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Defining geometry type in QgsDataSourceUri



Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Postgis - Converting 'geometry_dump' data type to a 'geometry' typePostgis geometry type = GeometryPostGIS change column type varchar => geometryCan QGIS import Multipatch geometry type?PostgreSQL geometry type invalidDefining features geometry in new created layer using PyQGIS?Debug QgsDataSourceUri as a QgsVectorLayerQGIS change geometry type in shapefileHow to backup QGIS (3.2.1) layer to independent PostGIS table?Invalid geometry vs. no geometry



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1















There is a PostgreSQL database containing some layers that have few types of geometry (multipolygons and linestrings). I need to distinguish these geometries in two separate layers.



My basic strings that make layer for QGIS are:



uri.setConnection('srs', '5432', 'db_name', 'postgres' , 'password') 
for i, name in enumerate(sorted(list_of_layers)): # list containing elements like ['schema_name', 'layer_name']
if name[1] in ['layer_with_geometry_collection']:
uri.setDataSource(name[0], name[1], 'SP_GEOMETRY')
vlayer=QgsVectorLayer (uri.uri(), name[1], "postgres")


Where do I need to set the geometry type?
If I check vlayer.wkbType(), it throws me 0.
Manual set uri.setWkbType() hadn't helped yet.










share|improve this question
























  • It is very dangerous to casually connect to an RDBMS as the administrative user. Doing so can often result in casual destruction of the database. Instead, create a login user to own your data, and another to manage read-only or read-mostly connections. The postgres user should never own spatial data.

    – Vince
    Apr 10 at 12:22












  • @Vince you mean not to use postgres as a username?

    – Pavel Pereverzev
    Apr 10 at 12:24












  • Yes. Only connect as postgres to create new login and group roles.

    – Vince
    Apr 10 at 12:49

















1















There is a PostgreSQL database containing some layers that have few types of geometry (multipolygons and linestrings). I need to distinguish these geometries in two separate layers.



My basic strings that make layer for QGIS are:



uri.setConnection('srs', '5432', 'db_name', 'postgres' , 'password') 
for i, name in enumerate(sorted(list_of_layers)): # list containing elements like ['schema_name', 'layer_name']
if name[1] in ['layer_with_geometry_collection']:
uri.setDataSource(name[0], name[1], 'SP_GEOMETRY')
vlayer=QgsVectorLayer (uri.uri(), name[1], "postgres")


Where do I need to set the geometry type?
If I check vlayer.wkbType(), it throws me 0.
Manual set uri.setWkbType() hadn't helped yet.










share|improve this question
























  • It is very dangerous to casually connect to an RDBMS as the administrative user. Doing so can often result in casual destruction of the database. Instead, create a login user to own your data, and another to manage read-only or read-mostly connections. The postgres user should never own spatial data.

    – Vince
    Apr 10 at 12:22












  • @Vince you mean not to use postgres as a username?

    – Pavel Pereverzev
    Apr 10 at 12:24












  • Yes. Only connect as postgres to create new login and group roles.

    – Vince
    Apr 10 at 12:49













1












1








1








There is a PostgreSQL database containing some layers that have few types of geometry (multipolygons and linestrings). I need to distinguish these geometries in two separate layers.



My basic strings that make layer for QGIS are:



uri.setConnection('srs', '5432', 'db_name', 'postgres' , 'password') 
for i, name in enumerate(sorted(list_of_layers)): # list containing elements like ['schema_name', 'layer_name']
if name[1] in ['layer_with_geometry_collection']:
uri.setDataSource(name[0], name[1], 'SP_GEOMETRY')
vlayer=QgsVectorLayer (uri.uri(), name[1], "postgres")


Where do I need to set the geometry type?
If I check vlayer.wkbType(), it throws me 0.
Manual set uri.setWkbType() hadn't helped yet.










share|improve this question
















There is a PostgreSQL database containing some layers that have few types of geometry (multipolygons and linestrings). I need to distinguish these geometries in two separate layers.



