How to query a specific layer from a Feature Service

Question

Hello everyone,
I am trying to get the list of feature service which has a specific layer (for example WETTBEWERBER (the layer id is 0)) from a feature service,

from arcgis.gis import GIS
gis = GIS(url='https://url/portal', username='arcgis_python', password='amazing_arcgis_123')

layer_items = gis.content.search(query="WETTBEWERBER", item_type="Feature Layer")
for layer in layer_items:
    if "WETTBEWERBER" in layer.title.upper():
        print(f"Layer name: {layer.title}, 	Owner: {layer.owner}, 	Layer ID: {layer.id}")

But no results are printed out, but the layer WETTBEWERBER is present in most of my feature services. What can be done in this case ??

I have tried to get the layers first using the following code to check whether I can get the layers using the API and I can get the layers.

layer_items = gis.content.search(query="5fg239sdsfs56sdfcs466s4f", item_type="Feature Service")

for item in self.api_query_result:
    layers = item.layers
    for lyr in layers:
        print(item)
        try:
            print(item.id, item.url, lyr.properties["id"], lyr.properties["name"])
        except Exception as e:
            print(e)

I can get all the layers including the layer WETTBEWERBER (0) in it.

But I cannot list all the feature service which contains this particular layer, What am I doing wrong ??

Thank you for the insights.

Aravinthkumar · Answer

Thank you for the detailed explanation, whether I should leave the query empty in the place of "some string" and kindly can you edit it with the exact syntax I cannot figure out from this, that will be really helpful.

jcarlson · Answer

Looking at it again, a layer doesn't have a title property. The parent item has a title, but the layers have properties, in which you can find a name, as shown in the other code block you quoted.

To get the feature service itself, you have to reference the parent item. This snippet would return a list of all Feature Layers in your search:

items = gis.content.search('some string', item_type='Feature Service')

[a.url for b in items for a in b.layers]

To filter this list, we can include a simple ternary operator in the same line.

[a for b in items for a in b.layers if a.properties['name'] == 'WETTBEWERBER']

I tested this on my own layers, and searching for the layer name "Records" across a bunch of layers yielded the following:

['https://maps.co.kendall.il.us/server/rest/services/Cadastral/Ordinance_Fabric/FeatureServer/1',
 'https://maps.co.kendall.il.us/server/rest/services/Hosted/ord_fabric_hosted/FeatureServer/0']

Aravinthkumar · Answer

Yes, sorry for the typo here it is "layer.title.upper()" I have updated the post.