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Login using HTTP on Protocol

Solved80 views6 days ago
3
Jose Araujo [DevOps Advocate]547 4th June 2025 0 Comments

Hi

I have issues when run this session:

<Sessionid="2"name="API Login">

<Connectionid="1"name="Credentials">

<Requestverb="POST"url="/api/v1/system/configuration/login">

<!--<Parameters>

<Parameter key="user" pid="135"></Parameter>

<Parameter key="password" pid="136"></Parameter>

</Parameters>-->

<Headers>

  <Header key="Content-Type">application/json</Header>

</Headers>

</Request>

<ResponsestatusCode="4">

<Contentpid="5"></Content>

</Response>

</Connection>

</Session>

in wireshark appear this session send post with this body "{"

for login need send this format in POST to endpoint

{
"user":"xxxxxx",
"password":"xxxxxxx"
}

Jose Araujo [DevOps Advocate] Selected answer as best 6 days ago

2 Answers

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0
Bram Devlaminck [SLC] [DevOps Advocate]320 Posted 5th June 2025 0 Comments

Hi Jose,

On this documentation page you can find an example of how DataMiner encodes the data from the "Parameter" tags when sending a POST (the example right above the "Response" section).
As you have noticed yourself, this is not in the format that you need.

Alternatively you can use the "data" tag, as shown here (the first example).
When using a POST the contents of the data tag is sent as the body, so in your case the contents of that data tag (set with pid='<parameterid>') will need to be the JSON data that you want to send.
The easiest way to put the data in that parameter in the right format will be with a QAction.

Hope this helps.

Kind regards,

Jose Araujo [DevOps Advocate] Selected answer as best 6 days ago
0
Jose Araujo [DevOps Advocate]547 Posted 6 days ago 0 Comments

Hi Bram,

I use this QAction and work:

using System;
using System.Text;
using Skyline.DataMiner.Scripting;

public static class QAction
{
public static void Run(SLProtocol protocol)
{
try
{
string username = Convert.ToString(protocol.GetParameter(10));
string password = Convert.ToString(protocol.GetParameter(11));

string json = $"{{\"user\":\"{username}\",\"password\":\"{password}\"}}";protocol.SetParameter(12, json);
}
catch (Exception ex)
{
protocol.Log($"QAction Error: {ex.Message}");
}
}
}

Jose Araujo [DevOps Advocate] Answered question 6 days ago
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