你的函数不能修改addressLine，因为JavaScript没有引用调用（而且字符串是不可变的）。如果你想修改那个字符串的值，你也可以返回它。
因此，返回addressLine的匹配值和新值，您可以将其作为一个对（数组）返回，调用者可以将该对重新构造为单独的变量。
注意：不需要await，因为不涉及异步API
更正：

function ok (){
    var arr = ['one', 'two', 'three'];
    var addressLine = 'This is one'

    // Get two returns by destructuring
    const [newAddressLine, ans] = matchAndRemove(arr, addressLine)

    console.log("newAddressLine:", newAddressLine) // 'This is'
    console.log("ans:", ans) // 'one'
}

const matchAndRemove = (arr, addressLine) => {
    for (var val of arr) {
        if (addressLine.toLowerCase().indexOf(val.toLowerCase()) != -1) {
            const match = addressLine.match(new RegExp(val, 'ig')).join(' ');
            addressLine = addressLine.replace(new RegExp(val, 'ig'), '');
            return [addressLine, match]; // Return both as a pair
        }
    }
}

ok()

这不是您的问题，但使用 * one * 正则表达式可能更有趣，而不是需要逐个检查的单词数组：

const arr = ['one', 'two', 'three'];
const regex = RegExp("\\b(" + arr.join("|") + ")\\b\\s*", "gi"); // One regex for all

var addressLine = 'This is one';

// One replace call will replace all words in arr
const result = addressLine.replace(regex, "").trim();
console.log("result:", result) // 'This is'

// If you wanted to know the matches:
const matched = addressLine.match(regex);
console.log("matched:", matched);

您的代码在for循环中返回match的值。这就是为什么找到第一个匹配项后函数会关闭的原因。您应该将匹配项存储在数组中，然后将它们连接起来以获得最终结果。请尝试以下操作：

async function ok() {
  var arr = ['one', 'two', 'three'];
  var addressLine = 'This is one';

  const ans = await matchAndRemove(arr, addressLine);

  console.log(`This is address ${addressLine}`); // Output is 'This is'
  console.log(`This is match ${ans}`); // Output is 'This is match one'
}

const matchAndRemove = (arr, addressLine) => {
  let matches = [];

  for (var val of arr) {
    if (addressLine.toLowerCase().indexOf(val.toLowerCase()) != -1) {
      matches.push(val);
      addressLine = addressLine.replace(new RegExp(val, 'ig'), '');
    }
  }

  return matches.join(' ');
};

ok();