Hello everyone I have my code almost done but I'm trying to add in some sort of check to avoid errors. But I'm not really understanding what statement would be better to use to test the code. I know there are a few options of either using a loop, if-statement, or try. But here is the code in regards to doing captcha. I need it to run the first set of code which if the captcha doesn't pop up I continue on. But if the captcha does pop up solve it then continue on.
Some times captcha doesnt appear and if I run the whole set of code I get an error because we are expecting captcha to pop up.
Or if the captcha does appear to solve it which would.
I would really appreciate any help please as I'm not sure the right statement to use.
try should be used when something would return an error and would otherwise cause your program to stop/crash. An example of this would be:
try:
import pandas
except:
print("Unable to import pandas. Module not installed")
In this example your program will attempt to import the pandas module. If this fails it will then print out a line of text and continue running.
if statements are used to decided when to do something or not based on the returned logic. The key difference is that logic IS returned and not an error.
if x > 10:
print("This is a large number")
else:
print("This is a small number")
With this example, if 'x' did not exist it would produce an error, no more code will be executed, and the program will crash. The main difference between IF and TRY is whether logic is returned as true/false or is something just plains fails.
With your specific example it is important to know if the captcha appearing or not will break your code. Does the logic boil down to captcha = false or does captcha not exist at all and logic fails entirely?
Q: How do you define sometimes captcha doesn't appear (1%, 20%, 50%, ...)?
A: Maybe 5% of the time captcha doesn't appear.
In this case, I prefer to use Exception handling: do stuff and if something goes wrong, fix it
try:
# > 95% of the time the code below works when the captcha appears
except SomeException:
# < 5% of the time the code is called when the captcha doesn't appear
IMHO, you have not really 2 different codes: you have one and a fallback solution, it's really different than:
if x % 2:
...
else:
...
Related
Currently I'm making a python bot for whatsapp manually without APIs or that sort because I am clueless. As such, I'm using Selenium to take in messages and auto reply. Currently, I'm noticing that every few messages, one message doesn't get picked up because the loops ran are too slow and my computer is already pretty fast. Here's the code:
def incoming_msges():
msges = driver.find_elements_by_class_name("message-in")
msgq = []
tq = []
try:
for msg in msges:
txt_msg = msg.find_elements_by_class_name("copyable-text")
time = msg.find_elements_by_class_name("_18lLQ")
for t in time:
tq.append(t.text.lower())
for txt in txt_msg:
msgq.append(txt.text.lower())
msgq = msgq[-1]
tq = tq[-1]
if len(msgq) > 0:
return (msgq, tq)
except StaleElementReferenceException:
pass
return False
Previously, I didn't add the time check thing, and the message sent would be saved, with this loop continuously running such that even if the other party sent the same thing again, the code would not recognise it as a new message because it thinks it's the same one as before. So now, the problem is that my code is super time consuming and I have no idea how to speed it up. I tried doing this:
def incoming_msges():
msges = browser.find_elements_by_class_name("message-in")
try:
msg = msges[-1]
txt_msg = msg.find_element_by_xpath("/span[#class=\"copyable-text\"]").text.lower()
time = msg.find_element_by_xpath("/span[#class=\"_18lLQ\"]").text.lower()
return (txt_msg, time)
except Exception:
pass
return False
However, like this, the code just doesn't find any messages. I have gotten the elements' types and classes correct according to the whatsapp web website but it just doesn't run. What's the correct way of rewriting my first code block as it is still correct? Thanks in advance.
First thing first ...
I definitely recommend using API ... Because what you are trying to do here is to reinvent the wheel. API has the power of telling you if there is a change in your status and you can queue these changes ... So I definitely recommend to use API ... It might be hard at the beginning, but trust me, its worth it ...
Next I would recommend you to use normal variable names. msges msgq tq (these are kindof unreadable and I still dont get what they are supposed to be after reading the code twice ...)
But to your speed problem ... "try - catch (aka except)" blocks are really heavy on a performance ... I would recommend to use safe programming if possible (20 if statements might be faster, but might not a same time) ... Also I think you are kind of unaware of a python language (atleast from what i can see here)
msgq = msgq[-1] # you are telling it to take the last element and change array variable to string .. to be more specific...
msgq ([1,2,3,4]) = msgq[-1] (4) will result to -> msgq = 4 (which in my option hits you performance as well)
tq = tq[-1] # same here
This would be better :)
if len(msgq[-1]) > 0:
return (msgq[-1], tq[-1])
If I understand your code correctly, you are trying to scrape the messages, but if its like you are saying that you want to make auto-reply bot, I would recommend you to eighter get ready for some JS magic or switch tool. I personally noticed that the selenium has a problem with dynamic content ... to be more specific ... once its at the end of the file it does not scrape it again ... so if you do not want to auto refresh every 5-10 seconds to get the latest HTML file, I recommend eighter to create this bot in JS (that will trigger everytime that an element changes) or use the API and use selenium just for responses. I was told that Selenium was created to simulate the common user to check if user interface works as it should (if buttons exists, if the website contains all what it should etc.) ... I think that selenium is for this job something like a flower small sponge for a car clean ... you can do it ... buts gonna cost you alot of time and you might miss some spots (like you missed those messages) ...
