I have no idea how to fix this. I've tried retyping the program.
I get an unexpected indentation error for the last main function.
resident = 81
nonresident = 162
def main():
# initialize counters and total tuition
resident_counter = 0
nonresident_counter = 0
total_tuition = 0
print("Name \tCode\tCredits\tTuition")
print
try:
# open the data file
infile = open('enroll.txt', 'r')
# read the first value from the file
student_name = infile.readline()
# continue reading from file until the end
while student_name != '':
# strip the new line character and print the student's name
student_name = student_name.rstrip('\n')
print(student_name, end='\t')
# read the code type, strip the new line, and print it
code = infile.readline()
code = code_type.rstrip('\n')
print(code_type, end='\t')
# read the number of credits, strip the new line, and print it
credits = infile.readline()
credits = int(credits)
print(format(credits, '3.0f'), end='\t')
# check the room type and compute the rental amount due
# increment the appropriate counter
if code_type == "R" or room_type == "r":
payment_due = credits * resident
resident_counter += 1
elif code_type == "N" or room_type == "n":
payment_due = credits * nonresident
nonresident_counter += 1
elif code_type != "R" or code_type != "r" or code_type != "N" or code_type != "n":
payment_due = 0
# accumulate the total room rent
tuition += payment_due
# print the appropriate detail line
if payment_due == 0:
print('invalid code')
else:
print('$', format(tuition, '8,.2f'))
# get the next studen't name
student_name = infile.readline()
# close the input file
infile.close()
# print the counters and payment total amount
print
print('total number of resident students: ', resident_counter)
print('total number of nonresident: ', nonresident_counter)
print
print('total students: ', end='')
print('$', format(tuition, ',.2f'))
# execute the main function
main()
You don't have an except clause to match the try.
Despite what everyone else is saying, if you have a try, you don't have to have an except: you must have an except or a finally.
That may seem nitpicky, but it's pretty plausible that the code you were copying was actually using finally (e.g., to make sure some cleanup code always gets run—given that it's doing C-/Java-style manual cleanup).
At any rate, adding an except clause is not the right answer. If you're actually looking to handle or ignore errors, then yes, add an except clause. If you're looking for somewhere to add cleanup code that gets run even on an error, add a finally clause instead. If you don't want either, just remove the try line.
you are mixing room type and code type in your if else statement. You are also missing your except statement. This should be your last two lines before calling main.
It should read:
except IOError:
print('an error occurred trying to open or read enroll.txt')
You are using the try statement. It means that you MUST have an except block
try:
some code
except:
pass
It is simplification, but will solve your problem.
The other answers have (correctly) pointed out that you need to have an except to go with your try: (or not use a try altogether). I'm going to try to explain how you could go about debugging similar issues in the future, since these can be very annoying to debug.
Your error message most likely looked something like this:
File "test.py", line 74
main()
^
IndentationError: unexpected unindent
At first glance, that doesn't seem too helpful, and it's definitely not very clear.
The fact that you "tried retyping the program" sounds like you saw IndentationError and thought, "this means my indents are messed up; maybe there's an extra space somewhere or a mixture of spaces and tabs or something." Those are definitely valid IndentationErrors, and retyping the program would fix them, if you didn't make the same typing error again. (On a side note, I believe that any text editor worth using will have the option to convert all of your tabs to 4 spaces to avoid a lot of indent headaches.)
However, looking at the error text, we see "unexpected unindent". To me this sounds kind of like gibberish, but what it means is that, when python was looking at your program, it ran into a line that it thought should be indented more than it was. (Unfortunately, this same kind of error can be called "unexpected unindent" or "expected an indented block", and it can even show up as a plain old SyntaxError: invalid syntax.)
So knowing what the error message means, we can look back through the code to see why python thought that main() was indented strangely- though we have to be careful because python isn't always right about where the problem actually is. Here would be something like my thought process:
There's a function definition right before main() is called, maybe that function wants main() to be inside it? No that isn't it, because python only really expects at least one line of code after a function definition.
What else "wants" code to be indented after it? if, elif, and while all do, but all of them have code after them as well.
The only other "indenter" is try:. At this point, you either know that try: needs to be followed by a except: or a finally: and you fix it, or you don't. If you don't know that, than you could still see that try: is followed by an indented block, and, given that it is a likely culprit for you error, decide that it's time to look up what try: does.
Well I hope that helps troubleshooting problems like this in the future, and check out abarnert's answer and my link for more on handling errors with try/except/else/finally statements.
As noted by LevLevitsky and david above, there's no except clause. The easiest way to solve this is adding a line like the following to your code before the call to main:
except: pass
The second comment I have to your code is more stylistic, but you could add the following, before your call to main and after the except clause:
if __name__ == '__main__':
main()
Besides the missing except or finally or the extra try... If you have a look at the edited code, you can see the strangely mixed color blocks in the indentation. It means that you are mixing tabs and spaces which can lead to problems like bad indentation. The reason is that the interpreter thinks differently about how the tabs should be interpreted (for example 8 vs. 4 column tab stops).
You must not mix tabs and spaces for the indentation. I personally would recommend to use only the spaces. Any decent editor is capable to expand TAB key to spaces and/or Untabify the existing mixture of tabs and spaces.
Related
I'm trying to familiarize myself with IDLE by executing the following code from Automate the Boring Stuff with Python by Al Sweigart:
name = ""
while name != "Mark":
print("What is your name?")
name = input();
print("Thank you")
Syntax Error
For some reason though, I get a syntax error when trying to type the last print statement. I dont know how to get around the indentation / how to be able to type again OUTSIDE of the loop. I understand that only one block of code is executed at a time but I can't seem to be able to incorporate the final print statement. Does anybody know how I can get around this? Thank you very much
In IDLE, put an empty, unindented line after entering this line:
name = input();
Press enter again to make the extra blank line to exit that indented block and your loop should execute after that.
