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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.

Python password alarm for five minutes

I have written this python program for a game:
print("CC ACTIVATED")
import lcddriver
import time
import subprocess
display = lcddriver.lcd()
try:
display.lcd_display_string("CC... ", 1)
time.sleep(2)
display.lcd_display_string("ONLINE", 2)
time.sleep(2)
display.lcd_clear()
except Exception:
print("SCREEN ERROR")
try:
display.lcd_display_string("SETUP A", 1)
display.lcd_display_string("PASSWORD? Y/N", 2)
except Exception:
print("SCREEN ERROR")
activate = input("")
if activate == 'y':
print("ACTIVATED")
try:
display.lcd_clear()
display.lcd_display_string("", 1)
time.sleep(2)
display.lcd_display_string("LOADING", 2)
time.sleep(2)
display.lcd_clear()
except Exception:
print("SCREEN ERROR")
else:
print("ABORT")
try:
display.lcd_clear()
display.lcd_display_string("", 1)
time.sleep(2)
display.lcd_display_string("ABORT", 2)
time.sleep(2)
display.lcd_clear()
subprocess.call(["sudo","halt"])
except Exception:
print("SCREEN ERROR")
subprocess.call(["sudo","halt"])
k = True
while k:
try:
display.lcd_clear()
display.lcd_display_string("ENTER PASWORD", 1)
display.lcd_display_string("----------------", 2)
except Exception:
print("SCREEN ERROR")
pasword = input("")
display.lcd_clear()
try:
display.lcd_clear()
display.lcd_display_string("YOU TYPED:", 1)
display.lcd_display_string(pasword, 2)
time.sleep(2)
display.lcd_display_string("CONFIRM? Y/N", 1)
except Exception:
print("SCREEN ERROR")
ok = input("")
if ok == 'y':
k = False
else:
display.lcd_clear()
try:
display.lcd_clear()
display.lcd_display_string("PASSWORD", 1)
display.lcd_display_string("SET", 2)
except Exception:
print("SCREEN ERROR")
time.sleep(2)
run = True
try:
display.lcd_clear()
display.lcd_display_string("STARTING ", 1)
display.lcd_display_string("GAME...", 2)
except Exception:
print("SCREEN ERROR")
time.sleep(2)
while run:
try:
display.lcd_clear()
display.lcd_display_string("ENTER PASSWORD ", 1)
display.lcd_display_string("TO DEACTIVATE", 2)
except Exception:
print("SCREEN ERROR")
pasword1 = input("")
if pasword1 == pasword:
try:
display.lcd_clear()
display.lcd_display_string("PASSWORD....", 1)
time.sleep(2)
display.lcd_display_string("ACCEPTED", 2)
time.sleep(2)
display.lcd_clear()
display.lcd_display_string("DEACTIVATED", 2)
subprocess.call(["sudo","halt"])
time.sleep(10)
except Exception:
print("SCREEN ERROR")
subprocess.call(["sudo","halt"])
else:
try:
display.lcd_clear()
display.lcd_display_string("PASSWORD....", 1)
time.sleep(2)
display.lcd_display_string("UNACCEPTED", 2)
time.sleep(2)
except Exception:
print("SCREEN ERROR")
It runs on a raspberry pi and works absolutely fine, but there is one thing I want to do that is a bit above my skill level. I need this program to give you three attempts to guess the password,then at the fourth attempt play an audio recording. Afterward it needs to respond to failed attempts with the audio recording for five minutes and then go back to the silent three chances. I know how to get it to do the audio recording, but I need help with the five minute part. Does anyone here know how to do this? Thanks.

