请问如何简化这段 if else

Question

if (type == 1) {
    if (state == 0) {
      this.aaa(activeId)
    } else if (state == 1) {
      this.qqq(type)
      this.clickNum = 1
    }
  } else if (type == 2) {
    if (state == 0) {
      this.bbb(hasFollow, shopId)
    } else if (state == 1 && hasFollow == true) {
      this.qqq(type)
      this.clickNum = 2
    }
  } else if (type == 3) {
    if (state == 0 || state == 1) {
      this.ccc()
    }
  } else if (type == 4) {
    if (state == 0) {
      this.ddd(skuId)
    } else if (state == 1) {
      this.qqq(type)
      this.clickNum = 4
    }
  } else if (type == 5) {
    if (state == 0) {
      this.$emit('shareId')
    }
  }

Answer 1

let a = 'this.qqq(type);this.clickNum = type'  
switch (type) {  
  case 1:  
    (state == 0 && this.aaa(activeId)) || (state == 1 && eval(a))  
    break  
 case 2:  
    (state == 0 && this.bbb(hasFollow, shopId)) || (state == 1 && hasFollow == true && eval(a))  
    break  
 case 3:  
    (state == 0 || state == 1) && this.ccc()  
    break  
 case 4:  
    (state == 0 && this.ddd(skuId)) || (state == 1 && eval(a))  
    break  
 case 5:  
    state == 0 && this.emit('shareId')  
}

Answer 2

switch (type) {
  case 1:
    m(state, {
      '0': {
        callback: this.aaa,
        par: [activeId]
      },
      '1': {
        callback: n,
        par: [type]
      }
    })
    break;
  case 2:
    if (state == 0) {
      this.bbb(hasFollow, shopId);
      break;
    }
    if (state == 1 && hasFollow) {
      this.qqq(type)
      this.clickNum = type
      break;
    }
    break;
  case 4:
    m(state, {
      '0': {
        callback: this.ddd,
        par: [skuId]
      },
      '1': {
        callback: n,
        par: [type]
      }
    })
    break;
  case 3:
    m(state, {
      '0': {
        callback: this.ccc,
        par: []
      },
      '1': {
        callback: this.ccc,
        par: []
      }
    })
    break;
  case 5:
    m(state, {
      '0': {
        callback: this.$emit,
        par: ['shareId']
      }
    })
    break;
  default:
    break;
}

function m(state, callbackObj) {
  if (state in callbackObj) {
    let Obj = callbackObj[state]
    Obj.callback(...Obj.par)
  }
}

function n(type) {
  this.qqq(type);
  this.clickNum = type;
}

Answer 3

不需要优化，这是人最易读的代码。简化代码的后果就是，后期看起来费劲

Answer 4

代码是给人看的，不是给机器看的，不需要多短多精炼，首要准则是便于快速阅读。
对你代码的唯一意见：else 看上去没有必要，直接用多个 if 即可。else if 相比于 if 更难阅读，因为阅读者需要去配合前面的 if 理解上下语境。