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

加入@Vianney的回答https://stackoverflow.com/a/44160941/6609678

我们可以导入node中的内置URL模块,如下所示

const { URL } = require('url');

例子:

Terminal $ node
> const { URL } = require('url');
undefined
> let url = new URL('', 'http://localhost:1989/v3/orders');
undefined
> url.href
'http://localhost:1989/v3/orders'
> let fetchAll=true, timePeriod = 30, b2b=false;
undefined
> url.href
'http://localhost:1989/v3/orders'
>  url.searchParams.append('fetchAll', fetchAll);
undefined
>  url.searchParams.append('timePeriod', timePeriod);
undefined
>  url.searchParams.append('b2b', b2b);
undefined
> url.href
'http://localhost:1989/v3/orders?fetchAll=true&timePeriod=30&b2b=false'
> url.toString()
'http://localhost:1989/v3/orders?fetchAll=true&timePeriod=30&b2b=false'

有用的链接:

https://developer.mozilla.org/en-US/docs/Web/API/URL https://developer.mozilla.org/en/docs/Web/API/URLSearchParams

其他回答

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;
    }
}

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;
}

重置所有查询字符串

Var params = {params:"val1", params:"val2"}; 让str = jQuery.param(参数); let uri = window.location. reff . tostring (); if (uri.indexOf("?") > 0) Uri = Uri。substring (0, uri.indexOf(“?”); console.log (uri +”?”+ str); / / window.location。Href = uri+"?"+str; < script src = " https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js " > < /脚本>

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。

试试这个。

// uses the URL class
function setParam(key, value) {
            let url = new URL(window.document.location);
            let params = new URLSearchParams(url.search.slice(1));

            if (params.has(key)) {
                params.set(key, value);
            }else {
                params.append(key, value);
            }
        }