#geodjango

Considering already installed:

Ubuntu 18.04

Python 3.6.8

Django==2.2.7

pkg-resources==0.0.0

pytz==2019.3

spatialite==0.0.3

sqlparse==0.3.0

As described at Techiediaries, firstly install the libraries GEOS, PROJ.4 and GDAL:

sudo apt-get install libgeos-dev sudo apt-get install binutils libproj-dev sudo apt-get install gdal-bin libgdal-dev sudo apt-get install python3-gdal

Follow the instructions…

Estou trabalhando, atualmente, em um sistema de rastreamento de veículos e cheguei à conclusão de que precisava gerar GeoJSON nas minhas views do Django.

Muito naturalmente, corri atrás de uma app para isso. Encontrei o django-geojson, da makinacorpus https://github.com/makinacorpus/django-geojson. Instalei e configurei.

Péssima ideia.

Acontece que GeoJSON não é difícil de se gerar. Especialmente quando se está trabalhando com GeoDjango. Vide o site oficial do formato http://geojson.org/.

Veja que os nomes das features são idênticos aos nomes das classes usadas no django.gis: Point, LineString, Polygon, MultiPoint, MultiLineString, e MultiPolygon. Logo, é bem simples de se fazer um gerador genérico de GeoJSON pronto para gerar os dados no formato certo para o retorno da view.

Primeiro, eu já tenho uma app com atalhos, que chamo de "asael". Ela me dá uma bela função "return_json":

def return_string (r, s, response = None):     if not response:         return h.HttpResponse(s)     else:         response.write(s)         return response

def return_json (r, dados, response = None):     return return_string(r, json.dumps(dados), response)

Então é fácil fazer a "return_geojson", considerando que eu já tenho as "features" salvas direitinho no banco de dados (GeoDjango + PostGIS).

def get_feature_dict (elemento):

    """Dado um elemento (Point, Polygon, etc), retorna um dicionário pronto para a geração de GeoJSON a partir dele."""     nomeclasse = elemento.__class__.__name__

    # No meu projeto, toda feature chama-se "ponto", a não ser que não seja possível. Daí eu chamo de "coords":     for nome in ('ponto', 'coords', 'geom'):         geometry = getattr(elemento, nome, None)         if geometry:             break     else:         return {}     return {         'type': nomeclasse, # "Point", por exemplo.         'coordinates': geometry     } def return_geojson (r, features):     dicionario = {         'type': 'FeatureCollection',         'features': [{             'type': 'Feature',             'geometry': get_feature_dict(x),             'properties': getattr(x, 'properties', {'type': 'unknown'})         } for x in features],     }     return return_json (r, dicionario)

So if you are using PostGIS 2.0 and get an error like this:

django.db.utils.ProgrammingError: type modifier is not allowed for type "geometry"

It's best when a dataset just gives you coordinates in WSG84 (SRID: 4326), but it's not the end of the world if they use a different coordinate system. To transform their coordinates to latitude and longitude, you just need to know the SRID of their coordinate system.

Unfortunately, your contact person for the dataset might not know the SRID. But if she knows the name of the coordinate system then there's still a way for you to obtain the SRID. The following is how I do it.

1. Get some reference points in their coordinate system. These are points for which you know the equivalent latitude and longitude coordinates. This is usually trivial, because most datasets have an address field. To get the latitude and longitude, you can use Google Maps: search for an address, and, on the result page, right-click the red pin and select "What's here?". The latitude and longitude will appear in the search bar.

2. Do a search on spatialreference.org. Entering all the keywords sometimes gives you nothing, because the actual name in their database will be spelled slightly differently than you expect, and their search engine only does exact matches. Pick the most important keyword and search on that first. You may get too many results, but from there you can start winnowing them down.

3. Write a program to try out all possible SRIDs. You might end up with several possible SRIDs. You are unlikely to be able to guess the correct one by reading their descriptions. Instead, write a simple program using GeoDjango to determine the most likely SRID. Here is an example:

from django.contrib.gis.geos import Point # Reference point. Note that the x coordinate comes first. COORDINATES = (397162, 140182) # Reference address: 3100 - 3299 BLOCK OF 14TH STREET NW, dc # Reference latitude, longitude from Google Maps: 38.929527,-77.032734 # Candidate SRIDs from spatialreference.org. At least one of them is right. SRIDS = [2248, 2804, 2893, 3559, 3583, 6620, 6654, 7058, 26785, 26985, 102285, 102685] for srid in SRIDS: pt = Point(*COORDINATES, srid=srid) try: pt.transform(4326) print '{}: {}, {}'.format(srid, pt.y, pt.x) except Exception: pass

The output will look something like this:

2248: 38.0078526925, -80.1766510551 2804: 38.9295130132, -77.0327305535 2893: 38.0078526925, -80.1766510551 3559: 38.9295130132, -77.0327305535 3583: 42.255458399, -71.7461411817 ERROR 6: EPSG PCS/GCS code 6620 not found in EPSG support files. Is this a valid EPSG coordinate system? ERROR 6: EPSG PCS/GCS code 6654 not found in EPSG support files. Is this a valid EPSG coordinate system? ERROR 6: EPSG PCS/GCS code 7058 not found in EPSG support files. Is this a valid EPSG coordinate system? 26785: 38.2098999912, -78.4016791905 26985: 38.9295130132, -77.0327305535 102285: 38.9295130132, -77.0327305535 102685: 38.0078526924, -80.1766510563

Great GeoDjango Presentation by Zain Memon from DjangoCon!