My basic strings that make layer for QGIS are:



uri.setConnection('srs', '5432', 'db_name', 'postgres' , 'password') 
for i, name in enumerate(sorted(list_of_layers)): # list containing elements like ['schema_name', 'layer_name']
if name[1] in ['layer_with_geometry_collection']:
uri.setDataSource(name[0], name[1], 'SP_GEOMETRY')
vlayer=QgsVectorLayer (uri.uri(), name[1], "postgres")


Where do I need to set the geometry type?
If I check vlayer.wkbType(), it throws me 0.
Manual set uri.setWkbType() hadn't helped yet.







qgis pyqgis geometry






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 10 at 10:56









Vince

14.8k32850




14.8k32850










asked Apr 10 at 9:32









Pavel PereverzevPavel Pereverzev

543211




543211












  • It is very dangerous to casually connect to an RDBMS as the administrative user. Doing so can often result in casual destruction of the database. Instead, create a login user to own your data, and another to manage read-only or read-mostly connections. The postgres user should never own spatial data.

    – Vince
    Apr 10 at 12:22












  • @Vince you mean not to use postgres as a username?

    – Pavel Pereverzev
    Apr 10 at 12:24












  • Yes. Only connect as postgres to create new login and group roles.

    – Vince
    Apr 10 at 12:49

















  • It is very dangerous to casually connect to an RDBMS as the administrative user. Doing so can often result in casual destruction of the database. Instead, create a login user to own your data, and another to manage read-only or read-mostly connections. The postgres user should never own spatial data.

    – Vince
    Apr 10 at 12:22












  • @Vince you mean not to use postgres as a username?

    – Pavel Pereverzev
    Apr 10 at 12:24












  • Yes. Only connect as postgres to create new login and group roles.

    – Vince
    Apr 10 at 12:49
















It is very dangerous to casually connect to an RDBMS as the administrative user. Doing so can often result in casual destruction of the database. Instead, create a login user to own your data, and another to manage read-only or read-mostly connections. The postgres user should never own spatial data.

– Vince
Apr 10 at 12:22






It is very dangerous to casually connect to an RDBMS as the administrative user. Doing so can often result in casual destruction of the database. Instead, create a login user to own your data, and another to manage read-only or read-mostly connections. The postgres user should never own spatial data.

– Vince
Apr 10 at 12:22














@Vince you mean not to use postgres as a username?

– Pavel Pereverzev
Apr 10 at 12:24






@Vince you mean not to use postgres as a username?

– Pavel Pereverzev
Apr 10 at 12:24














Yes. Only connect as postgres to create new login and group roles.

– Vince
Apr 10 at 12:49





Yes. Only connect as postgres to create new login and group roles.

– Vince
Apr 10 at 12:49










1 Answer
1






active

oldest

votes


















2














Found an answer in PosGIS reference guide.
According to types of geometry that can be stored in PostGIS database, a filter should be written to get the specific geometry.



For example, I need to get only polygon features. The full code will be:



filter = """ST_GeometryType("") LIKE ''""".format('SP_GEOMETRY', 'ST_Polygon')
uri.setConnection('srs', '5432', 'db_name', 'postgres' , 'password')
for i, name in enumerate(sorted(list_of_layers)): # list containing elements like ['schema_name', 'layer_name']
if name[1] in ['layer_with_geometry_collection']:
uri.setDataSource(name[0], name[1], 'SP_GEOMETRY', filter) # filter added
vlayer=QgsVectorLayer (uri.uri(), name[1], "postgres")





share|improve this answer























  • You'd have to build a covering index to make this query performant, and doing do would conflict with use of a spatial index in the application.

    – Vince
    Apr 10 at 12:51











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Found an answer in PosGIS reference guide.
According to types of geometry that can be stored in PostGIS database, a filter should be written to get the specific geometry.



