Variable assignment from unknown length list in Python

Your second attempt is probably more pythonic. Of course, you don't know in advance whether the result of .find_all will be a list of length ==3 (or more, or less). So you should use the try/except or other logic to control how/when the results are written to your database.

# create a dictionary of your database column names:
dbColumns = {0:'column1', 1:'column2', 2:'column3'}

# get all the results; there might be 0 or 3 or any number really, 
#     we'll deal with that later
results = [s.text if s.text else "" for s in soup.find_all('a')]

# iterate the items in the list, and put in corresponding DB
for col in range(len(results)):
    # use the dbColumns dict to insert to the desired column

    query = "Insert INTO [db_name].[" + dbColumns[col] + "]"
    query += "VALUES '" + results[i] + '"

    """
    db.insert(query)  # assumes a db object that has an "insert" function; modify as needed
    """

The point of this approach is that there seems to be nothing about this problem that technically would require hardcoding exactly three objects (var1, var2, var3) and trying to assign to these. Instead, just return the results of find_all and deal with them by their index within that resulting list.

You can use some simple list multiplication:

# use a constant at the top of your script in case the number of columns
# change in the future
COLUMNS = 3

# ... other code ...

all = [s.text for s in soup.find_all('a')]
all.extend(['']*(COLUMNS-len(all))) # append 1 empty string for each missing text field
var1, var2, var3 = all 

But as David Zemens has mentioned in the comments, there has got to be a better way to do this. I can't make any concrete suggestions without seeing the code that consumes your text variables, but you should seriously reconsider your design. Even if you use the constant like I suggested, having var1, var2, var3 = all is still going to make it difficult to maintain and modify this script in the future.

Based on your edit, I would suggest you use a dictionary instead. This will allow you to reference specific data by name, like you would reference a variable, but retains the flexibility of a list instead of restricting you to the number of variables you have hard coded.

For example:

all = [s.text for s in soup.find_all('a')]

d = {}
for i, field in enumerate(all): 
    d['var{}'.format(i)] = field

# later in your code that consumes this dictionary...

try:
    foo(d['var1']) # function to do something with the scraped string corresponding
                   # to var1
except KeyError:
    # do something else or pass when the expected data doesn't exist

If all is ['a', 'b'], then this code produces this:

{'var1': 'b', 'var0': 'a'}

Variable assignments are really nothing more than a mapping - your code knows the variable name, and it can look up the corresponding value. A dictionary lets your code build the mapping on the fly instead of you having to hard code it. Now we have built a dictionary where the varX variables are constructed dynamically. If you decide to add another column, you don't have to change this code at all. You just add your code that would use var4 and be ready to catch the exception if var4 doesn't exist in the dictionary. No more adding empty strings - your code is ready to handle the case where the data it's looking for doesn't exist.

Notes:

1. The enumerate() function iterates over an iterable object and increments a counter for you. In my code, i is the counter (so we can construct the 'var1', 'var2'... strings), and field is each item from the list.

How about:

all = soup.find_all('a')
var1 = all[0].text if len(all) > 0 else ""
var2 = all[1].text if len(all) > 1 else ""
var3 = all[2].text if len(all) > 2 else ""

The conditional expression x if y else z (often called a ternary operator) keeps the code simple and readable. It's not going to win any design awards though.

you can try this

#if list1 has uncertain number of values and you want to give them each variable 

#create random list2 with max number of possible veriables 
list2 = ['var1', 'var2', 'var3', 'var4' , . . . ]

for li1, li2 in zip(list1, list2):
    globals()[li2] = li1
    print(li2)

I am not pro in python, i just figured this out on my own it might not be very pythonic but it solves the problem