此函数的圈复杂度为 11，大于 10 授权

Question

if ((['ALL', ''].includes(this.accountnumber.value) ? true : ele.accountnumber === this.accountnumber.value) &&
        (['ALL', ''].includes(this.description.value) ? true : ele.description === this.description.value) &&
        (['ALL', ''].includes(this.country.value) ? true : ele.country === this.country.value) &&
        (!this.entryDate ? true : (this.entryDate === dateEntry)) && 
        (!this.editedDate ? true : (this.editedDate === dateEdited))) {
        return true;
      }

Answer 1

有时最好编写更多代码以便更好地理解正在发生的事情。

第一个快速更改是扩展您的 if 内容，例如：

请注意，三进制不是强制性的。一个简单的 || 运算符将具有相同的效果。

PS：因为我没有考虑过你为什么做你正在做的事情，所以我使用了错误的变量命名。如果您重复使用以下内容，请更改它们并添加有关您要实现的目标的评论。

const isA = [
 'ALL',
 '',
].includes(this.accountnumber.value) || ele.accountnumber === this.accountnumber.value;

const isB = [
  'ALL',
  '',
].includes(this.description.value) || ele.description === this.description.value;

const isC = [
  'ALL',
  '',
 ].includes(this.country.value) || ele.country === this.country.value;

const isD = !this.entryDate || this.entryDate === dateEntry;

const isE = !this.editedDate || this.editedDate === dateEdited;

if (isA && isB && isC && isD && isE) {
  return true;
}

扩展还使某些代码重复变得更加明显。

function checkCond(key, obj, arr = [
  'ALL',
  '',
]) {
  return arr.includes(this[key].value) || obj[key] === this[key].value; 
}

const isA = checkCond('accountnumber', ele);
const isB = checkCond('description', ele);
const isB = checkCond('country', ele);

const isD = !this.entryDate || this.entryDate === dateEntry;

const isE = !this.editedDate || this.editedDate === dateEdited;

if (isA && isB && isC && isD && isE) {
  return true;
}

更进一步：

function checkCond(key, obj, arr = [
  'ALL',
  '',
]) {
  return arr.includes(this[key].value) || obj[key] === this[key].value; 
}

const conditions = [
  checkCond('accountnumber', ele),
  checkCond('description', ele),
  checkCond('country', ele),

  !this.entryDate || this.entryDate === dateEntry,
  !this.editedDate || this.editedDate === dateEdited,
];

if (conditions.every(x => x)) {
  return true;
}

此函数的圈复杂度为 11，大于 10 授权

The Cyclomatic Complexity of this function is 11 which is greater than 10 authorized

typescript

angular6