在一个使用AJAX调用的web应用程序中,我需要提交一个请求,但在URL的末尾添加一个参数,例如:

原始URL:

http://server/myapp.php?id=10

导致的网址:

http://server/myapp.php?id=10&enabled=true

寻找一个JavaScript函数,该函数解析URL并查看每个参数,然后添加新参数或更新已经存在的值。


当前回答

如果你在链接或其他地方弄乱了url,你可能也必须考虑哈希。这里有一个相当容易理解的解决方案。可能不是最快的,因为它使用正则表达式…但在99.999%的情况下,这种差异真的不重要!

function addQueryParam( url, key, val ){
    var parts = url.match(/([^?#]+)(\?[^#]*)?(\#.*)?/);
    var url = parts[1];
    var qs = parts[2] || '';
    var hash = parts[3] || '';

    if ( !qs ) {
        return url + '?' + key + '=' + encodeURIComponent( val ) + hash;
    } else {
        var qs_parts = qs.substr(1).split("&");
        var i;
        for (i=0;i<qs_parts.length;i++) {
            var qs_pair = qs_parts[i].split("=");
            if ( qs_pair[0] == key ){
                qs_parts[ i ] = key + '=' + encodeURIComponent( val );
                break;
            }
        }
        if ( i == qs_parts.length ){
            qs_parts.push( key + '=' + encodeURIComponent( val ) );
        }
        return url + '?' + qs_parts.join('&') + hash;
    }
}

其他回答

我添加我的解决方案,因为它支持相对url除了绝对url。在其他方面,它与顶部的答案相同,后者也使用Web API。

/**
 * updates a relative or absolute
 * by setting the search query with
 * the passed key and value.
 */
export const setQueryParam = (url, key, value) => {
  const dummyBaseUrl = 'https://dummy-base-url.com';
  const result = new URL(url, dummyBaseUrl);
  result.searchParams.set(key, value);
  return result.toString().replace(dummyBaseUrl, '');
};

还有人开玩笑说:

// some jest tests
describe('setQueryParams', () => {
  it('sets param on relative url with base path', () => {
    // act
    const actual = setQueryParam(
      '/', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('/?ref=some-value');
  });
  it('sets param on relative url with no path', () => {
    // act
    const actual = setQueryParam(
      '', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('/?ref=some-value');
  });
  it('sets param on relative url with some path', () => {
    // act
    const actual = setQueryParam(
      '/some-path', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('/some-path?ref=some-value');
  });
  it('overwrites existing param', () => {
    // act
    const actual = setQueryParam(
      '/?ref=prev-value', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('/?ref=some-value');
  });
  it('sets param while another param exists', () => {
    // act
    const actual = setQueryParam(
      '/?other-param=other-value', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('/?other-param=other-value&ref=some-value');
  });
  it('honors existing base url', () => {
    // act
    const actual = setQueryParam(
      'https://base.com', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('https://base.com/?ref=some-value');
  });
  it('honors existing base url with some path', () => {
    // act
    const actual = setQueryParam(
      'https://base.com/some-path', 'ref', 'some-value',
    );
    // assert
    expect(actual).toEqual('https://base.com/some-path?ref=some-value');
  });
});

感谢大家的贡献。我使用annakata代码并修改为也包括url中根本没有查询字符串的情况。 希望这能有所帮助。

function insertParam(key, value) {
        key = escape(key); value = escape(value);

        var kvp = document.location.search.substr(1).split('&');
        if (kvp == '') {
            document.location.search = '?' + key + '=' + value;
        }
        else {

            var i = kvp.length; var x; while (i--) {
                x = kvp[i].split('=');

                if (x[0] == key) {
                    x[1] = value;
                    kvp[i] = x.join('=');
                    break;
                }
            }

            if (i < 0) { kvp[kvp.length] = [key, value].join('='); }

            //this will reload the page, it's likely better to store this until finished
            document.location.search = kvp.join('&');
        }
    }

我有一个'类',这是:

function QS(){
    this.qs = {};
    var s = location.search.replace( /^\?|#.*$/g, '' );
    if( s ) {
        var qsParts = s.split('&');
        var i, nv;
        for (i = 0; i < qsParts.length; i++) {
            nv = qsParts[i].split('=');
            this.qs[nv[0]] = nv[1];
        }
    }
}

QS.prototype.add = function( name, value ) {
    if( arguments.length == 1 && arguments[0].constructor == Object ) {
        this.addMany( arguments[0] );
        return;
    }
    this.qs[name] = value;
}

QS.prototype.addMany = function( newValues ) {
    for( nv in newValues ) {
        this.qs[nv] = newValues[nv];
    }
}

QS.prototype.remove = function( name ) {
    if( arguments.length == 1 && arguments[0].constructor == Array ) {
        this.removeMany( arguments[0] );
        return;
    }
    delete this.qs[name];
}

QS.prototype.removeMany = function( deleteNames ) {
    var i;
    for( i = 0; i < deleteNames.length; i++ ) {
        delete this.qs[deleteNames[i]];
    }
}

QS.prototype.getQueryString = function() {
    var nv, q = [];
    for( nv in this.qs ) {
        q[q.length] = nv+'='+this.qs[nv];
    }
    return q.join( '&' );
}

QS.prototype.toString = QS.prototype.getQueryString;

//examples
//instantiation
var qs = new QS;
alert( qs );

//add a sinle name/value
qs.add( 'new', 'true' );
alert( qs );

//add multiple key/values
qs.add( { x: 'X', y: 'Y' } );
alert( qs );

//remove single key
qs.remove( 'new' )
alert( qs );

//remove multiple keys
qs.remove( ['x', 'bogus'] )
alert( qs );

我已经重写了toString方法,所以不需要调用QS::getQueryString,你可以使用QS::toString,或者像我在示例中所做的那样,仅仅依赖于对象被强制转换为字符串。

这是一个添加查询参数的简单方法:

const query = new URLSearchParams(window.location.search);
query.append("enabled", "true");

这就是这里更多的内容。

请注意支持规格。

你需要适应的基本实现是这样的:

function insertParam(key, value) {
    key = encodeURIComponent(key);
    value = encodeURIComponent(value);

    // kvp looks like ['key1=value1', 'key2=value2', ...]
    var kvp = document.location.search.substr(1).split('&');
    let i=0;

    for(; i<kvp.length; i++){
        if (kvp[i].startsWith(key + '=')) {
            let pair = kvp[i].split('=');
            pair[1] = value;
            kvp[i] = pair.join('=');
            break;
        }
    }

    if(i >= kvp.length){
        kvp[kvp.length] = [key,value].join('=');
    }

    // can return this or...
    let params = kvp.join('&');

    // reload page with new params
    document.location.search = params;
}

这大约是正则表达式或基于搜索的解决方案的两倍,但这完全取决于查询字符串的长度和任何匹配的索引


为了完成起见,我以慢速regex方法为基准(大约慢了150%)

function insertParam2(key,value)
{
    key = encodeURIComponent(key); value = encodeURIComponent(value);

    var s = document.location.search;
    var kvp = key+"="+value;

    var r = new RegExp("(&|\\?)"+key+"=[^\&]*");

    s = s.replace(r,"$1"+kvp);

    if(!RegExp.$1) {s += (s.length>0 ? '&' : '?') + kvp;};

    //again, do what you will here
    document.location.search = s;
}