Lastly ... the work with strings in general is really costly. you are doing O(n^2) of operations in a try block ... which i can imagine can be really costly ... if its possible, I would reduce the number of inner for loops.
I wish you good luck in this project and I hope you find the answer you seek, while I hope my answer was at least a little helpful.
I'm struggling to troubleshoot a strange problem I've been having since starting to use Sikuli over multiple projects. I've been using the IDE and later tried to branch out due to having strange things happening with code. If I were to debug code earlier with popups, I can save the code, even restart my pc, even check the code in other text editors but the now non-existent popups (and old code) sometimes, well, pop up. In the end normally I end up ditching original files, and having to sometimes strangely comment out lines and re-add them one at a time (even though in the grand scale of things the end script was the same as before i did all that). I'm at a real loss for words.
It's making me struggle to differentiate between bad code and something going wrong elsewhere. Does anyone know what might cause this "phantom code"? Because I'm really at a loss.
And i would like advice as to what's going wrong with the while i < (inputvariable). I can't figure out what might be going wrong at all, am i over looking something?
I'm running all scripts through Sikuli IDE at the moment. I did want to learn how to write scripts and include sikuli hoping i could package it neatly but i couldn't seem to wrap my head around it.
For the while loop, where it's being compared to "SSLoops" i can't see why it's not breaking out of the loop when the criteria is met. (prints out above and beyond the number.)
I've had to do strange workarounds such as commenting out whole sections of code, trying to get it to work, and then slowly one by one reintroduce it till it matched the old script entirely. If I copied the script to a new file to make a cleaner copy, in hopes that if there is some sort of caching issue(?) it'd resolve, but I'd normally have to tinker around with it again.
BP = getBundlePath()
print(BP)
setBundlePath(BP + "\images")
BP2 = getBundlePath()
print(BP2)
# Regions
gameRegion = Region(230, 138, 1442, 875)
matchSpeedRegion = Region(1282, 920, 162, 91)
rewardRegion = Region()
def main():
SSLoops = input("How many times would you like to run Super Smash?")
SuperSmash(SSLoops)
def SuperSmash(SSLoops):
print(SSLoops)
i = 1
while i < SSLoops:
print(i)
print(SSLoops)
if exists("btnEnterSuperSmash.PNG"):
click("btnEnterSuperSmash.PNG")
while True:
if exists("btnReward.png"):
print("Completed! On to Rewards.")
#selectRewards()
break
else:
pass
if matchSpeedRegion.exists("btnStart.png"):
matchSpeedRegion.click("btnStart.png")
matchSpeedRegion.wait("btnRetreat.png", 3600)
if matchSpeedRegion.exists("btnSpeedUp.png"):
matchSpeedRegion.click("btnSpeedUp.png")
print("clicked x1")
print("clicking retreat")
matchSpeedRegion.click("btnRetreat.png")
matchSpeedRegion.wait(Pattern("btnRetreat.png").similar(0.65), 3600)
print("clicking okay")
gameRegion.click("btnOK.png")
wait(2)
gameRegion.wait("btnOK.png", 3600)
gameRegion.click("btnOK.png")
print("Completed!")
i = i + 1
if __name__ == '__main__':
main()
I have been getting popups saying "hey" because i had a loop in while true btnRewards to run a function to say "hey" - this would have hopefully on on to pick a reward out of 5 images in the end. But after removing it, as i'm trying to trouble shoot the main loop, it still pops up.
The loop that compares the user input variable to i just keeps on increasing. The indentation looks okay to me? but i must be wrong? or is something else making it go wrong?
I've been making the program run on a folder so the pictures to break the loop are immediately up, so it should have theoretically ran the amount of times inputted without anything else (1). Any help is deeply appreciated.
====
1
1
1
[log] CLICK on L[889,656]#S(0) (568 msec)
Completed! On to Rewards.
Completed!
2
1
[log] CLICK on L[889,656]#S(0) (565 msec)
Completed! On to Rewards.
Completed!
3
1
[log] CLICK on L[889,656]#S(0) (584 msec)
Completed! On to Rewards.
Completed!
