I would like for the loop to keep going even exception is generated at first iteration. How to do this?
mydict = {}
wl = ["test", "test1", "test2"]
try:
for i in wl:
a = mydict['sdf']
print(i)
except:
# I want the loop to continue and print all elements of list, instead of exiting it after exception
# exception will occur because mydict doesn't have 'sdf' key
pass
You can use dict.get(). It will return None if the key not exist. You can also specify default value in dict.get(key, default_value)
for i in wl:
a = mydict.get('sdf')
print(i)
The best I can advice is to move the try inside the loop as below:
mydict = {}
wl = ["test", "test1", "test2"]
for i in wl:
try:
a = mydict['sdf']
print(i)
except:
continue
print will be skipped.
print to be first.
except that catches all exceptions is bad practice. Only catch the exceptions that you know might be raised.This is my approach for your problem
Hope its work for you as you want
mydict = {}
wl = ["test", "test1", "test2"]
for i in wl:
try:
a = mydict['sdf']
except:
pass
print(i)
try/exceptinside loop body