To permit three password attempts, you will need a counter variable which will count up to three with each failed attempt.
Rather than just running while k == True, you want to check against both k == True and attempts <= 3. As far as waiting five minutes, you will want to create a timestamp of when the user last guessed.
from datetime import datetime
...
k = True
password_attempts = 0
while k and password_attempts < 4:
try:
display.lcd_clear()
display.lcd_display_string("ENTER PASWORD", 1)
display.lcd_display_string("----------------", 2)
except Exception:
print("SCREEN ERROR")
pasword = input("")
display.lcd_clear()
try:
display.lcd_clear()
display.lcd_display_string("YOU TYPED:", 1)
display.lcd_display_string(pasword, 2)
time.sleep(2)
display.lcd_display_string("CONFIRM? Y/N", 1)
except Exception:
print("SCREEN ERROR")
ok = input("")
if ok == "y":
k = False
else:
display.lcd_clear()
password_attempts += 1
if password_attempts == 4:
#Create a timestamp of last attempt
last_request = datetime.now()
current_request = datetime.now()
while((last_request - current_request).total_seconds() < 5 * 60):
#Code to play music
current_request = datetime.now()
Then it's up to you to re-route the player back to the point in the program where they try entering the passwords again.
But, the design choices you've made at the beginning of this project will really start to get in the way of what you want to do.
I suggest you consider a redesign of your program, in order to make working on it much easier and more pleasant.
One of the best parts of programming is code re-use; this is most easily understood with functions. In Python, you create a function with the syntax def function_name():
Then, when you want to run the code within that function, you simply call it by name: function_name().
What does this mean for your app?
You could wrap writing to the LCD in a method, so that you can avoid repeatedly typing out the same try:... except: block (by the way... when you write except Exception, you are catching ALL exceptions your program might throw. This is considered bad practice, because you may not want to do the same thing irrespective of what the exception was in the first place. Rather, try to find a specific exception you're trying to catch, and catch that one by name, specifically).
An example of rewriting the LCD printing:
def print_to_lcd(statements_to_print: [str]):
line_number = 1
try:
for statement in statements_to_print:
display.lcd_display_string(statement, line_number)
line_number += 1
except Exception:
print("SCREEN ERROR")
Then, you just produce a list of things for the LCD to print, and call the LCD function with them:
things_to_write = ["PASSWORD...", "UNACCEPTED"]
print_to_lcd(things_to_write)
The same can be said about how you check your password value, when you play the music, etc.
Any time you find yourself repeatedly re-writing the same information, take a moment and think if there's a way you can write it once, and just re-use it. It'll make your programs much easier to write, and much easier to use.
Particularly in this case, since you want to re-do the same thing after waiting for a period of time. Imagine, rather than having to write tons of if-statements, when the user enters the password wrong for the fourth time, you just write something like:
play_music_for_five_minutes()
prompt_for_password()
Let me know if there are any further questions I can answer!

handle the eval function exceptions

I'm using the eval method in my project and I have to handle all the possible exceptions it may raise. I thought that simple try and except would work but when I enter a symbol it doesn't familiar with it stuck.. like eval("5:2").
try:
max = abs(eval(game_function))
except:
while True:
x += (x_f_coordinate - x_0_coordinate) / 1050
try:
max = abs(eval(game_function))
break
except KeyboardInterrupt:
raise
except:
if x < x_f_coordinate:
continue
It's freezing in the 7th line.
Any suggestions?

Re-enter try statement after an except

In my code below, it stops executing when I hit an exception. How can I get it to re-enter the try statement from where the exception left off? Maybe I am looking for a different way to tackle this without a try-except statement?
import requests
from requests import exceptions
contains_analyst = []
try:
for x in data:
r = requests.get(str(x), timeout=10, verify=False)
if "analyst" in r.text:
contains_analyst.append("Analyst")
print "Analyst # %s" % x
else:
contains_analyst.append("NOPERS")
print "Nopers"
except exceptions.RequestException:
contains_analyst.append("COULD NOT CONNECT")

You should put the try/except around only the part whose error you want to trap. In your example, it looks like you want something more like this:
for x in data:
try:
r = requests.get(str(x), timeout=10, verify=False)
except exceptions.RequestException:
contains_analyst.append("COULD NOT CONNECT")
else:
if "analyst" in r.text:
contains_analyst.append("Analyst")
print "Analyst # %s" % x
else:
contains_analyst.append("NOPERS")
print "Nopers"
Here I use the else clause of the try block to handle the case where no exception is raised (see documentation). In many cases, if you don't need to do anything else after the exception, you could just return at that point and put the following no-exception code in the main function body, reducing indentation a bit:
for x in data:
try:
r = requests.get(str(x), timeout=10, verify=False)
except exceptions.RequestException:
contains_analyst.append("COULD NOT CONNECT")
return contains_analyst
# execution reaches here if no exception
if "analyst" in r.text:
contains_analyst.append("Analyst")
print "Analyst # %s" % x
else:
contains_analyst.append("NOPERS")
print "Nopers"
Of course, whether it makes sense to return at that point depends on the surrounding context of your code.