Javascript 需要帮助从深层数组中提取对象并合并到单个数组中吗

提取具有

target

属性的对象并将其组合到一个数组中的最佳方法是什么

app.js

const p1 = [[{ target1: 3 }, { target2: 1 }], [{ t: 2 }]];
const p2 = [[{ target1: 1 }, { target2: 2 }], [{ t: 2 }]];
const p3 = [[{ target1: 4 }, { target2: 1 }], [{ t: 2 }]];
const p4 = [[{ target1: 2 }, { target2: 2 }], [{ t: 2 }]];


Promise.all([p1, p2, p3, p4]).then((values) => {
    console.log(values);
    // Some way to extract the objects that have the target property in them 
    // and combine them into a single array
});

[ [ [ [Object], [Object] ], [ [Object] ] ],
  [ [ [Object], [Object] ], [ [Object] ] ],
  [ [ [Object], [Object] ], [ [Object] ] ],
  [ [ [Object], [Object] ], [ [Object] ] ] ]

结果

const p1 = [[{ target1: 3 }, { target2: 1 }], [{ t: 2 }]];
const p2 = [[{ target1: 1 }, { target2: 2 }], [{ t: 2 }]];
const p3 = [[{ target1: 4 }, { target2: 1 }], [{ t: 2 }]];
const p4 = [[{ target1: 2 }, { target2: 2 }], [{ t: 2 }]];


Promise.all([p1, p2, p3, p4]).then((values) => {
    console.log(values);
    // Some way to extract the objects that have the target property in them 
    // and combine them into a single array
});

[ [ [ [Object], [Object] ], [ [Object] ] ],
  [ [ [Object], [Object] ], [ [Object] ] ],
  [ [ [Object], [Object] ], [ [Object] ] ],
  [ [ [Object], [Object] ], [ [Object] ] ] ]

所需的

[{ target1: 3 }, { target2: 1 },
{ target1: 1 }, { target2: 2 },
{ target1: 4 }, { target2: 1 },
{ target1: 2 }, { target2: 2 }]

---

您可以使用下面的代码实现您的目标

const p1 = [[{ target1: 3 }, { target2: 1 }], [{ t: 2 }]];
const p2 = [[{ target1: 1 }, { target2: 2 }], [{ t: 2 }]];
const p3 = [[{ target1: 4 }, { target2: 1 }], [{ t: 2 }]];
const p4 = [[{ target1: 2 }, { target2: 2 }], [{ t: 2 }]];

Promise.all([p1, p2, p3, p4]).then((values) => {
   const [a1, a2, a3, a4] = values

   const [t1] = a1;
   const [t2] = a2;
   const [t3] = a3;
   const [t4] = a4;

   const result = [...t1, ...t2, ...t3, ...t4];

   console.log(result);
});

---

尝试在阵列上映射并使用解构：

let transformedValues = Array.prototype.concat( //poor man's one level flatten when combined with ...
  ...values.map(([[target1, target2]])=> [target1, target2])
)

---

我假设

p

值的数量不同，但是

p

值的结构是恒定的。

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