Javascript：如何在条件语句中使用运算符**

我想在字符串中找到一个运算符，如下面的

“3+7”

，识别该运算符并进行求和。对于运算符

+

-

*

/

，我有一个和平的代码，但对于

**

，我的代码返回布尔值

我想知道为什么会发生这种情况，以及如何使用具有条件的指数运算符

// To know if the string have a "+", "-", "*", "**" or "/" inside of him
let foundCharacterFunction = function(str, char) {
 for(var i = 0; i < str.length; i++) {
   if (str[i] == char) {
     return char;
   } 
 }
 return false;
};


// With this we can use the specific character founded
let characterFoundAlready = foundCharacterFunction("3+7", "+");


// "characterFoundAlready" always is a string but work with all, less with exponential operator
console.log(typeof characterFoundAlready);


// To know what we going to do if we have one character or other
let enterTheCondition = 
 (characterFoundAlready === "+" ) ? console.log("We going to sum!") :
 (characterFoundAlready === "-" ) ? console.log("We going to rest!") :
 (characterFoundAlready === "*" ) ? console.log("We going to multiply!") :
 (characterFoundAlready === "**" ) ? console.log("We going to power!!") :
 (characterFoundAlready === "/" ) ? console.log("We going to divide!") :
 console.log("That character it can't be procesed");

//知道字符串的内部是否有“+”、“-”、“*”、“***”或“/”
让foundCharacterFunction=函数（str，char）{
对于（变量i=0；i

提前感谢您的时间和关注。

我建议您阅读 Algo操作命令，而不是字符，并允许处理不同优先级的操作。你也应该这样

编辑：

下面是一些小示例，说明如何处理它：

const opertaionDescriptions = [
  { operation: '+', message: 'We going to sum!' },
  { operation: '-', message: 'We going to rest!' },
  { operation: '*', message: 'We going to multiply!' },
  { operation: '**', message: 'We going to power!' },
  { operation: '/', message: 'We going to divide!' },
];

let foundCharacterFunction = function (str) {
  const found = str.match(/[\d\s]([+*/-]+)[\s\d]/i);
  if (found && found[1]) {
    return found[1];
  }

  return false;
};

// With this we can use the specific character founded
let characterFoundAlready = foundCharacterFunction("3+7");

// To know what we going to do if we have one character or other
console.log(
  (opertaionDescriptions.find(({operation}) => operation === characterFoundAlready) 
    || { message: "That character it can't be procesed"})
  .message
); 

// With this we can use the specific character founded
characterFoundAlready = foundCharacterFunction("3 ** 7");

// To know what we going to do if we have one character or other
console.log(
  (opertaionDescriptions.find(({operation}) => operation === characterFoundAlready) 
    || { message: "That character it can't be procesed"})
  .message
); 

---

当然，您仍然可以通过手动创建Regexp来定义精确的搜索操作，例如，smth

/[\d\s]（char\u或string\u to\u search）[\s\d]/i

您希望一个字符变成两个字符的字符串。这是不可能的。

if（str[i]==char）

一次检查一个字符。如果要检查两个字符的标记（如

***

），则需要输入字符串。建议：考虑使用使用字符串不char，我在回答中添加了一个简单的例子，问题不是“中缀”与“RPN”。问题是OP的“解析”方法对于中缀表达式恰好有效，其中操作数是单个字符，如果操作数恰好是多个字符，则会失败。解决方案是更改他标识操作数的方式（即“以不同方式解析”）