[英]Is there a way to get all the querystring name/value pairs into a collection?
Is there a way to get all the querystring name/value pairs into a collection?有没有办法将所有查询字符串名称/值对放入一个集合中?
I'm looking for a built in way in .net, if not I can just split on the & and load a collection.我正在寻找 .net 中的内置方式,如果没有,我可以在 & 上拆分并加载一个集合。
Yes, use the HttpRequest.QueryString
collection: 是的,使用
HttpRequest.QueryString
集合:
Gets the collection of HTTP query string variables.
获取HTTP查询字符串变量的集合。
You can use it like this: 你可以像这样使用它:
foreach (String key in Request.QueryString.AllKeys)
{
Response.Write("Key: " + key + " Value: " + Request.QueryString[key]);
}
Well, Request.QueryString
already IS a collection. 好吧,
Request.QueryString
已经是一个集合。 Specifically, it's a NameValueCollection
. 具体来说,它是一个
NameValueCollection
。 If your code is running in ASP.NET, that's all you need. 如果您的代码在ASP.NET中运行,那就是您所需要的。
So to answer your question: Yes, there is. 所以回答你的问题:是的,有。
You can use LINQ to create a List of anonymous objects that you can access within an array: 您可以使用LINQ创建可以在数组中访问的匿名对象列表:
var qsArray = Request.QueryString.AllKeys
.Select(key => new { Name=key.ToString(), Value=Request.QueryString[key.ToString()]})
.ToArray();
If you have a querystring ONLY represented as a string, use HttpUtility.ParseQueryString to parse it into a NameValueCollection. 如果只有一个查询字符串表示为字符串,请使用HttpUtility.ParseQueryString将其解析为NameValueCollection。
However, if this is part of a HttpRequest, then use the already parsed QueryString-property of your HttpRequest. 但是,如果这是HttpRequest的一部分,那么使用已经解析的HttpRequest的QueryString属性。
QueryString
property in HttpRequest
class is actually NameValueCollection class. HttpRequest
类中的QueryString
属性实际上是NameValueCollection类。 All you need to do is 你需要做的就是
NameValueCollection col = Request.QueryString;
NameValueCollection col = Request.QueryString;
Create dictionary of parameters创建参数字典
Dictionary<string, string> parameters = new Dictionary<string, string>();
parameters = Request.QueryString.Keys.Cast<string>().ToDictionary(k => k, v => Request.QueryString[v]);
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