javascript - 返回这些字符之一之前的字符串

Question

我想在这些元素之一之前返回一个字符串。如果字符串仍然超过 80 个字符，我将返回它们直到 80 个字符。

"Hello! my name is Marcel. I am 30 years old -" => "Hello!"

"Hello, my name is Marcel. I am 30 years old," => "Hello, my name is Marcel."

"Hello my name is Marcel <br/> I am 30 years old -" => "Hello my name is Marcel"

这是我尝试过的，我觉得必须有更简单的东西:

 getStrWithMinOfElements(str){
              var arr=[str.indexOf('.'), str.indexOf('!'), str.indexOf('<br'), str.indexOf('-')];
              var index = arr.indexOf(-1);
              if (index > -1) {
                arr.splice(index, 1);
              }
              index = arr.indexOf(-1);
              if (index > -1) {
                arr.splice(index, 1);
              }
              index = arr.indexOf(-1);
              if (index > -1) {
                arr.splice(index, 1);
              }

              var min = Math.min(...arr);
              if(min<8 || min>80) return str.substring(0,80);
              return str.substring(0,min);
            }

Answer 1

您可以匹配要分隔的字符的第一个实例，并将其后面的任何内容替换为空字符串，并从替换的字符串中获取长度为 80 的子字符串

let strs = ["Hello! my name is Marcel. I am 30 years old -","Hello, my name is Marcel. I am 30 years old,","Hello my name is Marcel<br/> I am 30 years old -"]

strs.forEach(str=>{
  let strReplaced = str.replace(/(\.|!|<br\/>).*$/,"$1")
  let final = strReplaced.substring(0,80)
  console.log(final)
})