For example, I need to get only polygon features. The full code will be:



filter = """ST_GeometryType("") LIKE ''""".format('SP_GEOMETRY', 'ST_Polygon')
uri.setConnection('srs', '5432', 'db_name', 'postgres' , 'password')
for i, name in enumerate(sorted(list_of_layers)): # list containing elements like ['schema_name', 'layer_name']
if name[1] in ['layer_with_geometry_collection']:
uri.setDataSource(name[0], name[1], 'SP_GEOMETRY', filter) # filter added
vlayer=QgsVectorLayer (uri.uri(), name[1], "postgres")





share|improve this answer























  • You'd have to build a covering index to make this query performant, and doing do would conflict with use of a spatial index in the application.

    – Vince
    Apr 10 at 12:51















2














Found an answer in PosGIS reference guide.
According to types of geometry that can be stored in PostGIS database, a filter should be written to get the specific geometry.



For example, I need to get only polygon features. The full code will be:



filter = """ST_GeometryType("") LIKE ''""".format('SP_GEOMETRY', 'ST_Polygon')
uri.setConnection('srs', '5432', 'db_name', 'postgres' , 'password')
for i, name in enumerate(sorted(list_of_layers)): # list containing elements like ['schema_name', 'layer_name']
if name[1] in ['layer_with_geometry_collection']:
uri.setDataSource(name[0], name[1], 'SP_GEOMETRY', filter) # filter added
vlayer=QgsVectorLayer (uri.uri(), name[1], "postgres")





share|improve this answer























  • You'd have to build a covering index to make this query performant, and doing do would conflict with use of a spatial index in the application.

    – Vince
    Apr 10 at 12:51













2












2








2







Found an answer in PosGIS reference guide.
According to types of geometry that can be stored in PostGIS database, a filter should be written to get the specific geometry.



For example, I need to get only polygon features. The full code will be:



filter = """ST_GeometryType("") LIKE ''""".format('SP_GEOMETRY', 'ST_Polygon')
uri.setConnection('srs', '5432', 'db_name', 'postgres' , 'password')
for i, name in enumerate(sorted(list_of_layers)): # list containing elements like ['schema_name', 'layer_name']
if name[1] in ['layer_with_geometry_collection']:
uri.setDataSource(name[0], name[1], 'SP_GEOMETRY', filter) # filter added
vlayer=QgsVectorLayer (uri.uri(), name[1], "postgres")





share|improve this answer













Found an answer in PosGIS reference guide.
According to types of geometry that can be stored in PostGIS database, a filter should be written to get the specific geometry.



For example, I need to get only polygon features. The full code will be:



filter = """ST_GeometryType("") LIKE ''""".format('SP_GEOMETRY', 'ST_Polygon')
uri.setConnection('srs', '5432', 'db_name', 'postgres' , 'password')
for i, name in enumerate(sorted(list_of_layers)): # list containing elements like ['schema_name', 'layer_name']
if name[1] in ['layer_with_geometry_collection']:
uri.setDataSource(name[0], name[1], 'SP_GEOMETRY', filter) # filter added
vlayer=QgsVectorLayer (uri.uri(), name[1], "postgres")






share|improve this answer












share|improve this answer



share|improve this answer










answered Apr 10 at 11:11









Pavel PereverzevPavel Pereverzev

543211




543211












  • You'd have to build a covering index to make this query performant, and doing do would conflict with use of a spatial index in the application.

    – Vince
    Apr 10 at 12:51

















  • You'd have to build a covering index to make this query performant, and doing do would conflict with use of a spatial index in the application.

    – Vince
    Apr 10 at 12:51
















You'd have to build a covering index to make this query performant, and doing do would conflict with use of a spatial index in the application.

– Vince
Apr 10 at 12:51





You'd have to build a covering index to make this query performant, and doing do would conflict with use of a spatial index in the application.

– Vince
Apr 10 at 12:51

















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