placing file in SFTP
i created folder and changed the information for the folder to an SFTP .
Below is my code :
# time
import time
current_day = time.strftime("%Y%m%d")
name = "REGN_CONFIG_VALUES_"+current_day+".xlsx"
path = os.path.join(new_file_info["path"], name)
writer = pd.ExcelWriter(path)
study_config_values_stacked_distinct_df.to_excel(writer, sheet_name='Study Config Values',index=False, encoding='utf-8')
latest_file_data_df.to_excel(writer, sheet_name='Country Config Values',index=False, encoding='utf-8')
latest_file_data_df1.to_excel(writer, sheet_name='Site Config Values',index=False, encoding='utf-8')
latest_file_data_df2.to_excel(writer, sheet_name='Regn Codelist Values',index=False, encoding='utf-8')
writer.save()
I get this error each time i run the code :
--------------------------------------------------------------------------- KeyError Traceback (most recent call last) <ipython-input-8-76c9459ddcf4> in <module> 43 name = "REGN_CONFIG_VALUES_"+current_day+".xlsx" 44 ---> 45 path = os.path.join(new_file_info["path"], name) 46 47 writer = pd.ExcelWriter(path) KeyError: 'path'
Operating system used: windows
Answers
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Miguel Angel Dataiker, Dataiku DSS Core Designer, Dataiku DSS ML Practitioner, Dataiku DSS Adv Designer, Registered Posts: 118 Dataiker
Hi,
The "KeyError: 'path'" exception is telling us there was an issue retrieving the key you were looking for. This error normally occurs when trying to access a key that is not in the dictionary.
Can you doublecheck the "new_file_info" dictionary has a "path" key?
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how do i check that ?
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no it does not
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How can i have my sftp sharing same path information in the folder?
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Miguel Angel Dataiker, Dataiku DSS Core Designer, Dataiku DSS ML Practitioner, Dataiku DSS Adv Designer, Registered Posts: 118 Dataiker
Hi,
From the snippet shared we do not know what 'new_file_info' is or how it has been constructed. Since we have in the snippet 'new_file_info["path"]', we can make the assumption it is a dictionary that expects to have a 'path' key.
In python, you can print all the keys of a dictionary by running: print(new_file_info.keys())
If as I suspect the 'path' key does not exist, you'd need to modify your code or contact the code's author to fix this.