I want to exception handle 'list index out of range.'

Question

I am using BeautifulSoup and parsing some HTMLs.

I\'m getting a certain data from each HTML (using for loop) and adding that data to a cert

Answer 1

For anyone interested in a shorter way:

gotdata = len(dlist)>1 and dlist[1] or 'null'

But for best performance, I suggest using False instead of 'null', then a one line test will suffice:

gotdata = len(dlist)>1 and dlist[1]

Answer 2

Handling the exception is the way to go:

try:
    gotdata = dlist[1]
except IndexError:
    gotdata = 'null'

Of course you could also check the len() of dlist; but handling the exception is more intuitive.

Answer 3

Taking reference of ThiefMaster♦ sometimes we get an error with value given as '\n' or null and perform for that required to handle ValueError:

Handling the exception is the way to go

try:
    gotdata = dlist[1]
except (IndexError, ValueError):
    gotdata = 'null'

Answer 4

for i in range (1, len(list))
    try:
        print (list[i])

    except ValueError:
        print("Error Value.")
    except indexError:
        print("Erorr index")
    except :
        print('error ')

Answer 5

You have two options; either handle the exception or test the length:

if len(dlist) > 1:
    newlist.append(dlist[1])
    continue

or

try:
    newlist.append(dlist[1])
except IndexError:
    pass
continue

Use the first if there often is no second item, the second if there sometimes is no second item.

Answer 6

A ternary will suffice. change:

gotdata = dlist[1]

to

gotdata = dlist[1] if len(dlist) > 1 else 'null'

this is a shorter way of expressing

if len(dlist) > 1:
    gotdata = dlist[1]
else: 
    gotdata = 'null'