RestSharp android upload multiple lines

Question

I can use RestSharp very well sending "mono-line" things like:

"user, password, email, telephone, ... "

But I can't understand how to send multiple lines, for example, all the values in this table:

+-----------+-----------+-------+
|    Dog    |   Race    | user  |
+-----------+-----------+-------+
| Skitty    | Doberman  | User1 |
| Birillo   | Pinscher  | User2 |
| Fragolino | Corgi     | User3 |
| ...       | ...       | ...   |
+-----------+-----------+-------+

How am I supposed to format parameters for restsharp?

Thanks

Edit: as asked I show How I send "normal" data:

I take data from user input in EditText (like, string1, string2, string3), and send them to a class I use for RestSharp, and I send them with:

var client = new RestClient("x.x.x.x/app/");
var request = new RestRequest("/ServiceX", Method.POST);
request.AddParameter("Dog", string1);
request.AddParameter("Race", string2);
request.AddParameter("User", string3);
IRestResponse response = client.Execute(request);
var content = response.Content;

This is a single line in a table, I can add only 1 line per request in this way. But I want to upload more lines.

Thanks

yes I use post, I added code on how I normally upload data, but I really don't know how to expand this on multiple lines. I didn't find anything. Thanks — Raikoug
– Raikoug, Commented Sep 19, 2016 at 12:03

M. Wiśnicki · Accepted Answer · 2016-09-19 20:20:00Z

1

Try use this. JsonHelper 1.0.2 You can download by NuGet or here

           public class YourModel
            {
                public string Doggy { get; set; }
                public string Race { get; set; }
                public string User { get; set; }

            }
            public YourModel nYModel = new YourModel();
            nYModel.Doggy = string1;
            nYModel.Race  = string2;
            nYModel.User = string3;

    var client = new RestClient(ServiceUrl);

    var request = new RestRequest("/resource/", Method.POST);

    // Json to post.
    string jsonToSend = JsonHelper.ToJson(nYModel);

    request.AddParameter("application/json; charset=utf-8", jsonToSend, ParameterType.RequestBody);
    request.RequestFormat = DataFormat.Json;

    try
    {
        client.ExecuteAsync(request, response =>
        {
            if (response.StatusCode == HttpStatusCode.OK)
            {
                // OK
            }
            else
            {
                // NOK
            }
        });
    }
    catch (Exception error)
    {
        // Log
    }

edited Sep 19, 2016 at 20:20

answered Sep 19, 2016 at 12:11

M. Wiśnicki

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5 Comments

Raikoug Over a year ago

Thanks a lot for answering! In this way I can add multiple entry for "YourModel"? or I need to make an array of "YourModel"? Thanks again!

M. Wiśnicki Over a year ago

You can make array or list.

Raikoug Over a year ago

I hope last silly question. The packet for JsonHelper? really thanks for all!

Raikoug Over a year ago

I did everything, and instead of making an array of my custom model, I did a list, working very well! Thanks for all help, i will post my code as soon as I've time.

Raikoug Over a year ago

I last used Serialize just for easy php handling, after 4 hours of echoes!! haha thanks a lot!

Raikoug · Accepted Answer · 2016-09-21 09:15:51Z

This is all the code i came up with:

In the c# app, i've a class storing the data, like

public class MyClass
    {
        public string field1{ get; set; }
        public string field2 { get; set; }
        public string field3 { get; set; }
        public string field4 { get; set; }
    }

Then I populate a List with all my entries:

List<MyClass> list= new List<MyClass>();
// make new element of MyClass like "element1"
list.Add(element1);

Well, this method is awesome because list can be populated very easily with a foreach or a while, taking all edittext in the layout, even dynamics ones, awesome.

After this next part require "Newtonsoft.Json" (NutGet is you friend... your BEST friend!)

 string json = JsonConvert.SerializeObject(list);

Well now you have to send it, I use post like I wrote in first post.

In the php part i took the data and parsed with:

$data = (json_decode($data));

This worked like a charm, becouse I can access at elements with:

foreach ($data as &$object){
  $obj->field1
  $obj->field2
  $obj->field3
  $obj->field4
}

Well now just build the query string and it's over... (just for my own desire for completeness, i paste the last part)

$query = 'INSERT INTO `table_name`(`field1`, `field2`, `field3`, `field4`) VALUES ';
foreach ($data as &$object)
    {
        $query_parts[] = "('" . $object->field1 . "', '" . $object->field2 . "', '" . $object->field3 . "', '" . $object->field4 . "')";

    }
        $query .= implode(',', $query_parts);

DONE!

Thanks @mwisnicki for the help. MANY thanks!

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