我的提供:

public static Uri AddQuery(this Uri uri, string name, string value)
{
    // this actually returns HttpValueCollection : NameValueCollection
    // which uses unicode compliant encoding on ToString()
    var query = HttpUtility.ParseQueryString(uri.Query);

    query.Add(name, value);

    var uriBuilder = new UriBuilder(uri)
    {
        Query = query.ToString()
    };

    return uriBuilder.Uri;
}

用法:

var uri = new Uri("http://stackoverflow.com").AddQuery("such", "method")
                                             .AddQuery("wow", "soFluent");

// http://stackoverflow.com?such=method&wow=soFluent

我有一个扩展方法的Uri:

接受匿名对象:uri。WithQuery(new {name = "value"}) 接受字符串/字符串对的集合(例如Dictionary ' 2)。 接受字符串/对象对的集合(例如RouteValueDictionary)。 接受NameValueCollections。 按键对查询值进行排序，以便相同的值产生相同的uri。 每个键支持多个值，保持其原始顺序。

文档版本可以在这里找到。

扩展:

public static Uri WithQuery(this Uri uri, object values)
{
    if (uri == null)
        throw new ArgumentNullException(nameof(uri));

    if (values != null)
    {
        var query = string.Join(
            "&", from p in ParseQueryValues(values)
                 where !string.IsNullOrWhiteSpace(p.Key)
                 let k = HttpUtility.UrlEncode(p.Key.Trim())
                 let v = HttpUtility.UrlEncode(p.Value)
                 orderby k
                 select string.IsNullOrEmpty(v) ? k : $"{k}={v}");

        if (query.Length != 0 || uri.Query.Length != 0)
            uri = new UriBuilder(uri) { Query = query }.Uri;
    }

    return uri;
}

查询解析器:

private static IEnumerable<KeyValuePair<string, string>> ParseQueryValues(object values)
{
    // Check if a name/value collection.
    var nvc = values as NameValueCollection;
    if (nvc != null)
        return from key in nvc.AllKeys
               from val in nvc.GetValues(key)
               select new KeyValuePair<string, string>(key, val);

    // Check if a string/string dictionary.
    var ssd = values as IEnumerable<KeyValuePair<string, string>>;
    if (ssd != null)
        return ssd;

    // Check if a string/object dictionary.
    var sod = values as IEnumerable<KeyValuePair<string, object>>;
    if (sod == null)
    {
        // Check if a non-generic dictionary.
        var ngd = values as IDictionary;
        if (ngd != null)
            sod = ngd.Cast<dynamic>().ToDictionary<dynamic, string, object>(
                p => p.Key.ToString(), p => p.Value as object);

        // Convert object properties to dictionary.
        if (sod == null)
            sod = new RouteValueDictionary(values);
    }

    // Normalize and return the values.
    return from pair in sod
           from val in pair.Value as IEnumerable<string>
            ?? new[] { pair.Value?.ToString() }
           select new KeyValuePair<string, string>(pair.Key, val);
}

下面是测试:

var uri = new Uri("https://stackoverflow.com/yo?oldKey=oldValue");

// Test with a string/string dictionary.
var q = uri.WithQuery(new Dictionary<string, string>
{
    ["k1"] = string.Empty,
    ["k2"] = null,
    ["k3"] = "v3"
});

Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?k1&k2&k3=v3"));

// Test with a string/object dictionary.
q = uri.WithQuery(new Dictionary<string, object>
{
    ["k1"] = "v1",
    ["k2"] = new[] { "v2a", "v2b" },
    ["k3"] = null
});

Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?k1=v1&k2=v2a&k2=v2b&k3"));

// Test with a name/value collection.
var nvc = new NameValueCollection()
{
    ["k1"] = string.Empty,
    ["k2"] = "v2a"
};

nvc.Add("k2", "v2b");

q = uri.WithQuery(nvc);
Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?k1&k2=v2a&k2=v2b"));

// Test with any dictionary.
q = uri.WithQuery(new Dictionary<int, HashSet<string>>
{
    [1] = new HashSet<string> { "v1" },
    [2] = new HashSet<string> { "v2a", "v2b" },
    [3] = null
});

Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?1=v1&2=v2a&2=v2b&3"));

// Test with an anonymous object.
q = uri.WithQuery(new
{
    k1 = "v1",
    k2 = new[] { "v2a", "v2b" },
    k3 = new List<string> { "v3" },
    k4 = true,
    k5 = null as Queue<string>
});

Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?k1=v1&k2=v2a&k2=v2b&k3=v3&k4=True&k5"));

// Keep existing query using a name/value collection.
nvc = HttpUtility.ParseQueryString(uri.Query);
nvc.Add("newKey", "newValue");

q = uri.WithQuery(nvc);
Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?newKey=newValue&oldKey=oldValue"));

// Merge two query objects using the RouteValueDictionary.
var an1 = new { k1 = "v1" };
var an2 = new { k2 = "v2" };

q = uri.WithQuery(
    new RouteValueDictionary(an1).Concat(
        new RouteValueDictionary(an2)));

Debug.Assert(q == new Uri(
    "https://stackoverflow.com/yo?k1=v1&k2=v2"));

这与公认的答案完全相同，只是稍微更紧凑:

private string ToQueryString(NameValueCollection nvc)
{
    return "?" + string.Join("&", nvc.AllKeys.Select(k => string.Format("{0}={1}", 
        HttpUtility.UrlEncode(k), 
        HttpUtility.UrlEncode(nvc[k]))));
}

我需要为我正在开发的可移植类库(PCL)解决同样的问题。在这种情况下，我没有访问系统的权限。所以我不能用ParseQueryString。

相反，我使用System.Net.Http.FormUrlEncodedContent像这样:

var url = new UriBuilder("http://example.com");

url.Query = new FormUrlEncodedContent(new Dictionary<string,string>()
{
    {"param1", "val1"},
    {"param2", "val2"},
    {"param3", "val3"},
}).ReadAsStringAsync().Result;

在dotnet核心QueryHelpers.AddQueryString()将接受一个IDictionary<string,string>的键值对。为了节省一些内存分配和CPU周期，您可以使用SortedList<，>而不是Dictionary<，>，并按照排序顺序添加适当的容量和项…

var queryParams = new SortedList<string,string>(2);
queryParams.Add("abc", "val1");
queryParams.Add("def", "val2");

string requestUri = QueryHelpers.AddQueryString("https://localhost/api", queryParams);

我为我的剃刀项目写了一个助手，使用了其他答案的一些提示。

ParseQueryString业务是必要的，因为我们不允许篡改当前请求的QueryString对象。

@helper GetQueryStringWithValue(string key, string value) {
    var queryString = System.Web.HttpUtility.ParseQueryString(HttpContext.Current.Request.QueryString.ToString());
    queryString[key] = value;
    @Html.Raw(queryString.ToString())
}

我是这样使用的:

location.search = '?@Helpers.GetQueryStringWithValue("var-name", "var-value")';

如果希望它接受多个值，只需将参数更改为Dictionary，并将对添加到查询字符串中。