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feat(projects): #1594 filtered the public projects having > 1 repo #1598

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feat(projects): #1594 filtered the public projects having > 1 repo #1598

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7ce1111
feat(projects): #1579 filtered the public projects having more than 1…
webkhushboo Nov 4, 2019
4ec8479
feat(projects): #1594 filtered the public projects having more than 1…
webkhushboo Nov 4, 2019
22ed567
Merge branch 'issue-1594' of https://github.com/DashboardHub/Pipeline…
webkhushboo Nov 4, 2019
5074f4d
fix(projects): #1594 fixed logic for showing project if having atleas…
webkhushboo Nov 5, 2019
cbc713b
fix(config): reverting changes of environment.ts file
webkhushboo Nov 5, 2019
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3 changes: 2 additions & 1 deletion web/src/app/shared/components/private-public-project/private-public-project.component.ts
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Original file line number Diff line number Diff line change
Expand Up @@ -40,7 +40,8 @@ export class PrivatePublicProjectComponent implements OnInit, OnDestroy {
if (this.router.url === '/') {
this.projectSubscription = this.projectService
.findPublicProjects()
.subscribe((projects: IProject[]) => this.projects = projects.map((project: IProject) => new ProjectModel(project)));
.subscribe((projects: IProject[]) => this.projects = projects.map((project: IProject) => new ProjectModel(project))
.filter((project: ProjectModel) => project.repositories.length > 0));
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@eddiejaoude eddiejaoude Nov 5, 2019 •
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When the list get longer in the future it would be more efficient to get the database to do the filtering if possible.

Currently the query limits to 10 results, if we remove items after the query we could end up with less than 10

ref.where('type', '==', 'public')
              .orderBy('updatedOn', 'desc').limit(10)

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It does not look like it is possible to do a repositories.length > 0 query in Firestore. We would need to store the repositories count to do this.

Best to store the repository length, so we can do a query to Firestore.

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Okay Do I have to make this change in this issue ? or we can do it in another ?

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I think do it in this PR, as this PR is tiny

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okay I will do it here

} else {
this.showTitle = true;
this.projectSubscription = this.projectService
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