用javascript我怎么能添加一个查询字符串参数的url,如果不存在或如果它存在,更新当前值?我使用jquery为我的客户端开发。


当前回答

这样做的目的是:

function updateQueryString(url, key, value) {
    var arr =  url.split("#");
    var url = arr[0];
    var fragmentId = arr[1];
    var updatedQS = "";
    if (url.indexOf("?") == -1) {
        updatedQS = encodeURIComponent(key) + "=" + encodeURIComponent(value);
    }
    else {
        updatedQS = addOrModifyQS(url.substring(url.indexOf("?") + 1), key, value); 
    }
    url = url.substring(0, url.indexOf("?")) + "?" + updatedQS;
    if (typeof fragmentId !== 'undefined') {
        url = url + "#" + fragmentId;
    }
    return url;
}

function addOrModifyQS(queryStrings, key, value) {
    var oldQueryStrings = queryStrings.split("&");
    var newQueryStrings = new Array();
    var isNewKey = true;
    for (var i in oldQueryStrings) {
        var currItem = oldQueryStrings[i];
        var searchKey = key + "=";
        if (currItem.indexOf(searchKey) != -1) {
            currItem = encodeURIComponent(key) + "=" + encodeURIComponent(value);
            isNewKey = false;
        }
        newQueryStrings.push(currItem);
    }
    if (isNewKey) {
        newQueryStrings.push(encodeURIComponent(key) + "=" + encodeURIComponent(value));
    }
    return newQueryStrings.join("&");
}   

其他回答

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()));
});

如果你想一次设置多个参数:

function updateQueryStringParameters(uri, params) {
    for(key in params){
      var value = params[key],
          re = new RegExp("([?&])" + key + "=.*?(&|$)", "i"),
          separator = uri.indexOf('?') !== -1 ? "&" : "?";
      if (uri.match(re)) {
        uri = uri.replace(re, '$1' + key + "=" + value + '$2');
      }
      else {
        uri = uri + separator + key + "=" + value;
      }
    }
    return uri;
}

与@amateur的函数相同

如果jslint给出了一个错误,在for循环之后添加这个

if(params.hasOwnProperty(key))

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] || '');
  }
}

下面是我对此略有不同的方法,作为练习写的

function addOrChangeParameters( url, params )
{
  let splitParams = {};
  let splitPath = (/(.*)[?](.*)/).exec(url);
  if ( splitPath && splitPath[2] )
    splitPath[2].split("&").forEach( k => { let d = k.split("="); splitParams[d[0]] = d[1]; } );
  let newParams = Object.assign( splitParams, params );
  let finalParams = Object.keys(newParams).map( (a) => a+"="+newParams[a] ).join("&");
  return splitPath ? (splitPath[1] + "?" + finalParams) : (url + "?" + finalParams);
}

用法:

const url = "http://testing.com/path?empty&value1=test&id=3";

addOrChangeParameters( url, {value1:1, empty:"empty", new:0} )

"http://testing.com/path?empty=empty&value1=1&id=3&new=0"

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