如何合并到对象并将空值替换为新值，如果新值为空则保留旧值

Question

我每次选择一个值时都有一个对象数组，我将其推入数组中，我想合并具有相同键（即键“code”）的两个对象，并删除空值

(4) [{…}, {…}, {…}, {…}]
0: {code: "abc", value: "", access: "A"}
1: {code: "abc", value: "ok", access: ""}
2: {code: "def", value: "xyz", access: ""}
3: {code: "def", value: "", access: "B"}

the output should be
(2) [{…},{…}]
0: { code: "abc", value: "ok", access: "A"}
1: {code: "def", value: "xyz", access: "B"}

我尝试过使用展开运算符，但它不起作用。请帮帮我

Answer 1

您可以使用 .reduce(..) 这是一个示例：

const input = [
  {code: "abc", value: "", access: "A"},
  {code: "abc", value: "ok", access: ""},
  {code: "def", value: "xyz", access: ""},
  {code: "def", value: "", access: "B"}
];

const output = input.reduce((a, c) => {
  const found = a.find(({ code }) => c.code === code);
  if (found) {
    Object.entries(c).forEach(([key, value]) => {
      if (!found[key]) {
        found[key] = value;
      }
    });
  } else {
    a.push(c);
  }
  return a;
}, []);

console.log(output);

即使这对于示例输入正确有效，但在某些情况下也会失败，因为它替换了所有 falsy 值（布尔值

false

数字

0

等），这可能不是您的意图。

为防止出现这种情况，请将

!found[key]

语句中的

if

条件替换为适合您需要的任何检查。