JavaScript根據另一個對象數組的屬性過濾對象數組

Question

我有2 array of objects

Tasks
[{
 name: "test",
 type: "one"
},
{
 name: "test2",
 type: 'two'
}]

活動

[{
 name: "test",
 type: "one"
},
{
 name: "test2",
 type: 'two'
},
{
 name: "test3",
 type: "three"
 }]

如何通過任務數組的屬性過濾活動數組？ 我想要這個結果。

活動

[{
 name: "test",
 type: "one"
},
{
 name: "test2",
 type: 'two'
}]

我嘗試使用.filter（），但是返回一個空集合。

let tests = activities.filter(x => tasks.includes(x.type));

如何在Javascript中實現？

Answer 1

filter是正確的工具，但includes不是。 tasks中的條目是對象，而不是字符串。 它們都不匹配tests中的任何類型。

相反，簡單版本使用some來查看tasks中是否有任何條目與被測試活動的類型匹配：

let tests = activities.filter(a => tasks.some(t => t.type == a.type));

現場示例：

 const tasks = [ { name: "test", type: "one" }, { name: "test2", type: 'two' } ]; const activities = [ { name: "test", type: "one" }, { name: "test2", type: 'two' }, { name: "test3", type: "three" } ]; let tests = activities.filter(a => tasks.some(t => t.type == a.type)); console.log(tests); 

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

自然地，這意味着對於activities每個條目至少部分地重新遍歷tasks 。 對於您顯示的示例，這絕對很好； 如果tasks是成千上萬個條目，或者是在一個緊密的循環中完成的，那么就不那么多了。 :-)

在這種情況下，您可以預先為自己提供一組已知類型的tasks ，然后僅對該組進行測試：

const knownTypes = new Set();
for (const task of tasks) {
    knownTypes.add(task.type);
}
let tests = activities.filter(a => knownTypes.has(a.type));

現場示例：

 const tasks = [ { name: "test", type: "one" }, { name: "test2", type: 'two' } ]; const activities = [ { name: "test", type: "one" }, { name: "test2", type: 'two' }, { name: "test3", type: "three" } ]; const knownTypes = new Set(); for (const task of tasks) { knownTypes.add(task.type); } let tests = activities.filter(a => knownTypes.has(a.type)); console.log(tests); 

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

或者，除了Set您還可以使用一個對象用於knownTypes ，通過Object.create(null)在可能的情況下通過Object.create(null)創建它，您的任何類型恰好與Object.prototype上的屬性Object.prototype ：

const knownTypes = Object.create(null);
for (const task of tasks) {
    knownTypes[task.type] = true;
}
let tests = activities.filter(a => knownTypes[a.type]);

現場示例：

 const tasks = [ { name: "test", type: "one" }, { name: "test2", type: 'two' } ]; const activities = [ { name: "test", type: "one" }, { name: "test2", type: 'two' }, { name: "test3", type: "three" } ]; const knownTypes = Object.create(null); for (const task of tasks) { knownTypes[task.type] = true; } let tests = activities.filter(a => knownTypes[a.type]); console.log(tests); 

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

Answer 2

快好了。 嘗試使用.some數組方法檢查該項目是否存在，因為includes對於嵌套屬性不起作用。

 const tasks = [{ name: "test", type: "one" }, { name: "test2", type: 'two' }] const activities = [{ name: "test", type: "one" }, { name: "test2", type: 'two' }, { name: "test3", type: "three" }] let tests = activities.filter(x => tasks.some(t => t.type === x.type)); console.log(tests)