Jira - Create new issue V2

[MFL_UK]

Hi all,

I'm having an issue with Create a new issue V2. My Jira is connected correctly via API code and I do not receive any error messages.

When the list of Projects pulls through it only has a small proportion of those on the Jira instance and does not have the Project I am looking for. Does anyone know of a way around this or a specific criteria that needs to be entered in Jira or in Power Automate so that it will show on the Projects list?

Please see snapshot below Thanks in advance

[prameelaj]

I too have the same issue.  I am using v3.   I do not know what are the issue type fields it is looking for that needs to be added into Body section and what is the format for it.

"message": "Issue fields are empty. Please ensure that all required fields is provided.", power automate

[Thai_Nguyen_Dev]

Looking at the Jira Power Automate connector documentation (https://learn.microsoft.com/en-us/connectors/jira/) it seems to be a limitation of the Jira API

In the section titled "Known Issues and Limitations" it provides the following line. 

• As per the Jira API documentation, jira API returns maximum 50 projects, so top 50 projects will be listed under dropdown in actions/triggers.

Is there an option to "Add a custom item" where you can then enter the project key? The project key is usually a unique string that you typically see as a prefix to the issues you create in that project. For example, for a project called account, Jira would typically automatically assign the project a key labelled "AC". When you create issues in this project you would typically see it as AC-1. 

[prameelaj]

I tried by using custom item.  Below is the body content i am using

"fields": {
"issuetype": { "id": "@{items('Apply_to_each')?['ID']}" },
"summary": "@{items('Apply_to_each')?['Title']}",
"description": "@{items('Apply_to_each')?['Description']}",
"reporter": { "id": "@{items('Apply_to_each')?['Author']}" },
"priority": { "id": "@{items('Apply_to_each')?['khvl']}" },
"labels": "Epic_Test",
"assignee": { "id": "@{items('Apply_to_each')?['Assigned_x0020_SA']}" },
"customfield_40000": { "type": "doc", "version": 1, "content": [ { "type": "paragraph", "content": [ { "text": "Occurs on all orders", "type": "text" } ] } ] }}

I am getting below error:

"message": "Payload is not a valid json.",

[prameelaj]

Below is my updated Body.

{"fields": {
"issuetype": { "id": "@{items('Apply_to_each')?['ID']}" },
"summary": "@{items('Apply_to_each')?['Title']}",
"description": "@{items('Apply_to_each')?['Description']}",
"reporter": { "id": "@{items('Apply_to_each')?['Author']}" },
"priority": { "id": "@{items('Apply_to_each')?['khvl']}" },
"labels": "Epic_Test",
"assignee": { "id": "@{items('Apply_to_each')?['Assigned_x0020_SA']}" }
}
}

Now I am getting below error:

{

  "errorMessages": [],

  "errors": {

    "project": "Specify a valid project ID or key"

  }

}

[Thai_Nguyen_Dev]

I don't follow. We were discussing the limitations where Jira doesn't give a full list of its projects unless you use the Get Projects connector. 

I'm new to this Jira connector as well. Are you still at the part where they ask for the project and issue type id?

[prameelaj]

Thanks for letting me know about the limitations.  I am facing issue with the parameters to be assigned in create a new issue (v3)(preview) in my flow.

I am getting below error;  Invalid issue type id.

When I googled, I found below information.

  id   |             pname
-------+--------------------------------
 10001 | Story
 10002 | Task
 10003 | Sub-task
 10004 | Bug
 10100 | Initiative
 10101 | Incident
 10102 | Service Request
 10103 | Service Request with Approvals
 10104 | Change
 10105 | Problem
 10000 | Epic
 10200 | New Feature
 10201 | Support

I gave 10000 as Issue Type ID. But getting below error.

{

  "errorMessages": [],

  "errors": {

    "issuetype": "The issue type selected is invalid."

  }

}

[prameelaj]

Finally fixed it.  Worked with the JIRA admin to find the exact issue type id and applied it..  It worked fine.  Thanks to everyone whoever provided the information above.