我有两个嵌套对象数组。我想根据PermissionObj筛选数据。
这来自数据库。这是permissionObj中的子数组的数组。
我需要在reduce函数中做另一个条件。例如，如果Pubsidebar的值是token是public，我想保留静态内容{label：“ test”，value：“ public”}而不用PermissionObj过滤，并且如果其他键和值与PermissionObj匹配，则将其推送内部令牌。
let permissionObj = [
{
'OA deal': [
{
label: 'can view',
value: 'can_view',
},
],
},
{
Deposit: [
{
label: 'can_view',
value: 'can_view',
},
],
},
{
Journals: [
{
label: 'can create',
value: 'can_create',
},
],
},
{
Dashboard: [
{
label: 'can view',
value: 'can_view',
},
],
},
{
token: [
{
label: 'can view',
value: 'can_create',
},
],
},
]
const PubSidebar = [
{
label: 'Dashboard',
value: 'can_view',
},{
label: 'transction',
value:"public",content:[{label:"test2",value:"public"}]
},
{
label: 'token',
value: 'public',
content: [
{
key: 'token',
value: 'can_create',
},
{
key: 'test',
value: 'public',
},
],
},
{
label: 'OA deal',
content: [
{
label: 'view oadeal',
key: 'OA deal',
value: 'can_view',
},
{
label: 'Deposit',
key: 'Deposit',
value: 'can_view',
},
{
label: 'Corrections',
key: 'Corrections',
value: 'can_edit',
},
],
},
{
label: 'Journals',
content: [
{
label: 'Add Journal',
key: 'Journals',
value: 'can_create',
},
],
},
]
const filterObject = permissionObj.reduce((a, c) => {
for (const key in c) {
a[key] = c[key]
}
return a
}, {})
const result = PubSidebar.reduce((a, c) => {
if (
filterObject[c.label] &&
c.value &&
filterObject[c.label].some(s => s.value === c.value)
) {
a.push(c)
} else if (c.value === 'public' && c.label === 'token') {
if (
(c.content = c.content.filter(
f =>
filterObject[f.key] &&
filterObject[f.key].some(s => s.value == f.value)
))
) {
c.content = c.content.filter(
f =>
filterObject[f.key] &&
filterObject[f.key].some(s => s.value == f.value)
)
a.push(c)
}
} else if (c.content.some(s => filterObject[s.key]) && c.content) {
c.content = c.content.filter(
f =>
filterObject[f.key] && filterObject[f.key].some(s => s.value == f.value)
)
a.push(c)
}
return a
}, [])
console.log(result)
我正在尝试从侧边栏获取公共数据，而无需使用PermissionObj进行过滤。
我的预期输出将是：
[
{
"label": "Dashboard",
"value": "can_view"
},
{
label: 'transction',
value:"public",content:[{label:"test2",value:"public"}]
},
{
"label": "token",
"value": "public",
"content": [{
"key": "test",
"value": "public"
}
{
"key": "token",
"value": "can_create"
}
]
},
{
"label": "OA deal",
"content": [
{
"label": "view oadeal",
"key": "OA deal",
"value": "can_view"
},
{
"label": "Deposit",
"key": "Deposit",
"value": "can_view"
}
]
},
{
"label": "Journals",
"content": [
{
"label": "Add Journal",
"key": "Journals",
"value": "can_create"
}
]
}
]
答案 0 :(得分：1)
您可以在reduce方法内添加条件，并且您的content属性将不会被过滤：
if (c.value ==='public' && c.content.some(f => filterObject[f.key] &&
filterObject[f.key].some(s => s.value == f.value))) {
a.push(c);
}
并尝试避免在filter条件内if使用您的数据。
所以代码看起来像这样：
let permissionObj = [
{
'OA deal': [
{
label: 'can view',
value: 'can_view',
},
],
},
{
Deposit: [
{
label: 'can_view',
value: 'can_view',
},
],
},
{
Journals: [
{
label: 'can create',
value: 'can_create',
},
],
},
{
Dashboard: [
{
label: 'can view',
value: 'can_view',
},
],
},
{
token: [
{
label: 'can view',
value: 'can_create',
},
],
},
]
const PubSidebar = [
{
label: 'Dashboard',
value: 'can_view',
},
{
label: 'token',
value: 'public',
content: [
{
key: 'token',
value: 'can_create',
},
{
key: 'test',
value: 'public',
},
],
},
{
label: 'OA deal',
content: [
{
label: 'view oadeal',
key: 'OA deal',
value: 'can_view',
},
{
label: 'Deposit',
key: 'Deposit',
value: 'can_view',
},
{
label: 'Corrections',
key: 'Corrections',
value: 'can_edit',
},
],
},
{
label: 'Journals',
content: [
{
label: 'Add Journal',
key: 'Journals',
value: 'can_create',
},
],
},
]
const filterObject = permissionObj.reduce((a, c) => {
for (const key in c) {
a[key] = c[key]
}
return a
}, {})
const result = PubSidebar.reduce((a, c) => {
if (filterObject[c.label] && c.value && filterObject[c.label].some(s => s.value === c.value)) {
a.push(c)
} else if (c.value === 'public' && c.label === 'token') {
if (c.value ==='public' && c.content.some(f => filterObject[f.key] && filterObject[f.key].some(s => s.value == f.value))) {
a.push(c);
}
else {
c.content = c.content.filter(f => filterObject[f.key]
&& filterObject[f.key].some(s => s.value == f.value));
a.push(c);
}
} else if (c.content.some(s => filterObject[s.key]) && c.content) {
c.content = c.content.filter(
f =>
filterObject[f.key] && filterObject[f.key].some(s => s.value == f.value)
)
a.push(c)
}
return a
}, [])
console.log(result)
答案 1 :(得分：-1)
我尝试使用您在问题（permissionObj）中提供的对象
permissionObj.map(
p =>{
return {
label: Object.keys(p)[0], value:p[Object.keys(p)[0]][0].value
}
}
)
遍历键，并以第一个作为标签和它的第一个子数组对象来获取值
结果：
[{"label":"OA deal","value":"can_view"},{"label":"Deposit","value":"can_view"},{"label":"Journals","value":"can_create"},{"label":"Dashboard","value":"can_view"},{"label":"token","value":"can_create"}]
您需要的某些数据不在PermissionObj中，所以我无法添加它-因为我不知道那是哪里来