Avoiding eval() function when using custom function

Question

CONTEXT

I have several raster layers which overlap. I created a point layer with labels that represent different values for each of these raster layers. I created a function that

lists all layer names in the canvas
if the layer in the canvas is checked, its name is compared to a list string representing layer names that I would like to label.
if they match, then the function returns a string that represents a string representing the field name.

When I use this function for labeling, I still need to use eval() to get the value I want to show as a label.

Is there a way to skip from using eval()?

The function is shown below:

from qgis.core import *
from qgis.gui import *

@qgsfunction(args='auto', group='Custom', referenced_columns=[])
def raster_label(feature, parent):
    field_list = ["LAYER_A", "LAYER_B", "LAYER_C", "LAYER_D"]
    raster_layer_name_list = ["A", "B", "C", "D"]
    field_to_show = ""

    # create a list with all layers in the canvas
    layers_list = []
    root = QgsProject.instance().layerTreeRoot()
    for child in root.children():
        if isinstance(child, QgsLayerTreeGroup):
            get_group_layers(child)
        elif isinstance(child, QgsLayerTreeLayer):
            layers_list.append(child.name())
        
    # iterate over all items in layer list ...
    for item in range(0, len(layers_list)):
        l = QgsProject().instance().mapLayersByName(layers_list[item])
        node = root.findLayer(l[-1])
        # then identify if the layer is checked ...
        if node.isVisible() == True:
            # to compare agains the raster_layer_name_list ...
            for raster in range(0, len(raster_layer_name_list)):
                # returning then the corresponding field
                if l[0].name() == raster_layer_name_list[raster]:
                    field_to_show = field_list[raster]
                    break
            if field_to_show != "":
                break
        if field_to_show != "":
            break
    
    return field_to_show

Kadir Şahbaz · Accepted Answer · 2022-12-17 17:52:42Z

6

You cannot avoid using eval as long as you return a field name (it is a string) to the expression area. It will be considered as a string in Expression Dialog.

But you can use the conditional return instead of return field_to_show as follows:

from qgis.core import *
from qgis.gui import *

@qgsfunction(args='auto', group='Custom', referenced_columns=[])
def raster_label(feature, parent):
    
    #
    # previous lines
    #

    if field_to_show = "":
        return ""
    else:
        return feature[field_to_show]

answered Dec 17, 2022 at 17:52

Kadir Şahbaz

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