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IDProjectCategoryView StatusLast Update
0010267mantisbtbugtrackerpublic2009-04-06 17:55
Reportermiki000 Assigned To 
PrioritynormalSeverityminorReproducibilityhave not tried
Status newResolutionopen 
Summary0010267: Report and assign one bug to more project
Description

Can I report one bug to more than one project?
For example I have:

Main project

Subproject1
Subproject2
Subproject3

I want to report bug from Main project to Subproject1 and Subproject2.
I want users of Subproject1 and Subproject2 to be able notice into this bug but I don't want users of Subproject3 to see into this bug.
I don't want to create 2 separate bugs from Subproject1 and Subproject2.

Thanks and sorry for my english (I hope you understand what I mean :).

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mravcek

mravcek

2009-04-06 17:55

reporter   ~0021408

Last edited: 2009-04-07 03:42

I would like to have this feature too :)
For now I have done small change on Roadmap page and this helps me.

Basically:
Show children in Roadmap under Project and Version based on rule:
List the child also in the case when PARENT is in the project and version.

This means .. if you have:
0000001 Bug Project A, Version A,
0000002 Bug Project A, Version A, parent of 0000003
0000003 Bug Project B, Version B,
0000004 Bug Project B, Version B,

You will see in roadmap:
ProjectA Version A

ProjectB Version B

The change is in roadmap_page.php
Replace (line cca 103):
$t_version = $t_version_row['version'];
$c_version = db_prepare_string( $t_version );
//original
/
$query = "SELECT sbt.
, $t_relation_table.source_bug_id, dbt.target_version as parent_version FROM $t_bug_table AS sbt
LEFT JOIN $t_relation_table ON sbt.id=$t_relation_table.destination_bug_id AND $t_relation_table.relationship_type=2
LEFT JOIN $t_bug_table AS dbt ON dbt.id=$t_relation_table.source_bug_id
WHERE sbt.project_id='$c_project_id' AND sbt.target_version='$c_version' ORDER BY sbt.status ASC, sbt.last_updated DESC";
*/

//show children (even if they are not in project, but parent is in project)
$query = "SELECT sbt.*, $t_relation_table.source_bug_id, dbt.target_version as parent_version FROM $t_bug_table AS sbt
LEFT JOIN $t_relation_table ON sbt.id=$t_relation_table.destination_bug_id AND $t_relation_table.relationship_type=2
LEFT JOIN $t_bug_table AS dbt ON dbt.id=$t_relation_table.source_bug_id
WHERE (sbt.project_id='$c_project_id' AND sbt.target_version='$c_version')
OR (dbt.project_id='$c_project_id' AND dbt.target_version='$c_version')
ORDER BY sbt.status ASC, sbt.last_updated DESC";

        $t_description = $t_version_row['description'];

Note: Permisions for the child should be check too. I may add it later.