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[processing] Handle case of folder of shapefiles in build vrt alg #10083

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[processing] Handle case of folder of shapefiles in build vrt alg #10083

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volaya
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@volaya volaya commented May 23, 2019

#fixes 21519

Description

If a shapefile (i.e /my/data/points.shp) was loaded by loading a directory of shapefiles, its source wont be the filepath, but instead a uri-like thing such as /my/data|layername=points.

Although that file can be read by the external tool that creates the virtual vector, that source string won't work. I added a patch to try to resolve the uri-like source to a regular filepath.

It's just a patch for this tool, but this might affect other tools as well (basically, all the ones with external apps, like SAGA, etc), which wont work with shpafiles loaded as part of a folder.

Ideally, there should be a method to resolve the source to a valid filepath if possible, but I couldnt find it. I am making the PR in any case, to get some discussion about this.

I will be happy to improve this if someone points me in the direction of such method of workflow, in case it exists.

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@volaya volaya requested a review from nyalldawson May 23, 2019 12:58
#might be a shapefile loaded as a directory of shapefiles. We try to resolve the uri to a filepath
sourceParts = QgsProviderRegistry.instance().decodeUri('ogr', inFile)
if sourceParts.get("layerName"):
inFile = os.path.join(sourceParts.get("path"), sourceParts.get("layerName") + ".shp")
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Won't this break if a geopackage layer (or similar) was loaded? In that case the layerName parameter will be present, but the file name generated here will be broken.

This is really quite tricky to handle -- I think we'd need a check in here to first see if sourceParts['path'] is actually a directory (and if not, there's nothing required). Then, if it IS a directory, we'd need to check inside that path for files which start with the layername and have known vector data format extensions (because the same situation may arise for say KML files), and use that to determine the actual name of the file. Or alternatively, maybe we could use OGR api to try to load the file using just the path and layername, and let it run whatever internal logic it has to resolve the layername to a file, and then retrieve this actual file name from the OGR layer...

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It wont actually break anything, since gpkgs dont work now. Any layer with a source that is not an actual filepath cannot be used with that algorithm at the moment. However, I agree that the trick i added is not very general and only fixes the case of shapefiles.

I improved it a bit and put it in a separate function in the processing.tools.vector module. From there, it can be moved to C++ if someone takes care of that.

Still, some cases are not handled. It works now for gpkgs, but as long as the layer is the only one in the gpkg file.

In short, it's an improvement over the previous status of the algorithm, but it wont work with all input layers, like other Processing algorithms do. Other algorithms might be affected of this issue, so it's worth checking and maybe adding a similar mechanism.

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Ideally, there should be a method to resolve the source to a valid filepath if possible, but I couldnt find it. I am making the PR in any case, to get some discussion about this.

Agreed - we need a central method for handling this. Maybe in QgsOgrUtils.cpp?

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@stale stale bot added the stale Uh oh! Seems this work is abandoned, and the PR is about to close. label Jun 7, 2019
@nyalldawson nyalldawson removed the stale Uh oh! Seems this work is abandoned, and the PR is about to close. label Jun 11, 2019
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@volaya I've asked here https://lists.osgeo.org/pipermail/gdal-dev/2019-June/050380.html to see if there's a cleaner way we can approach this

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volaya commented Jun 12, 2019

cool. Good idea. Let's wait to see what they say

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m-kuhn commented Jun 22, 2019

Feedback @rouault

My question is -- GetName() here returns just the file base name (i.e.
"lines" for "/path/lines.shp"). Is there anyway to retrieve the actual
layer file path for these?

No, for most formats, there is no direct mapping from a layer to file(s)
containing layer data, so there is nothing in the OGR API for that. So you
have to use the knowledge that this is opened by the shapefile or mapinfo +
that the datasource name is a directory, and then you can form a filename from
that and the layer name.

if os.path.exists(folder):
if os.path.isdir(folder):
for ext in QgsVectorFileWriter.supportedFormatExtensions():
f = os.path.join(folder, "%s.%s" % (layerName, ext))
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Suggested change
f = os.path.join(folder, "%s.%s" % (layerName, ext))
f = os.path.join(folder, "{basename}.{extension}".format(basename=layerName, extension=ext))

Just because ;)

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@stale stale bot added the stale Uh oh! Seems this work is abandoned, and the PR is about to close. label Jul 6, 2019
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3 participants