Preparing requests
Create a request
Before making a request using RestClient
, you need to create a request instance:
var request = new RestRequest(resource); // resource is the sub-path of the client base path
The default request type is GET
and you can override it by setting the Method
property. You can also set the method using the constructor overload:
var request = new RestRequest(resource, Method.Post);
After you've created a RestRequest
, you can add parameters to it. Below, you can find all the parameter types supported by RestSharp.
Request headers
Adds the header parameter as an HTTP header that is sent along with the request. The header name is the parameter's name and the header value is the value.
You can use one of the following request methods to add a header parameter:
AddHeader(string name, string value);
AddHeader<T>(string name, T value); // value will be converted to string
AddOrUpdateHeader(string name, string value); // replaces the header if it already exists
For example:
var request = new RestRequest("/path").AddHeader("X-Key", someKey);
You can also add header parameters to the client, and they will be added to every request made by the client. This is useful for adding authentication headers, for example.
client.AddDefaultHeader(string name, string value);
RestSharp will use the correct content type by default. Avoid adding the Content-Type
header manually to your requests unless you are absolutely sure it is required. You can add a custom content type to the body parameter itself.
Get or Post parameters
The default RestSharp parameter type is GetOrPostParameter
. You can add GetOrPost
parameter to the request using the AddParameter
function:
request
.AddParameter("name1", "value1")
.AddParameter("name2", "value2");
GetOrPost
behaves differently based on the HTTP method. If you execute a GET
call, RestSharp will append the parameters to the URL in the form url?name1=value1&name2=value2
.
On a POST
or PUT
requests, it depends on whether you have files attached to a request.
If not, the parameters will be sent as the body of the request in the form name1=value1&name2=value2
. Also, the request will be sent as application/x-www-form-urlencoded
.
In both cases, name and value will automatically be URL-encoded, unless specified otherwise:
request.AddParameter("name", "Væ üé", false); // don't encode the value
If you have files, RestSharp will send a multipart/form-data
request. Your parameters will be part of this request in the form:
Content-Type: text/plain; charset=utf-8
Content-Disposition: form-data; name="parameterName"
ParameterValue
Sometimes, you need to override the default content type for the parameter when making a multipart form call. It's possible to do by setting the ContentType
property of the parameter object. As an example, the code below will create a POST parameter with JSON value, and set the appropriate content type:
var parameter = new GetOrPostParameter("someJson", "{\"attributeFormat\":\"pdf\"}") {
ContentType = "application/json"
};
request.AddParameter(parameter);
When the request is set to use multipart content, the parameter will be sent as part of the request with the specified content type:
Content-Type: application/json; charset=utf-8
Content-Disposition: form-data; name="someJson"
{"attributeFormat":"pdf"}
You can also add GetOrPost
parameter as a default parameter to the client. This will add the parameter to every request made by the client.
client.AddDefaultParameter("foo", "bar");
It will work the same way as request parameters, except that it will be added to every request.
Query string
QueryString
works like GetOrPost
, except that it always appends the parameters to the url in the form url?name1=value1&name2=value2
, regardless of the request method.
Example:
var client = new RestClient("https://search.me");
var request = new RestRequest("search")
.AddParameter("foo", "bar");
var response = await client.GetAsync<SearchResponse>(request);
It will send a GET
request to https://search.me/search?foo=bar
.
For POST
-style requests you need to add the query string parameter explicitly:
request.AddQueryParameter("foo", "bar");
In some cases, you might need to prevent RestSharp from encoding the query string parameter.
To do so, set the encode
argument to false
when adding the parameter:
request.AddQueryParameter("foo", "bar/fox", false);
You can also add a query string parameter as a default parameter to the client. This will add the parameter to every request made by the client.
client.AddDefaultQueryParameter("foo", "bar");
The line above will result in all the requests made by that client instance to have foo=bar
in the query string for all the requests made by that client.
Using AddObject
You can avoid calling AddParameter
multiple times if you collect all the parameters in an object, and then use AddObject
.
For example, this code:
var params = new {
status = 1,
priority = "high",
ids = new [] { "123", "456" }
};
request.AddObject(params);
is equivalent to:
request.AddParameter("status", 1);
request.AddParameter("priority", "high");
request.AddParameter("ids", "123,456");
Remember that AddObject
only works if your properties have primitive types. It also works with collections of primitive types as shown above.
If you need to override the property name or format, you can do it using the RequestProperty
attribute. For example:
public class RequestModel {
// override the name and the format
[RequestProperty(Name = "from_date", Format = "d")]
public DateTime FromDate { get; set; }
}
// add it to the request
request.AddObject(new RequestModel { FromDate = DateTime.Now });
In this case, the request will get a GET or POST parameter named from_date
and its value would be the current date in short date format.
Using AddObjectStatic
Request function AddObjectStatic<T>(...)
allows using pre-compiled expressions for getting property values. Compared to AddObject
that uses reflections for each call, AddObjectStatic
caches functions to retrieve properties from an object of type T
, so it works much faster.
You can instruct AddObjectStatic
to use custom parameter names and formats, as well as supply the list of properties than need to be used as parameters. The last option could be useful if the type T
has properties that don't need to be sent with HTTP call.