4
1
====
Your problem: input() returns a string like so "4"
you then compare it using
while i < SSLoops:
which is always True and hence the loop does not end.
using
SSLoops = int(input("How many times would you like to run Super Smash?")) will solve your problem.
Be aware: this will crash if the given input cannot be converted to an integer value.
Suggestion: debug prints should look like so:
print "SSLoops =", SSLoops
so the output is better readable.
RaiMan from SikuliX (greetings to your cat ;-)
I am doing a python course where they suggested a try and except block in a while loop in order to keep asking for input until the condition is satisfied. Intuitively I feel it is shorter to just call the function again in the "except" block like this:
def exceptiontest():
try:
print(int(input("number 1: "))+int(input("number 2:")))
except:
print("a mistake happened")
exceptiontest()
exceptiontest()
When asking on the forum on the course I got the reply that it is not the same. I am a bit confused now. Anyone that can clarify for me? Thanks in advance!
Calling the function in the except will eventually raise a RecursionError: maximum recursion depth exceeded error if you keep entering bad inputs. Generally must humans won't be entering that many bad data to hit the error before they give up, but you are unnecessarily putting function calls on a stack.
A while loop is better since it's one function call, waiting for a valid input. IT doesn't waste any more resources than it needs.
while loop, for two reasons
it's clearer to read: while not success, try again
recursion is not free. It leaves the previous function stack open. it could run out of memory (probably won't, in this case, but in principle, avoid it)
Another reason to use the while loop which has not yet been mentioned is that you could leverage the assignment expressions coming with Python 3.8.
The function add encapsulates getting two numbers and trying to add them.
def add():
'try to add two numbers from user input, return None on failure'
x = input('number 1: ')
y = input('number 2: ')
try:
return float(x) + float(y)
except TypeError, ValueError:
return None
The following while loop runs as long as there is no result.
while (result := add()) is None:
print('you made a mistake, make sure to input two numbers!')
# use result
An Emirp is a prime number whose reversal is also a prime number. For example, 17 is a prime and 71 is a prime, so 17 and 71 are emirps.
The following code compiles and accepts input. The input validation is correct but the program keeps running and does not output anything. I also wanted to know how it's possible to find errors in python. Please ignore indentation errors if there are any.
class negerror(UserWarning):
pass
while True:
prompt = input('Please enter a positive number: ')
try:
number=int(prompt)
if number<=0:
raise negerror
break
except ValueError:
print('You did not enter an integer. Please try again. ')
except negerror:
print('You entered a negative number. Please make sure to enter a positive number')
def isPrime(value):
count=0
for i in range(1,value+1):
if value%i==0:
count=count+1
if count<=2:
return True
else:
return False
def reverse(value):
val=str(value)
val=val[::-1]
val=int(val)
return val
Test=2
countemirps=0
numberinoneline=0
while countemirps<number:
if isPrime(Test) and isPrime(reverse(Test)):
print('%6s'%Test, end = ' ')
countemirps=countemirps+1
Test=Test+1
numberinoneline=numberinoneline+1
if numberinoneline%5==0:
print('\n')
Your isPrime function is off. It counts the number of divisors of value and stores it in variable count. However, you return True if count>2 and False otherwise. It should be the other way around: a prime number has two divisors and composite numbers have more then two. So change the test to count <= 2. Even better and more pythonic, replace the last lines of that function with
return count <= 2
or perhaps
return count == 2
Do you see why that works?
(I see that you have now corrected this error and edited your question but the program still does not work.)
Another error is that in your main loop you have the test
if isPrime(Test) and isPrime(reverse(Test)):
If that test is passed, you print the number and update your variables including Test--all is well. However, if the test fails you do nothing, and in particular the value of Test is not changed. The loop repeats and you do exactly the same test, and nothing changes. The program is stuck in an infinite loop.
You can fix this by moving the line that updates Test out of the if test and place it at the end of the loop so it is executed on each loop. Your loop then becomes
while countemirps<number:
if isPrime(Test) and isPrime(reverse(Test)):
print('%6s'%Test, end = ' ')
countemirps=countemirps+1
numberinoneline=numberinoneline+1
if numberinoneline%5==0:
print('\n')
Test=Test+1
When I test your program now, it seems to work.
There may be other errors as well that I do not see. You should test function isPrime separately from the rest of your code. When that works well, then test function reverse. Then test sections of your code. Running your code all at once makes it difficult to localize and find errors.
