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How to try-except block exit when try is working

The code I have works. The last for loop gets reached and choice.click() gets clicked on. The question I have is why the except block gets executed even though the:
if choice.text == call_resource:
choice.click()
break
piece of the code is reached and the choice.click() portion works?
def select_choice(driver, resource_tag):
try:
call_resource = None
access_data = common.retrieve_tag_access_data()
for row in access_data["access_data"]:
if str(row["id"]) == resource_tag:
call_resource = row["row_tag"]["name"]
break
if call_resource is None:
access_data = common.retrieve_fixture_access_data()
for row in access_data["access_data"]:
if str(row["id"]) == resource_tag:
call_resource = row["row_tag"]["name"]
break
menu = driver.find_element_by_css_selector("ul[role='listgrid']")
choices = menu.find_elements_by_css_selector("li[role='choices']")
for choice in choices:
if choice.text == call_resource:
choice.click()
break
except:
error(logger, "unable to select choice")
pass
Because the last for loop works, shouldn't it break entirely out of the function after choice.click() without executing the except: portion of the code?

The except: portion will run only if an Exception occurred inside the try block. You should change the except: line to something like except Exception as e: and print the variable e in some way so you can discover what the problem is.

choices = menu.find_elements_by_css_selector("li[role='choices']")
for choice in choices:
if choice.text == call_resource:
choice.click()
break
Could be replaced with since your only trying to click one element with a certain text why not just try to find it. If it errors out it will go to the exception you provided.
driver.find_element(By.XPATH,f"//li[#role='choices' and contains(text(),{call_resource})]").click()
Also use to find errors use the following.
except Exception as e:
print(str(e))

Exception In Python Doesn't Do Anything

I am beginner to python and I have doubt in one program when using try except part
I try to take input from user and add that number by 1 like if you enter 5 then 6 will be printed but if you enter string then except would come in role and print enter only number
I've written code but except is not working there, code is working like if I enter 5 then 6 gets printed but except statement is not working
Here is the code that I've written
def increment(num):
try:
return num + 1
except Exception as e:
print("ufff error occured :", e)
n = int(input("enter your num: "))
a = increment(n)
print(a)

You would have to change to the following:
def increment(num):
try:
return int(num) + 1
except Exception as e:
print("ufff error occured :", e)
n = input("enter your num: ")
a = increment(n)
print(a)
The reason for that is that the exception was raised by the int() functions, because it couldn't convert your input to int. The call to int() was before the increment function. Therefore the exception was unhandled.

You tried converting the input to int before going into the try/except block. If you just put it in the try block it will work.
def increment(num):
try:
num = int(num)
return num + 1
except Exception as e:
print("ufff error occured :", e)
n = input("enter your num: ")
a = increment(n)
print(a)

You have to use the type conversion within the try block so that if num is not a number, the controller would move to expect portion.
The problem here is you are using type conversion outside of try-catch block.
Another point to note, convert e to string as a good practice.
Third point to note,the last line print(a) will itself cause trouble in cases of exception as you have returned nothing from the exception part and directly printed the error. Use a return statement in there else it would return None type and which itself will mess up the flow.
def increment(num):
try:
return int(num) + 1 #Updated Line
except Exception as e:
return "ufff error occurred :" + str(e) #Updated Line
n = input("enter your num: ")
a = increment(n)
print(a)

Might I recommend the following?
def increment(num):
try:
return int(num) + 1
except Exception as e:
print("ufff error occurred:", e)
n = input("enter your num: ")
a = increment(n)
print(a)
The only difference here is that the attempt to convert the user's input to an int occurs within the try-except block, rather than outside. This allows the type conversion error to be caught and your message ("ufff error occurred: ...") to be displayed.
Edit: regarding your comment:
yes it worked out thanks everyone, but can anyone tell what should i have to do if i use int with input line what changes i've to make
Optionally, you could do the following:
def increment(num):
return num + 1
try:
n = int(input("enter your num: "))
a = increment(n)
print(a)
except Exception as e:
print("ufff error occurred:", e)
In this case, I would recommend removing the try-except block in the increment function, because you can safely assume that all values passed to that function will be numeric.

The only code that would raise an exception is the int(...) conversion - and it is not protected under a try/except block. You should put it under the try:
def increment():
try:
return int(input("enter your num: ")) + 1
except ValueError as e:
return f"ufff error occured: {e}"
print(increment())
You should always try to catch the narrowest exception. In this case, a ValueError.

how to catch error(200)

I'm trying to get input from user, until he press ctrl-c. yet, I can't catch the
error, I think it has something to do with sklearn (I imported it for the rest of the code)
this is the code:
try:
while(True):
i+=1
put = input("\tEnter name of feature number " + str(i) +":\t")
features.append(put)
except KeyboardInterrupt:
print("\n\tFeatures Added!")
sleep(SLEEP)
return None
except:
exit("\nError has occurred, please come back later...")`

Fix your indentation as the following:
try:
while(True):
i+=1
put = input("\tEnter name of feature number " + str(i) +":\t")
features.append(put)
except KeyboardInterrupt:
print("\n\tFeatures Added!")
sleep(SLEEP)
return None
except:
exit("\nError has occurred, please come back later...")

