Hi (again) !
I figured out (after 2 days of research !) that in a standalone app, to be able
to open a custom layer form when identifying a feature, I have to implement a
specific QgsPythonRunner child class (code below).
And then I have to set the QgsPythonRunner.
The problem is that I can't get evalCommand() to work because the variable
result is a C++ reference, and python can't deal with it, can it ?
Maybe there is a workaround I didn't think through, or there is a need to pass
result as a return value ?
Thanks !
BR
class PythonRunnerStandalone(QgsPythonRunner):
def __init__(self):
super().__init__()
def runCommand(self, command, messageOnError):
try:
exec(command, globals())
return True
except Exception as e:
if messageOnError == "":
print("Error with command:\n")
else:
print(messageOnError + "\n")
print(command)
print("Catched exception:\n", e)
return False
def evalCommand(self, command, result):
try:
result = eval(command) # problem here
return True
except Exception as e:
print("Error with command:\n")
print(command)
print("Catched exception:\n", e)
return False
qgis.core.QgsPythonRunner.setInstance(PythonRunnerStandalone())
In QGIS source code qgis-latest/qgis-3.8.2/src/gui/qgsattributeform.cpp lines
1610-1625:
QString numArgs;
// Check for eval result
if ( QgsPythonRunner::eval( QStringLiteral(
"len(inspect.getargspec(%1)[0])" ).arg( initFunction ), numArgs ) ) // Here
numArgs is passed as reference to my python implementation
[...]
if ( numArgs == QLatin1String( "3" ) ) // always fails here
Jacky Volpes
HR Team pour la Société du Canal de Provence | 2SI - QGIS
Le Tholonet, CS70064
[email protected]<mailto:[email protected]>
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