Finally, you ask "how it's possible to find errors in Python". This is too broad a question for this site--you should read a book chapter or a tutorial on debugging. But in brief, there are two main approaches for beginners. The first is to liberally put print statements in your code, showing the flow of the program execution and the values of your key variables. If you had place the statement
print(Test)
at the beginning of your loop, you would have seen that the loop was repeating indefinitely and the value of Test was not changing. When the errors seem to be gone, you can remove or comment-out the print statements. Logging can do this a bit more easily
Another, better approach is to use a debugger. I do most of my Python programming in Spyder, a development environment which includes a debugger. I used that to execute your program one line at a time, and one window in Spyder showed my the values of your variables. There also are debuggers that work outside a development environment. I encourage you to find, learn, and use an Integrated Development Environment that includes a debugger. Spyder is a free, excellent one that specializes in scientific programming. I use it for my mathematics programming.
I'm writing a custom backup script in Python. Sometimes the mkdir function or the print function or whatever function are failing for diverse reasons. Such exceptions stop the whole script and halt the backup in the middle, which is very frustrating. So far, I've managed these problems by adding try: ... except: ... statements and properly managing these exceptions. However, one day some other statement or function might raise an exception as well because of some other reason that hasn't been triggered yet.
Is there a way to tell a script to proceed anyway? An equivalent of wrapping every single statement of the code in a try: ... except: pass clause? A log would be better of course.
I've noticed that when programming with GUI toolkits like Tkinter, the application keeps running even if exceptions are raised. Is it possible to accomplish this type of thing with the console?
There actually is a module that is supposed to do exactly that: https://github.com/ajalt/fuckitpy.
Although it was obviously written as a joke. I cannot imagine a situation where doing something like that is a good idea. God, I can't believe I'm even suggesting that as a solution.
What you should do instead is identify what lines of code can produce what kind of errors, and handle those errors properly. There is only so much places where errors can actually happen - mostly while interfacing with outside systems, including filesystem, network, user input etc. And remember that actually failing is often better than continuing "work" and messing up your data, files and so on. Exceptions are there for a reason, they are not a result of Guido's malice.
Python has no way of doing that, and for good reasons.
It seems you're confused about what does it mean to write "robust" software: a robust program is not a program that is hard to kill and that will keep running no matter what, but a program that will handle edge cases properly. Keeping running is NOT enough... keeping running doing sensible things is the key point.
Unfortunately there's no way to do reasonable things automatically and you've to think on a case-by-case basis how to handle the error.
Beware that if a program has a lot of catch it's rarely a good program. Exceptions are meant to be raised in a lot of places and caught almost nowhere.
Note also that every catch is potentially a source of bugs... for example:
try:
print my_dict[foo()]
except KeyError:
...
cannot distinguish if the KeyError is coming for accessing a non-existing key in my_dict or if instead escaped from foo(). Rarely the two cases should be handled the same way...
Better is to write:
key = foo()
if key in my_dict:
print my_dict[key]
else:
...
so that only the side case of missing key in my_dict is handled and instead a KeyError exception will stop the program (stopping a program when you're not sure of what it's doing is the only reasonable thing to do).
Python has 'BaseException' as the base class for Exception classes. You can catch and ignore the base class Exception and that will cover all exceptions.
try
... your code here ...
except BaseException as exp:
print "A General Exception Occurred"
try:
# code segment
except:
pass
pass keyword will ignore all the exceptions
Normally, this one should catch everything:
try:
....
except:
pass
Only problem is, that you don't get the exception object with this syntax, but that was not asked for in this case.
You can add a general except block like #Kanwar Saad proposed. The question is, can you continue with your program in a valid state after the exception has been raised?
From the Zen of Python:
Errors should never pass silently.
Unless explicitly silenced.
Trying to catch all exceptions you know is in my opinion is the best way to go here. If you can not explicitly catch an exception you should not try to work around it. You (and your users) should know what exactly went wrong, otherwise your code might become a nightmare to debug.
If you are worried about losing backup data maybe you could do something like this:
def save_unfinished_backup():
# try to find a graceful exit without losing any data
try:
# some code
except OSError:
# handle oS Errors
except Exception:
save_unfinished_backup()
raise
This way you get both: A chance to fend of data loss and the exact error to debug it.
I hope this helps!
On a funny note: You could also use the fuckit module. Which silences ALL errors, including syntax errors. Do not, ever, use this in productive code though.
This should work perfectly. It will not print the "foo", but you will reach the print("bar") without a crash.
import fuckit
with fuckit:
prnt("foo")
print("bar")
New answer for new Gen...
Python now ships with suppress(), this tells the interpreter to suppress the indicated exceptions while running.
Can be easily imported used as below
from contextlib import suppress
with suppress(ValueError):
int('this wont catch')
print('yea')
The above will work, wont raise the ValueError exception of changing invalid int string to int...
It's more clean than third party libraries.
Happy Hacking