how to pull google drive file id's from list python?

Question

Hello everyone I have multiple files in my google drive that I need to get the file id for to share download links to people. But I cant seem to get but 1 file id at a time. Is there a way to pull all the file ids and match them with a list?

In my list A,B C

each letter represents an order number I was trying to keep it simple but A would actually be an order number 3472834, B= 3293881, C =,3498249.

In google drive example: File1,File2,File3

The files that are in my google drive are pdf printouts of the orders. They are actually named the order numbers File 1= 3472834, File 2= 3293881, File 3 =3498249

So my goal is to iterate over the ListIDS each one and get the corresponding Google file id. But I cant seem to get it to do but the first google file id each time.

Code:

ListIDS= ['A','B','C']
for eachId in ListIDS:

service = create_service(CLIENT_SECRET_FILE, API_NAME, API_VERSION, SCOPES)
filename= f'{ListID}'
query = "name contains " + "'"+ filename +"' and trashed = false"
response = service.files().list(q=query, fields= 'files(id,name)').execute()
files = response.get('files')
if files:
  print("File Found!")
  google_file_name=  files[0].get('name')
  google_file_id = files[0].get('id')
  print (google_file_id)
  print(google_file_name)
else:
  print("FileNot Found")

FileURL= "https://drive.google.com/file/d/"+ google_file_id + "/view?usp=sharing"

But again the problem is I only get the 2nd value B for all 3 lists id's printed out for example

output:

https://drive.google.com/file/d/FileB_ID_Path/view?usp=sharing

But my expected outcome would be:

https://drive.google.com/file/d/FileA_ID_Path/view?usp=sharing
https://drive.google.com/file/d/FileB_ID_Path/view?usp=sharing
https://drive.google.com/file/d/FileC_ID_Path/view?usp=sharing

Any help please.

Answer 1

I believe your goal is as follows.

• You have the filenames like ListIDS= ['A','B','C']. The files of those filenames are put in your Google Drive.
• You want to retrieve the file IDs of the files using the filenames.
• You want to achieve this using googleapis for python.

In this case, how about the following modification?

Pattern 1:

In this pattern, using ListIDS= ['A','B','C'], one search query is used.

ListIDS = ['A','B','C']  # filenames

service = create_service(CLIENT_SECRET_FILE, API_NAME, API_VERSION, SCOPES)
qFiles = ["name contains " + "'" + e + "'" for e in ListIDS]
query = "(" + " or ".join(qFiles) + ") and trashed = false"
response = service.files().list(q=query, fields='files(id,name)').execute()
files = response.get('files')
values = {}
for e in files:
    for f in ListIDS:
        if f in e.get('name'):
            id = e.get('id')
            v = "https://drive.google.com/file/d/" + id + "/view?usp=sharing"
            if values.get(f):
                values[f].append(v)
            else:
                values[f] = [v]
print(values)

• When this script is run, the following JSON object is returned.

    {
      'A': ["https://drive.google.com/file/d/###/view?usp=sharing"],
      'B': ["https://drive.google.com/file/d/###/view?usp=sharing"],
      'C': ["https://drive.google.com/file/d/###/view?usp=sharing", "https://drive.google.com/file/d/###/view?usp=sharing"]
    }
  

• From your showing script, name contains '###' is used as the search query. So, when there are multiple files are returned by one filename, those file IDs are put in the array.

Pattern 2:

If ListIDS is large, the above pattern might not be able to be used. So, when pattern 1 cannot be used, please use this pattern 2. In this pattern, using ListIDS= ['A','B','C'], the multiple search queries are used.

ListIDS = ['A','B','C']  # filenames

service = create_service(CLIENT_SECRET_FILE, API_NAME, API_VERSION, SCOPES)
files = []
for filename in ListIDS:
    query = "name contains " + "'" + filename + "' and trashed = false"
    response = service.files().list(q=query, fields='files(id,name)').execute()
    files.extend(response.get('files'))
values = {}
for e in files:
    for f in ListIDS:
        if f in e.get('name'):
            id = e.get('id')
            v = "https://drive.google.com/file/d/" + id + "/view?usp=sharing"
            if values.get(f):
                values[f].append(v)
            else:
                values[f] = [v]
print(values)

• In this case, the same result as the above one is obtained.