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Get PortalItem status

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3
09-23-2022 05:49 AM
TomGeo
by
Occasional Contributor III

I recently experienced that while trying to create a layer from one of our portal services my AGP crashed. Turned out that the service I requested to create the layer from was stopped on the server.

To avoid that, I would like to check on the PortalItem (Item) status, but I do not seem to have the possibility to do so (Item Class Members). Fingers crossed I am only looking at the wrong place.

Can I check the PortalItem status?

- We are living in the 21st century.
GIS moved on and nobody needs a format consisting out of at least three files! No, nobody needs shapefiles, not even for the sake of an exchange format. Folks, use GeoPackage to exchange data with other GIS!
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3 Replies
GKmieliauskas
Esri Regular Contributor

Hi,

Have you tried CanCreateLayer from LayerFactory?

					if (LayerFactory.Instance.CanCreateLayerFrom(currentItem))
						LayerFactory.Instance.CreateLayer(currentItem, MapView.Active.Map);
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TomGeo
by
Occasional Contributor III

Yes, I tried that, but it's still causing an exception right after

Item curItem = ItemFactory.Instance.Create(curService);
if(LayerFactory.Instance.CanCreateLayerFrom(curItem))
{
    Layer newLayer = LayerFactory.Instance.CreateLayer(new Uri(curService, UriKind.Absolute), activeMapView.Map, 0);
}

 

Also when trying to show the service as a layer in the Map Viewer at the portal, the error message returned by the viewer is: The layer cannot be added to the map.

- We are living in the 21st century.
GIS moved on and nobody needs a format consisting out of at least three files! No, nobody needs shapefiles, not even for the sake of an exchange format. Folks, use GeoPackage to exchange data with other GIS!
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GKmieliauskas
Esri Regular Contributor

It worked some time ago. So you could place CreateLayer in try/catch block and make return with message on exception:

Item curItem = ItemFactory.Instance.Create(curService);

    try {
    Layer newLayer = LayerFactory.Instance.CreateLayer(new Uri(curService, UriKind.Absolute), activeMapView.Map, 0);
    }
    catch(Exception ex) {
        MessageBox.Show(ex.Message);
        return;
    }

It is not elegant way but it works

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