将对象数组中两个属性的唯一值存储到单个数组中

Question

如何获取requesterPractitionerId和performerOrganizationId的唯一值。 我需要在一个数组中使用requesterPractitionerId和performerOrganizationId的唯一值。

[ 
    { 
        id: '1043120',
        requesterPractitionerId: '1043119',
    },
    { 
        id: '1043081',
        requesterPractitionerId: '1043080'
    },
    { 
        id: 'e1dceebe-c5ba-46a5-a63a-bff709896af4',
        requesterPractitionerId: 'e0a844e4-6c8a-489a-8bd6-1d62267d311e',
        performerOrganizationId: '05D0889009',
    },
    { 
        id: '2709842f-41e3-4193-8607-fc34d3d24ec1',
        requesterPractitionerId: 'e0a844e4-6c8a-489a-8bd6-1d62267d311e',
        performerOrganizationId: '05D0889009'
    } 
]

预期产出：

1043119
1043080
e0a844e4-6c8a-489a-8bd6-1d62267d311e
05D0889009

我是Javascript的新手，并在几个小时内苦苦挣扎。 任何帮助都会得到很好的回应。

Answer 1

使用Set ， Array.reduce和Array.from

• 使用reduce创建具有所述字段的唯一值的集合
• 然后使用Array.from将集转换为数组

 let arr = [{id:'1043120',requesterPractitionerId:'1043119'},{id:'1043081',requesterPractitionerId:'1043080'},{id:'e1dceebe-c5ba-46a5-a63a-bff709896af4',requesterPractitionerId:'e0a844e4-6c8a-489a-8bd6-1d62267d311e',performerOrganizationId:'05D0889009'},{id:'2709842f-41e3-4193-8607-fc34d3d24ec1',requesterPractitionerId:'e0a844e4-6c8a-489a-8bd6-1d62267d311e',performerOrganizationId:'05D0889009'}]; let result = Array.from(arr.reduce((a,c) => { if(c.requesterPractitionerId) a.add(c.requesterPractitionerId); if(c.performerOrganizationId) a.add(c.performerOrganizationId); return a; }, new Set())); console.log(result); 

Answer 2

一种解决方案是使用Array.reduce（）从您需要的属性生成新的Set 。 之后，您可以将集合的项目spread到数组中。

 const input = [{id:'1043120',requesterPractitionerId:'1043119'},{id:'1043081',requesterPractitionerId:'1043080'},{id:'e1dceebe-c5ba-46a5-a63a-bff709896af4',requesterPractitionerId:'e0a844e4-6c8a-489a-8bd6-1d62267d311e',performerOrganizationId:'05D0889009',},{id:'2709842f-41e3-4193-8607-fc34d3d24ec1',requesterPractitionerId:'e0a844e4-6c8a-489a-8bd6-1d62267d311e',performerOrganizationId:'05D0889009'}]; let res = input.reduce((acc, o) => { if (o.hasOwnProperty("requesterPractitionerId")) acc.add(o.requesterPractitionerId); if (o.hasOwnProperty("performerOrganizationId")) acc.add(o.performerOrganizationId); return acc; }, new Set()) console.log([...res]); 

 .as-console {background-color:black !important; color:lime;} .as-console-wrapper {max-height:100% !important; top:0;} 

Answer 3

您可以使用forEach迭代data ，然后使用Set删除重复项

 let data = [{ id: '1043120', requesterPractitionerId: '1043119', }, { id: '1043081', requesterPractitionerId: '1043080' }, { id: 'e1dceebe-c5ba-46a5-a63a-bff709896af4', requesterPractitionerId: 'e0a844e4-6c8a-489a-8bd6-1d62267d311e', performerOrganizationId: '05D0889009', }, { id: '2709842f-41e3-4193-8607-fc34d3d24ec1', requesterPractitionerId: 'e0a844e4-6c8a-489a-8bd6-1d62267d311e', performerOrganizationId: '05D0889009' } ] let newArry = []; data.forEach(function(item) { newArry.push(item.requesterPractitionerId); if (item.hasOwnProperty('performerOrganizationId')) { newArry.push(item.performerOrganizationId) } }) let k = Array.from(new Set(newArry)); console.log(k) 

Answer 4

真的很简单 - Set和map ， reduce和filter ：

 const arr = [{id:'1043120',requesterPractitionerId:'1043119',},{id:'1043081',requesterPractitionerId:'1043080'},{id:'e1dceebe-c5ba-46a5-a63a-bff709896af4',requesterPractitionerId:'e0a844e4-6c8a-489a-8bd6-1d62267d311e',performerOrganizationId:'05D0889009',},{id:'2709842f-41e3-4193-8607-fc34d3d24ec1',requesterPractitionerId:'e0a844e4-6c8a-489a-8bd6-1d62267d311e',performerOrganizationId:'05D0889009'}]; const res = [...new Set(arr.map(({ requesterPractitionerId, performerOrganizationId }) => [requesterPractitionerId, performerOrganizationId]).reduce((acc, curr) => acc.concat(curr)).filter(Boolean))]; console.log(res); 

 .as-console-wrapper { max-height: 100% !important; top: auto; }