I want to sort the above given array so that the output array will be as given in output section. I have tried some code which is given below. I am using javascript for sorting. In angular I am using this To display menu according to user role. I have googled a lot but not getting solution
this.items = [
{
label: 'Home', routerLink: ['Home']
},
{
label: 'menu1',
items: [
{
label: 'submenu1',
routerLink: '/submenu1'
},
{
label: 'submenu2'
, routerLink: '/submenu2'
},
{
label: 'submenu3',
routerLink: ['/submenu3']
}
]
},
{
label: 'menu2',
items: [
{
label: 'submenu5',
routerLink: ['/submenu5']
},
]
},
{
label: 'menu3',
items: [
{
label: 'submenu6',
routerLink: ['/submenu6'],
}
]
},
];
output:
this.items = [
{
label: 'Home', routerLink: ['Home']
},
{
label: 'menu1',
items: [
{
label: 'submenu1',
routerLink: '/submenu1'
}
]
},
{
label: 'menu3',
items: [
{
label: 'submenu6',
routerLink: ['/submenu6'],
}
]
},
];
code for sorting:
let filterArr = this.filteredArray
.filter(x => x.label == "Home" && x.label == "menu3")
.map(y => y.items.filter(z => z.label == 'submenu6'));
You could move the wanted menu and submenu labels into arrays and filter the objects by creating new object with filtered menus
.
This approach does not mutate the data.
It uses Array#flatMap
for the outer array and Array#filter
for getting the wanted parts of the nested array.
If the nested array does not have any item, then take the original object.
var items = [{ label: 'Home', routerLink: ['Home'] }, { label: 'menu1', items: [{ label: 'submenu1', routerLink: '/submenu1' }, { label: 'submenu2', routerLink: '/submenu2' }, { label: 'submenu3', routerLink: ['/submenu3'] }] }, { label: 'menu2', items: [{ label: 'submenu5', routerLink: ['/submenu5'] }] }, { label: 'menu3', items: [{ label: 'submenu6', routerLink: ['/submenu6'] }] }], menu = ['Home', 'menu1', 'menu3'], submenu = ['submenu1', 'submenu6'], result = items.flatMap(o => { if (!menu.includes(o.label)) return []; var items = (o.items || []).filter(({ label }) => submenu.includes(label)); return items.length ? { ...o, items } : o; }); console.log(result);
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