To use custom parameter name or format, use the RequestProperty
attribute. For example:
class TestObject {
[RequestProperty(Name = "some_data")]
public string SomeData { get; set; }
[RequestProperty(Format = "d")]
public DateTime SomeDate { get; set; }
[RequestProperty(Name = "dates", Format = "d")]
public DateTime[] DatesArray { get; set; }
public int Plain { get; set; }
public DateTime[] PlainArray { get; set; }
}
URL segment parameter
Unlike GetOrPost
, URL segment parameter replaces placeholder values in the request URL:
var request = new RestRequest("health/{entity}/status")
.AddUrlSegment("entity", "s2");
When the request executes, RestSharp will try to match any {placeholder}
with a parameter of that name (without the {}
) and replace it with the value. So the above code results in health/s2/status
being the URL.
You can also add UrlSegment
parameter as a default parameter to the client. This will add the parameter to every request made by the client.
client.AddDefaultUrlSegment("foo", "bar");
Cookies
You can add cookies to a request using the AddCookie
method:
request.AddCookie("foo", "bar");
RestSharp will add cookies from the request as cookie headers and then extract the matching cookies from the response. You can observe and extract response cookies using the RestResponse.Cookies
properties, which has the CookieCollection
type.
However, the usage of a default URL segment parameter is questionable as you can just include the parameter value to the base URL of the client. There is, however, a CookieContainer
instance on the request level. You can either assign the pre-populated container to request.CookieContainer
, or let the container be created by the request when you call AddCookie
. Still, the container is only used to extract all the cookies from it and create cookie headers for the request instead of using the container directly. It's because the cookie container is normally configured on the HttpClientHandler
level and cookies are shared between requests made by the same client. In most of the cases this behaviour can be harmful.
If your use case requires sharing cookies between requests made by the client instance, you can use the client-level CookieContainer
, which you must provide as the options' property. You can add cookies to the container using the container API. No response cookies, however, would be auto-added to the container, but you can do it in code by getting cookies from the Cookes
property of the response and adding them to the client-level container available via IRestClient.Options.CookieContainer
property.
Request Body
RestSharp supports multiple ways to add a request body:
AddJsonBody
for JSON payloadsAddXmlBody
for XML payloadsAddStringBody
for pre-serialized payloads
We recommend using AddJsonBody
or AddXmlBody
methods instead of AddParameter
with type BodyParameter
. Those methods will set the proper request type and do the serialization work for you.
When you make a POST
, PUT
or PATCH
request and added GetOrPost
parameters, RestSharp will send them as a URL-encoded form request body by default. When a request also has files, it will send a multipart/form-data
request. You can also instruct RestSharp to send the body as multipart/form-data
by setting the AlwaysMultipartFormData
property to true
.
You can specify a custom body content type if necessary. The contentType
argument is available in all the overloads that add a request body.
It is not possible to add client-level default body parameters.
String body
If you have a pre-serialized payload like a JSON string, you can use AddStringBody
to add it as a body parameter. You need to specify the content type, so the remote endpoint knows what to do with the request body. For example:
const json = "{ data: { foo: \"bar\" } }";
request.AddStringBody(json, ContentType.Json);
JSON body
When you call AddJsonBody
, it does the following for you:
- Instructs the RestClient to serialize the object parameter as JSON when making a request
- Sets the content type to
application/json
- Sets the internal data type of the request body to
DataType.Json
Here is the example:
var param = new MyClass { IntData = 1, StringData = "test123" };
request.AddJsonBody(param);
It is possible to override the default content type by supplying the contentType
argument. For example:
request.AddJsonBody(param, "text/x-json");
If you use a pre-serialized string with AddJsonBody
, it will be sent as-is. The AddJsonBody
will detect if the parameter is a string and will add it as a string body with JSON content type.
Essentially, it means that top-level strings won't be serialized as JSON when you use AddJsonBody
. To overcome this issue, you can use an overload of AddJsonBody
, which allows you to tell RestSharp to serialize the string as JSON:
const string payload = @"
""requestBody"": {
""content"": {
""application/json"": {
""schema"": {
""type"": ""string""
}
}
}
},";
request.AddJsonBody(payload, forceSerialize: true); // the string will be serialized
request.AddJsonBody(payload); // the string will NOT be serialized and will be sent as-is
XML body
When you call AddXmlBody
, it does the following for you:
- Instructs the RestClient to serialize the object parameter as XML when making a request
- Sets the content type to
application/xml
- Sets the internal data type of the request body to
DataType.Xml
Do not send XML string to AddXmlBody
; it won't work!
Uploading files
To add a file to the request you can use the RestRequest
function called AddFile
. The main function accepts the FileParameter
argument:
request.AddFile(fileParameter);
You can instantiate the file parameter using FileParameter.Create
that accepts a bytes array, or FileParameter.FromFile
, which will load the file from disk.
There are also extension functions that wrap the creation of FileParameter
inside:
// Adds a file from disk
AddFile(parameterName, filePath, contentType);
// Adds an array of bytes
AddFile(parameterName, bytes, fileName, contentType);
// Adds a stream returned by the getFile function
AddFile(parameterName, getFile, fileName, contentType);
Remember that AddFile
will set all the necessary headers, so please don't try to set content headers manually.
You can also provide file upload options to the AddFile
call. The options are:
DisableFilenameEncoding
(defaultfalse
): if set totrue
, RestSharp will not encode the file name in theContent-Disposition
headerDisableFilenameStar
(defaulttrue
): if set totrue
, RestSharp will not add thefilename*
parameter to theContent-Disposition
header
Example of using the options:
var options = new FileParameterOptions {
DisableFilenameEncoding = true,
DisableFilenameStar = false
};
request.AddFile("file", filePath, options: options);
The options specified in the snippet above usually help when you upload files with non-ASCII characters in their names.