[QGIS-Developer] ProcessingAlgorithm: get filename of the INPUT param

[info]

Hi,

l.source() perhaps?

Cheers,
Benoit


On 2019-06-19 14:20, Richard Duivenvoorde wrote:
> Hi,
> 
> got a question from somebody who needed to find the filename of an 
> input
> param in a ProcessingAlgorithm
> (because he is going to read that file again or something like that).
> 
> He adds the INPUT via the initAlgorithm function:
> 
>         self.addParameter(
>             QgsProcessingParameterFeatureSource(
>                 self.INPUT,
>                 self.tr('Input layer (.geojson)'),
>                 [QgsProcessing.TypeVectorLine]
>             )
>         )
> 
> And then tried to get the filename in the
> processAlgorithm-implementation from the source via
> self.parameterAsSource, but then could not find an uri/filename.
> 
> Via:
> https://gis.stackexchange.com/questions/280667/using-a-selected-layer-with-processing-in-qgis-3
> I gave him:
> l = self.parameterAsVectorLayer(parameters, 'INPUT', context)
> feedback.pushInfo('l = {}'.format(l))
> feedback.pushInfo('l.dataProvider() = {}'.format(l.dataProvider()))
> feedback.pushInfo('l.dataProvider().uri() =
> {}'.format(l.dataProvider().uri()))
> feedback.pushInfo('l.dataProvider().uri().uri() =
> {}'.format(l.dataProvider().uri().uri()))
> 
> which at least gives him the dataprovider uri WITH the filename.
> 
> Question:
> 1) isn't there an easier way, or is this the preferred way?
> 2) if the filename contains a space, then the uri().uri() only shows 
> the
> part AFTER the space (as if it was a relative path, but then the wrong
> path...). Using a filename from a dir WITHOUT a space the uri().uri()
> shows you the full file path.
> 
> TIA & Regards,
> 
> Richard Duivenvoorde
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