Actually, your solution might be very simple.
name = input();
You don't need the ending semi-colon in Python, remove it so the line is just:
name = input()
When I hit enter after typing else: to move to the next line, it gives error indentation, no matter how I align it, i tried 4 spaces, everything still not working. I made sure that else is aligned perfectly with if as in the book, but still error.
Can someone please explain to me how indentation works? I'm using Python 2.7
Code:
if x%2 == 0:
print "Even"
else:
print "Odd"
print "done with conditional"
Assume as if python has an reverse hierarchical structure like an inverted traingle.
when ever you want to write function/loop/conditions in 1st level, we write whatever the code we are writing inside those three sections in 2nd level.
In your code "if" and "else" comes in 1st level. so dont give any spaces infront of these.
print statements inside 'if' and 'else' should have same tab space or white space as they come under 2nd level.
outer print statement will not have a space because it again comes to 1st level.
Hope this helps
In Python console, everything works fine with a positive number of spaces, or a Tab;
so any of these example will work:
if True:
print('Hello')
else:
print('Not hello')
if True:
print('Hello')
else:
print('Not hello')
if True:
print('Hello')
else:
print('Not hello')
EDIT
For searching where is the problem, try this answer; call
python -m tabnanny <your_script>.py
and you will get an output telling you which lines are the problem.
Example output:
'test.py': Indentation Error: unindent does not match any outer indentation level (<tokenize>, line 12)
Started learning python and trying to learn try blocks. However I am utterly confused to why this program creates an unident error (does not match other identation levels).
while True:
try:
yes=(int(input("how smart is neervesh")))
if neervesh == 0:
print("yes is smart")
elif neervesh ==1:
print("yeas")
else:
print("kill me now")
except ValueError:
print("dont be retarded like yes")
Your try is indented by 3 spaces, your except by 4. You should stick to one indentation width (usually 4).
I always use the program 'SublimeText'. This text editor gives you the possibility to change your indentation on the fly. You can either convert your indentation to tabs or spaces. This way, you'll likely prevent future
indentation errors.
With this code I am trying to generate simple multiplication tables. The program should ask for input and multiple that number in a range up to 15 and the generate the multiplication table for the number. After the if_name_ == 'main': line I end up with a syntax error after the colon. I normally program in python 2, so python 3 is a bit new to me but I'm not sure what the difference is. Below I have listed the short but full code. Any help would be much appreciated.
'''Multiplication Table'''
def multi_table(a):
for i in range(1,16):
print(' {0} x {1} = {2} '.format(a, i, a*i))
if_name_ == '_main_':
a = input('Enter a number: ')
multi_table(float(a))
if_name_ == '_main_':
a = input('Enter a number: ')
multi_table(float(a))
should be :
if __name__ == "__main__":
a = input('Enter a number: ')
multi_table(float(a))
Notice that both variable __name__ and __main__ has two underscores around them and that there must be a space between the if keyword and the start of the condition.
As #Maroun Maroun said right, it has to be if __name__ == "__main__" . But you wont need it. Just write it at the bottom :
'''Multiplication Table'''
def multi_table(a):
for i in range(1,16):
print(' {0} x {1} = {2} '.format(a, i, a*i))
a = input('Enter a number: ')
multi_table(float(a))
Should work, too.
EDIT: In the official docs :
https://docs.python.org/3/library/main.html
if __name__ == "__main__":
In the example provided, (i) there's a space missing after if, and (ii) the symbol's name should have two pairs of underscores: __name__. If these two corrections resolve your problem, great.
But in case you still have another more mysterious unresolved syntax error, then read on...
Recently I encountered a very similar error message, and it was hard to root-cause, since the error message seems to point in the wrong direction. The error seems to indicate that the syntax error lies at the end of the if statement. However, with Python 3.x I noticed that in case a parentheses mismatch error happens in a python source file, the error might be reported at the beginning of the next block rather than where the mismatch actually occurs.
It's worth checking for mismatched parentheses in the entire source code file, prior to the if statement. Check if all your (s are matched by a closing ).
Hello I am a fairly novice programmer. And I have a simple code
name=raw_input("Hello I am Bob. What is your name?")
print("It is very nice to meet you,", name)
response=raw_input("what do you want to do today?")
if response == "price match":
However on the fourth line I get SyntaxError: unexpected EOF while parsing error and I did look into it and I found that using the raw_input for inputted strings is much better than using the input function.I don't know why the error keeps popping up. Could I get some help and perhaps some suggestions as to how I can improve the code.
You have to do something in the if statement. For example, print a price:
if response == "price match":
print "Yes, we can do that for you".
But, you can't just leave a block (the stuff that is indented after a :) empty, or Python will give you an error.
In rare cases (and not in your case here), you may want to do absolutely nothing, but still have a block (e.g. if required to by an except:). In that case, you still have to put something in the block, but you can use the word pass which does nothing. (This is almost never used in an if block).
I'm no Python expert.. But it looks like your problem is here:
response=raw_input("what do you want to do today?") if response == "price match":
You are defining an if statement, but if this value is true, how do you handle it?
I'm guessing you would need something like:
if response == "price match":
print('Match')
else:
print('Did not match')
Hope this helps a little
If you need a placeholder for just having valid syntax before putting in the body of your if/else/functions/etc, use pass.
http://docs.python.org/release/2.5.2/ref/pass.html
As for your code, a valid if statement must always have a body. So put some code there or use pass.