Javascript过滤嵌套数组
我正在尝试在Angular应用程序中的对象数组中的嵌套数组上过滤。下面是一段组件代码-Javascript过滤嵌套数组,javascript,angular,typescript,Javascript,Angular,Typescript,我正在尝试在Angular应用程序中的对象数组中的嵌套数组上过滤。下面是一段组件代码- var teams = [ { name: 'Team1', members: [{ name: 'm1' }, { name: 'm2' }, { name: 'm3' }] }, { name: 'Team2', members: [{ name: 'm4' }, { name: 'm5' }, { name: 'm6' }] }, { name: 'Team3', members: [{
var teams = [
{ name: 'Team1', members: [{ name: 'm1' }, { name: 'm2' }, { name: 'm3' }] },
{ name: 'Team2', members: [{ name: 'm4' }, { name: 'm5' }, { name: 'm6' }] },
{ name: 'Team3', members: [{ name: 'm7' }, { name: 'm8' }, { name: 'm9' }] }
];
我试图实现的是,如果我搜索m5
,例如,我的结果应该是-
var teams = [
{ name: 'Team1', members: [] },
{ name: 'Team2', members: [{ name: 'm5' }] },
{ name: 'Team3', members: [] }
];
所以我有团队
和过滤团队
属性,在我的搜索功能中我正在做-
onSearchChange(event: any): void {
let value = event.target.value;
this.filteredTeams = this.teams.map(t => {
t.members = t.members.filter(d => d.name.toLowerCase().includes(value));
return t;
})
}
不过,这在某种程度上是可行的,因为我正在替换成员,这会在每次调用时破坏数组(如果这样做有意义的话)。我理解为什么会发生这种情况,但我的问题是实现此过滤器的最佳方法是什么?检查此项。而不是硬编码的
m5
传递您的值
const团队=[
{name:'Team1',成员:[{name:'m1'},{name:'m2'},{name:'m3'}]},
{name:'Team2',成员:[{name:'m4'},{name:'m5'},{name:'m6'}]},
{name:'Team3',成员:[{name:'m7'},{name:'m8'},{name:'m9'}]}
];
const filteredTeams=teams.map(team=>({name:team.name,members:team.members.filter(member=>member.name.includes('m5'))});
console.log(filteredTeams)代码>检查此项。而不是硬编码的m5
传递您的值
const团队=[
{name:'Team1',成员:[{name:'m1'},{name:'m2'},{name:'m3'}]},
{name:'Team2',成员:[{name:'m4'},{name:'m5'},{name:'m6'}]},
{name:'Team3',成员:[{name:'m7'},{name:'m8'},{name:'m9'}]}
];
const filteredTeams=teams.map(team=>({name:team.name,members:team.members.filter(member=>member.name.includes('m5'))});
console.log(filteredTeams)代码>您非常接近,您唯一做错的事情就是在团队中对源对象进行变异
基本上,您可以使用spread运算符生成一个新条目,然后返回一个包含新值的全新数组
const团队=[
{name:'Team1',成员:[{name:'m1'},{name:'m2'},{name:'m3'}]},
{name:'Team2',成员:[{name:'m4'},{name:'m5'},{name:'m6'}]},
{name:'Team3',成员:[{name:'m7'},{name:'m8'},{name:'m9'}]}
];
常量值='m5';
const result=this.teams.map(t=>{
const members=t.members.filter(d=>d.name.toLowerCase().includes(value));
返回{…t,成员};
})
console.log(result)
您非常接近,唯一做错的事情就是在团队中对源对象进行了变异
基本上,您可以使用spread运算符生成一个新条目,然后返回一个包含新值的全新数组
const团队=[
{name:'Team1',成员:[{name:'m1'},{name:'m2'},{name:'m3'}]},
{name:'Team2',成员:[{name:'m4'},{name:'m5'},{name:'m6'}]},
{name:'Team3',成员:[{name:'m7'},{name:'m8'},{name:'m9'}]}
];
常量值='m5';
const result=this.teams.map(t=>{
const members=t.members.filter(d=>d.name.toLowerCase().includes(value));
返回{…t,成员};
})
log(result)
您正在修改原始对象,但可以将新属性分配给result对象以进行映射
var-teams=[{name:'Team1',成员:[{name:'m1'},{name:'m2'},{name:'m3'}]},{name:'Team2',成员:[{name:'m4'},{name:'m5'},{name:'m6'},{name:'Team3',成员:[{name:'m7'},{name:'m8'},{name:'m9'}]}],
结果=teams.map(o=>Object.assign(
{},
啊,,
{members:o.members.filter(({name})=>name==='m5')}
));
控制台日志(结果);
控制台日志(团队)代码>
.as console wrapper{max height:100%!important;top:0;}
您正在修改原始对象,但可以将新属性分配给结果对象以进行映射
var-teams=[{name:'Team1',成员:[{name:'m1'},{name:'m2'},{name:'m3'}]},{name:'Team2',成员:[{name:'m4'},{name:'m5'},{name:'m6'},{name:'Team3',成员:[{name:'m7'},{name:'m8'},{name:'m9'}]}],
结果=teams.map(o=>Object.assign(
{},
啊,,
{members:o.members.filter(({name})=>name==='m5')}
));
控制台日志(结果);
控制台日志(团队)代码>
。作为控制台包装{max height:100%!important;top:0;}
首先尝试分离您的筛选函数:
const filterTeamMembers = (teams, filterArr) => {
const useFilter = filterArr.map(x => x.toLowerCase());
return teams.map(team => ({
...team,
members: team.members.filter(member => useFilter.includes(member.name))
}))
};
// =========== And then:
onSearchChange(event: any): void {
let value = event.target.value;
this.filteredTeams = filterTeamMembers(this.teams, [value]);
}
首先尝试分离过滤功能:
const filterTeamMembers = (teams, filterArr) => {
const useFilter = filterArr.map(x => x.toLowerCase());
return teams.map(team => ({
...team,
members: team.members.filter(member => useFilter.includes(member.name))
}))
};
// =========== And then:
onSearchChange(event: any): void {
let value = event.target.value;
this.filteredTeams = filterTeamMembers(this.teams, [value]);
}