Python: run "try" again after exception caught and worked out

Is there any way to enter try statement once again after catching the exception in the first try?
Now I'm using "while" and "if" statements and it is making the code messy.
Any ideas?
Will try to simplify it as possible, sorry that have no logic...
run = True
tryAgain = True
a=0
while run:
try:
2/a
except Exception:
if tryAgain:
tryAgain = False
a = 1
else:
run = False

You could try using a break statement in your try block:
while True:
try:
# try code
break # quit the loop if successful
except:
# error handling

Considering you are doing this in a while, then you can make use of continue to just continue back to the beginning of the while loop:
tryAgain = True
a=0
while True:
try:
2/a
break # if it worked then just break out of the loop
except Exception:
if tryAgain:
continue
else:
# whatever extra logic you nee to do here

I like using a for loop so that the trying and trying doesn't go on forever. Then the loop's else clause is a place to put the "I give up" code. Here is a general form that will support 'n' retries > 1:
a=0
num_tries = 5
for try_ in range(0,num_tries):
try:
2/a
except Exception:
print("failed, but we can try %d more time(s)" % (num_tries - try_ - 1))
if try_ == num_tries-2:
a = 1
else:
print("YESS!!! Success...")
break
else:
# if we got here, then never reached 'break' statement
print("tried and tried, but never succeeded")
prints:
failed, but we can try 4 more time(s)
failed, but we can try 3 more time(s)
failed, but we can try 2 more time(s)
failed, but we can try 1 more time(s)
YESS!!! Success...

I'm new to Python, so this may not be best practice. I returned to the try statement after triggering the exception by lumping everything into a single function, then recalling that function in the except statement.
def attempts():
while True:
try:
some code
break #breaks the loop when sucessful
except ValueError:
attempts() #recalls this function, starting back at the try statement
break
attempts()
Hope this addresses your question.

Repeat an iteration in loop if error occurs

Is there a command such as break and continue which could repeat recent iteration?
For example, when exception is thrown.
for i in range(0,500):
try:
conn = getConnection(url+str(i))
doSomething(conn)
except:
repeat
Let's have an iteration where i variable's value is 6. During this iteration some connection error occurred. I want to repeat this iteration.
Is there a command which can do that?
Of course I can do this:
i=0
while i!=500:
try:
conn = getConnection(url+str(i))
doSomething(conn)
i+=1
except:
pass

No, there is no command to "rewind" a for-loop in Python.
You could use a while True: loop inside the for-loop:
for i in range(500):
while True:
try:
conn = getConnection(url+str(i))
doSomething(conn)
except Exception: # Replace Exception with something more specific.
continue
else:
break
or without the else::
for i in range(500):
while True:
try:
conn = getConnection(url+str(i))
doSomething(conn)
break
except Exception: # Replace Exception with something more specific.
continue
But I personally think that your proposed solution is better because it avoids an indentation level.

for i in range(500):
while True
try:
conn = getConnection(url+str(i))
break
except Exception: # still allows to quit with KeyboardInterrupt
continue
do_your_stuff()
This looks bit risky, however, you should at least enable some logging inside a while block.
If you expect to use it in more places, you might write a simple decorator:
def keep_trying(fn, *args, **kwargs):
def inner(*args, **kwargs):
while True:
try:
return fn(*args, **kwargs)
except Exception:
continue
return inner
# later you can use it simple like this:
for i in range(500):
conn = keep_trying(getConnection)(url+str(i))

You can use generators :
def process_connections(n_connections, url, max_tries=50):
i = 0
try_count = 0
while i < n_connections:
try:
conn = getConnection(url+str(i))
yield conn
except:
try_count += 1
if try_count > max_tries:
raise Exception("Unable to connect after %s tries" % max_tries)
else:
i += 1 # increments only if no exception
And you perform your operations :
for conn in process_connections(500, url):
do_something(conn)

You can use nested for loops to put a cap on the number of times you retry the operation. This is bascially the sam as #PierreAlex's generator answer but without the extra function definition.
for i in range(500):
for retry in range(10):
try:
conn = getConnection(url+str(i))
doSomething(conn)
except Exception: # Replace Exception with something more specific.
time.sleep(1)
else:
print "iteration", i, "failed"

Why not just use an if statement?
n=6
i=0
while i!=500:
failed = False;
try:
conn = getConnection(url+str(i))
doSomething(conn)
i+=1
except:
#handle error
failed = True;
#try again if n-th case failed first time
if(i == n and failed):
try:
conn = getConnection(url+str(i))
doSomething(conn)
except:
#handle error

Here is one. You would need to add a logging or alert system to let you know that something is stuck:
state = "" #state of the loop
# If there is no error continue. If there is error, remain in loop
while True:
if state != "error":
try:
1/0 # command
break # no error so break out of loop
except:
state = "error" #declare error so maintain loop
continue
elif state == "error": # maintain loop
continue