用对象数组中匹配的值替换部分字符串？

Question

我试图弄清楚如何从字符串中的对象中填充一些值，以便最终执行一些数学运算。我可以从括号之间获取数据和匹配值，但我被困在从哪里开始，或者是否有另一种更好的做法。这是我正在使用的示例：

var array = [{name: M, value: 45},{name: F, value: 23},{name: L, value: 9}]
var string = '( 2 * [M] ) * ( 5 / [F] ) * ( 100 + [L] )'

为了简化我的解释，我需要采用那个字符串公式，并将“M”的值插入“[M]”，将“F”的值插入“[F]”，依此类推。 ..

Answer 1

您可以使用nerdamer。它是用来做公式和东西的。

let vars = {M: 45, F: 23, L: 9};
let equation = '( 2 * M ) * ( 5 / F ) * ( 100 + L )';

console.log(
  equation,
  '=',
  nerdamer(
    equation,
    vars
  )
  .evaluate()
  .text()
);

<script src="https://cdn.jsdelivr.net/npm/nerdamer@latest/nerdamer.core.js"></script>

如果您想要更简单的方法，只需：

let vars = {M: 45, F: 23, L: 9};
let equation = '( 2 * [M] ) * ( 5 / [F] ) * ( 100 + [L] )';

function calculate(equation, vars)
{
  let newEquation = equation.replaceAll(/\[[A-Z]\]/gi, function(v)
  {
    return vars[v[1]];
  });
  
  return eval(newEquation);
}

console.log(calculate(equation, vars));

Answer 2

这是一种解决方案：

const array = [{name: 'M', value: 45},{name: 'F', value: 23},{name: 'L', value: 9}]
const string = "( 2 * [M] ) * ( 5 / [F] ) * ( 100 + [L] )";

const calculate = (array) => array.reduce((accumulator, currentValue)=>{ return accumulator.replace(`[${currentValue.name}]`, currentValue.value)}, string);

console.log('Result: ', eval(calculate(array)));

Answer 3

你可以试试这个。谢谢。

var variables = [{name: "M", value: 45}, {name: "F", value: 23}, {name: "L", value: 9}];
var equation = '( 2 * [M] ) * ( 5 / [F] ) * ( 100 + [L] )';

const getfinalEquation = (array, equationWithVars) => {
  let finalEquation = equationWithVars;
  
  array.forEach(item => {
    finalEquation = finalEquation.replaceAll(`[${item.name}]`, item.value);
  })
  
  return finalEquation;
}

console.log(getfinalEquation(variables, equation));