I'm supposed to publish each file as a uniquely names feature service, but despite my approach I keep having issues where my csv files aren't found... a friend advised me to add my initials to each CSV because sometimes arcgis doesn't like when the names aren't unique enough, but I keep getting the same errors. any ideas?

from arcgis.gis import GIS
import os
import pandas as pd
from arcgis.gis import GIS
from arcgis.features.analysis import *
from arcgis.features import FeatureLayer

gis = GIS("http://slustl.maps.arcgis.com/home", "SLysy_SLUSTL")

data_dir = r"C:\Users\Stephanie\NOTEBOOKS\FINAL\FSI"
file_list = os.listdir(data_dir)

csv_file_list = [x for x in file_list if x.endswith(".csv")]
print("Number of .csv files found: " + str(len(csv_file_list)))

for csv in csv_file_list:
    item = gis.content.add({}, data_dir + csv)   # .sd file is uploaded and a .sd file item is created
    published_item = item.publish()   


RuntimeError: File(C:\Users\Stephanie\NOTEBOOKS\FINAL\FSIFSI-2006sl.csv) not found.

Print output

  • csv_file_list = [os.path.join(data_dir,x) for x in file_list if x.endswith(".csv")]. Or is each csv name starting with FSI-? Then remove \FSI from data_dir – BERA May 4 at 5:29

The path you're building up is probably NOT correct per:

data_dir = r"C:\Users\Stephanie\NOTEBOOKS\FINAL\FSI"
item = gis.content.add({}, data_dir + csv) 

I'd expect this to create a path like: C:\Users\Stephanie\NOTEBOOKS\FINAL\FSImycsv.csv (note the lack of \ separator between path and file name) I'd suggest doing your path like: item = gis.content.add({}, os.path.join(data_dir, csv))

Alternatively, you may want to consider glob, as it handles a lot of the listing of files for you. For example:

import glob
csv_list = [f for f in glob.glob("c:/temp/*.csv")]       
['c:/temp\\zzzz.csv', 'c:/temp\\aaaaa.csv', 'c:/temp\\bbbb.csv', 'c:/temp\\foo.csv']

Or better yet, just let glob run your loop and skip building the list.

for f in glob.glob("c:/temp/*.csv"):
   item = gis.content.add({}, f)   
   published_item = item.publish()  
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