[ { id: 1, name: 'Erik', yearsCompleted: 2, status: 'student' },
{ id: 2, name: 'Carol', yearsCompleted: 1, status: 'student' },
{ id: 3, name: 'Sarah', yearsCompleted: 4, status: 'wow' } ] }
{ actions:
[ { id: 1, name: 'Erik', yearsCompleted: 2, status: 'student' },
{ id: 2, name: 'Carol', yearsCompleted: 1, status: 'student' },
{ id: 3, name: 'Sarah', yearsCompleted: 4, status: 'foo' } ] }
{ actions:
[ { id: 1, name: 'Erik', yearsCompleted: 2, status: 'student' },
{ id: 2, name: 'Carol', yearsCompleted: 1, status: 'student' },
{ id: 3, name: 'Sarah', yearsCompleted: 4, status: 'foo' } ] }
为什么在这种情况下温度会受到影响?将 temp 的操作 属性 设置为 objectsArray.map 函数时,我们是否获得了对 objectsArray 的引用?我认为在这种情况下温度不会受到影响。谢谢。
这是因为 student.status = 'foo'
更改了 temp
数组中的实际项目。您需要使用 spread operator(...
) 制作副本并 return ,如下所示。
const objectsArray = [{
'id': 1,
'name': 'Erik',
'yearsCompleted': 2,
'status': 'student'
},
{
'id': 2,
'name': 'Carol',
'yearsCompleted': 1,
'status': 'student'
},
{
'id': 3,
'name': 'Sarah',
'yearsCompleted': 4,
'status': 'student'
}
];
temp = {
actions: objectsArray.map(student => {
if (student.yearsCompleted === 4) {
return { ...student,
status: 'wow'
}
};
return student
})
}
console.log(temp)
temp2 = {
actions: objectsArray.map(student => {
if (student.yearsCompleted === 4) {
return { ...student,
status: 'foo'
}
};
return student
})
}
console.log(temp)
console.log(temp2)