以有效的方式从字符串中删除特定单词

Question

我试图从一系列字符串中删除几个匹配的单词。 字符串的数组

let string = ["select from table order by asc limit 10 no binding"]

以下是我试图摆脱有订单和限制及其价值的所有 ，并保留其余的。我已经尝试了以下内容，但它们都不优雅/高效。

let splitString = string.split(' ');
let str1 = 'limit';
let str2 = 'order';
let str3 = 'by';
let str4 = 'asc';
let str5 = '10';

splitString = splitString.filter(x => x !== str1);
splitString = splitString.filter(x => x !== str2);
splitString = splitString.filter(x => x !== str3);
splitString = splitString.filter(x => x !== str4);
splitString = splitString.filter(x => x !== str5);

是否有适当的方法可以从字符串中摆脱这些单词？ tia

Answer 1

创建一个要删除的字符串的数组或集合，然后根据要迭代的单词是否在集合中进行过滤。

const input = ["select from table order by asc limit 10 no binding"]
const wordsToExclude = new Set(['limit', 'order', 'by', 'asc', '10']);
const words = input[0].split(' ').filter(word => !wordsToExclude.has(word));
console.log(words);

如果您实际上不想删除所有这些单词，而只想删除这样一个序列，请使用正则表达式来匹配limit，直到遇到数字。

const input = ["select from table order by asc limit 10 no binding"];
const result = input[0].replace(/order\b.*?\d+ /, '');
console.log(result);

如果序列之间可以出现其他数字，请在末尾匹配 limit \d+，而不仅仅是 \d+

const input = ["select from table order by 1 asc limit 33 no binding"];
const result = input[0].replace(/order\b.*?limit \d+ /, '');
console.log(result);

Answer 2

使用过滤器和包含

const [string] = ["select from table order by asc limit 10 no binding"];
const exclude = ["limit", "order", "by", "asc", "10"];

const res = string.split(" ").filter(wd => !exclude.includes(wd));

console.log(res)

或者使用 replaceAll

const [string] = ["select from table order by asc limit 10 no binding"];
const exclude = ['limit', 'order', 'by', 'asc', '10'];

const res = exclude.reduce((str, word) => 
  str.replaceAll(word, ''), string).replaceAll(/\s+/g, ' ');
  
console.log(res)