如何避免 JavaScript 中使用不同的、计算成本高昂的条件变量深层嵌套 if-else？

Question

我正在编写代码来找到合适的目标并对其进行后续操作。我当前的代码将首先尝试查找类型 1。如果它不存在，我将尝试查找类型 2，依此类推。因为 find 函数消耗 CPU 太多，所以我想避免一开始就使用它来查找所有类型。以下是我当前的代码。我怎样才能简化它们？

const target1 = find1(Type1);
if(target1){
    operate1(target1);
}else{

    const target2 = find2(Type2);
    if(target2){
        operate2(target2);
    }else{

        const target3 = find3(Type3);
        if(target3){
            operate3(target3);
        }else{
            ...
        }
    }
}        

Answer 1

尝试创建一个检查列表并循环遍历它们。 这将删除嵌套，但会添加循环。

const procedures = [
  procedure1,
  procedure2,
  procedure3,
  () => {
    console.log("nothing...");
  },
];

for (let i = 0; i < procedures.length; i++) {
  if (procedures[i]()) break; // stop if procedure is done
}

function procedure1() {
  const target1 = find1(Type1);

  if (target1) {
    operate1(target1);

    return true; // notify that procedure is done
  }
}

function procedure2() {
  const target2 = find2(Type2);

  if (target2) {
    const res = operate2(target2);

    if (res === null) return false; // continue with next procedure

    // notify that procedure is done
    return true;
  }
}

function procedure3() {
  const target3 = find2(Type3);

  if (target3) {
    const res = operate3(target3);

    return true;
  }
}

Answer 2

创建一个具有所有参数的对象。 如果找到则循环并中断。

const targets = [
    { type: Type1, find: find1, operate: operate1 },
    { type: Type2, find: find2, operate: operate2 },
    { type: Type3, find: find3, operate: operate3 },
    // Add more types as needed
];

let found = false;
for (const { type, find, operate } of targets) {
    const target = find(type);
    if (target) {
        operate(target);
        found = true;
        break; // Exit loop if target is found
    }
}

if (!found) {
    // Handle case when no target is found
    console.log("No target found");
}

Answer 3

您可以将此代码移至单独的函数并利用提前返回：

function findTarget()
{
    const target1 = find1(Type1);

    if(target1){
        operate1(target1);

        return;
    }

    const target2 = find2(Type2);

    if(target2){
        operate2(target2);

        return;
    }

    const target3 = find3(Type3);

    if(target3){
        operate3(target3);

        return;
    }

    ...
}

---

或者如果你的变量/函数确实是这样命名的，请使用简单的循环：

for (let i = 1; i <= possibleTypes; i++) {
    let target = window['find' + i](window['Type' + i])

    if (!target) {
        continue;
    }

    window['operate' + i](target);

    break;
}