给出Gal和tradyblix建议的修改window.location.search的代码示例:

var qs = window.location.search || "?";
var param = key + "=" + value; // remember to URI encode your parameters
if (qs.length > 1) {
    // more than just the question mark, so append with ampersand
    qs = qs + "&";
}
qs = qs + param;
window.location.search = qs;

我想要的是:

使用浏览器的本地URL API 可以添加、更新、获取或删除吗 期望在散列之后的查询字符串，例如对于单页应用程序

function queryParam(options = {}) { var defaults = { method: 'set', url: window.location.href, key: undefined, value: undefined, } for (var prop in defaults) { options[prop] = typeof options[prop] !== 'undefined' ? options[prop] : defaults[prop] } const existing = (options.url.lastIndexOf('?') > options.url.lastIndexOf('#')) ? options.url.substr(options.url.lastIndexOf('?') + 1) : '' const query = new URLSearchParams(existing) if (options.method === 'set') { query.set(options.key, options.value) return `${options.url.replace(`?${existing}`, '')}?${query.toString()}` } else if (options.method === 'get') { const val = query.get(options.key) let result = val === null ? val : val.toString() return result } else if (options.method === 'delete') { query.delete(options.key) let result = `${options.url.replace(`?${existing}`, '')}?${query.toString()}` const lastChar = result.charAt(result.length - 1) if (lastChar === '?') { result = `${options.url.replace(`?${existing}`, '')}` } return result } } // Usage: let url = 'https://example.com/sandbox/#page/' url = queryParam({ url, method: 'set', key: 'my-first-param', value: 'me' }) console.log(url) url = queryParam({ url, method: 'set', key: 'my-second-param', value: 'you' }) console.log(url) url = queryParam({ url, method: 'set', key: 'my-second-param', value: 'whomever' }) console.log(url) url = queryParam({ url, method: 'delete', key: 'my-first-param' }) console.log(url) const mySecondParam = queryParam({ url, method: 'get', key: 'my-second-param', }) console.log(mySecondParam) url = queryParam({ url, method: 'delete', key: 'my-second-param' }) console.log(url)

感谢现代javascript, node.js和浏览器的支持，我们可以摆脱第三方库的漩涡(jquery, query-string等)和DRY自己。

下面是javascript(node.js)和typescript版本的函数，用于添加或更新给定url的查询参数:

Javascript

const getUriWithParam = (baseUrl, params) => { const Url = new URL(baseUrl); const urlParams = new URLSearchParams(Url.search); for (const key in params) { if (params[key] !== undefined) { urlParams.set(key, params[key]); } } Url.search = urlParams.toString(); return Url.toString(); }; console.info('expected: https://example.com/?foo=bar'); console.log(getUriWithParam("https://example.com", {foo: "bar"})); console.info('expected: https://example.com/slug?foo=bar#hash'); console.log(getUriWithParam("https://example.com/slug#hash", {foo: "bar"})); console.info('expected: https://example.com/?bar=baz&foo=bar'); console.log(getUriWithParam("https://example.com?bar=baz", {foo: "bar"})); console.info('expected: https://example.com/?foo=baz&bar=baz'); console.log(getUriWithParam("https://example.com?foo=bar&bar=baz", {foo: "baz"}));

打印稿

const getUriWithParam = (
  baseUrl: string,
  params: Record<string, any>
): string => {
  const Url = new URL(baseUrl);
  const urlParams: URLSearchParams = new URLSearchParams(Url.search);
  for (const key in params) {
    if (params[key] !== undefined) {
      urlParams.set(key, params[key]);
    }
  }
  Url.search = urlParams.toString();
  return Url.toString();
};

对于React Native

URL在React Native中没有实现。所以你必须事先安装react-native-url-polyfill。

对于对象参数

请看这个答案中的第二个解决方案

下面是我的库:https://github.com/Mikhus/jsurl

var u = new Url;
u.query.param='value'; // adds or replaces the param
alert(u)

我从这里得出的结论(与“使用严格”相容;并不真正使用jQuery):

function decodeURIParams(query) {
  if (query == null)
    query = window.location.search;
  if (query[0] == '?')
    query = query.substring(1);

  var params = query.split('&');
  var result = {};
  for (var i = 0; i < params.length; i++) {
    var param = params[i];
    var pos = param.indexOf('=');
    if (pos >= 0) {
        var key = decodeURIComponent(param.substring(0, pos));
        var val = decodeURIComponent(param.substring(pos + 1));
        result[key] = val;
    } else {
        var key = decodeURIComponent(param);
        result[key] = true;
    }
  }
  return result;
}

function encodeURIParams(params, addQuestionMark) {
  var pairs = [];
  for (var key in params) if (params.hasOwnProperty(key)) {
    var value = params[key];
    if (value != null) /* matches null and undefined */ {
      pairs.push(encodeURIComponent(key) + '=' + encodeURIComponent(value))
    }
  }
  if (pairs.length == 0)
    return '';
  return (addQuestionMark ? '?' : '') + pairs.join('&');
}

//// alternative to $.extend if not using jQuery:
// function mergeObjects(destination, source) {
//   for (var key in source) if (source.hasOwnProperty(key)) {
//     destination[key] = source[key];
//   }
//   return destination;
// }

function navigateWithURIParams(newParams) {
  window.location.search = encodeURIParams($.extend(decodeURIParams(), newParams), true);
}

使用示例:

// add/update parameters
navigateWithURIParams({ foo: 'bar', boz: 42 });

// remove parameter
navigateWithURIParams({ foo: null });

// submit the given form by adding/replacing URI parameters (with jQuery)
$('.filter-form').submit(function(e) {
  e.preventDefault();
  navigateWithURIParams(decodeURIParams($(this).serialize()));
});

在这个页面上有很多尴尬和不必要的复杂答案。评分最高的@amateur's相当不错，尽管它在RegExp中有一些不必要的错误。下面是一个稍微更优的解决方案，使用了更清洁的RegExp和更清洁的replace调用:

function updateQueryStringParamsNoHash(uri, key, value) {
  var re = new RegExp("([?&])" + key + "=[^&]*", "i");
  return re.test(uri)
       ? uri.replace(re, '$1' + key + "=" + value)
       : uri + separator + key + "=" + value
  ;
}

作为一个额外的好处，如果uri不是一个字符串，你不会得到错误，试图调用match或replace的东西可能没有实现这些方法。

如果你想处理散列的情况(并且你已经检查了正确格式化的HTML)，你可以利用现有的函数，而不是写一个包含相同逻辑的新函数:

function updateQueryStringParams(url, key, value) {
    var splitURL = url.split('#');
    var hash = splitURL[1];
    var uri = updateQueryStringParamsNoHash(splitURL[0]);
    return hash == null ? uri : uri + '#' + hash;
}

或者你可以对@Adam的精彩回答做一些小小的修改:

function updateQueryStringParameter(uri, key, value) {
  var re = new RegExp("([?&])" + key + "=[^&#]*", "i");
  if (re.test(uri)) {
    return uri.replace(re, '$1' + key + "=" + value);
  } else {
    var matchData = uri.match(/^([^#]*)(#.*)?$/);
    var separator = /\?/.test(uri) ? "&" : "?";    
    return matchData[0] + separator + key + "=" + value + (matchData[1] || '');
  }
}