我如何从 "complex" 条件与 && 或 || 的组合中编写一个干净的条件

Question

比如a有两个语句条件：

首先，类似于

const condition1 = value === '' && value1 === '' && value2 === '' && valu3e === '' && value4 === '' && value5 === ''

第二个条件更简单像

const condition2 = value === 'error'

我的 if 语句有两个这样的：

if (condition1 || condition2) {
....
}

有什么方法可以使它比实际实施更好？

Answer 1

第一个条件：

const value = '',
  value1 = '',
  value2 = '',
  value3 = '',
  value4 = '',
  value5 = '';

const condition1 = [
  value,
  value1,
  value2,
  value3,
  value4,
  value5,
].every(x => x === '');

console.log(`Condition1 : ${condition1}`);

如果你能将这些值重构为一个数组，那就更好了。

const values = [
  '',
  '',
  '',
  '',
  '',
  '',
];

const condition1 = values.every(x => x === '');

console.log(`Condition1 : ${condition1}`);

一共

const value = '',
  value1 = '',
  value2 = '',
  value3 = '',
  value4 = '',
  value5 = '';
  
if ([
  value,
  value1,
  value2,
  value3,
  value4,
  value5,
].every(x => x === '') || value === 'error') {
  console.log(`Condition : true`);
} else {
  console.log(`Condition : false`);
}

Answer 2

为什么不简单:

if ( (value === '' && value1 === '' && value2 === '' && valu3e === '') 
   || value === 'error' 
   ) {
  /// ...
}

?

我如何从 "complex" 条件与 && 或 || 的组合中编写一个干净的条件

How I write a clean condition from a "complex" combinations of conditions with && or ||

javascript

refactoring

coding-style

conditional-statements

typescript