Since there are a bunch of questions here on Stack Overflow that deal with SyntaxError in Python, we might want to know:
How do we tackle a SyntaxError? Are there strategies that can generally be applied?
0. Before the Error appears: Syntax Highlighting and Code Formatting
Even before running into a SyntaxError, there are important measurements to deal with SyntaxErrors, because the best way to deal with SyntaxErrors is to avoid them in the first place. This can be done first and foremost by using an editor or an Integrated Development Environment (IDE) which has syntax highlighting for Python.
Besides that, we can decrease the risk of running into a SyntaxError by good code and formatting style. There is a formal definition of the term "good formatting style", PEP 8 -- Style Guide for Python Code. Proper formatting makes our code much more readable which decreases the risk writing code that leads to a SyntaxError.
A very good way to apply good formatting to our code is to use an automatic code formatting tool. A code formatter has multiple advantages, amongst which are the following: Its code formatting is consistent. It applies best practices you might not even have thought of yet. It is very convenient.
For Python, black is a great code formatting tool.
1. Understand the Error Message
The Syntax Error indicates in which file and in which line the interpreter came across a problem in our code. We should use this information to find the bug.
We should be aware that the Python interpreter sometimes indicates a SyntaxError in the line after the actual problem. This is because the parser expects something in the erroneous line and can recognise that this is missing only when the whole line has been parsed. The prototypic example for that kind of SyntaxError is a missing parenthesis. So for instance, the following code raises a SyntaxError in line 2, even though the bug is in line 1:
bar = foo(
baz()
EOL stands for "End Of Line". This helps understanding the very common SyntaxError: EOL while scanning string literal. This is usually raised when you did not properly close a string definition with closing quotation marks, such as in the following example:
foo = "bar
2. Simplify the Code
Generally, a good strategy of bug fixing is to reduce any code that throws an Error or an Exception (or that does not return the expected output) to a minimal example. (This is a requirement for questions here on Stack Overflow, but much more than this, it is a good technique for pinning down a bug.)
In case of a SyntaxError, producing a minimal example is usually very easy, because a SyntaxError does not depend on any values of a variable, or any state of an object or any other semantics of your code. That's why the source of a SyntaxError is usually one line of code.
So, to identify the bug, we remove all the code besides the line that we think is the source of the Error. If the Error vanishes, it has been in a different line. If the Error persists, we try to simplify this line. For instance, we replace nested parentheses, by defining intermediate variables that hold the values:
Instead of
bar = foo(foo(baz(foo()))
the following (logically equivalent) code:
first = foo()
second = baz(first)
third = foo(second)
bar = foo(third
makes it much easier for us to identify the missing closing parenthesis.
Refer to documentation. Syntax Errors unfortunately cannot be captured in a Try: Except: block, so the only way to deal with them is to read the message returned, and if that doesn't help, following up with the python documentation:
https://docs.python.org/3/
Related
i am trying to execute this very short code that took out from my bigger code because i was having problems with it. I was able to reproduce the problem (it gave the exact same error).
BACKGROUND INFO: i am just trying to get the first sentence out of this wikipedia search. but because the word i am searching for (kip, means chicken in Dutch) has a wide variety of meanings or something i get an error. i want to bypass this error using try: except: but it keeps displaying the error message anyway.
here is the code that just doesnt seem to work:
import wikipedia
wikipedia.set_lang('nl')
try:
summry = wikipedia.summary('kip', sentences=1)
print(summry + "\n")
except:
print("error")
i have tried replacing except: with this
except wikipedia.exceptions.DisambiguationError:
but it still doesnt work :( it always displays the error code regradless and prints "error" afterwards
/opt/virtualenvs/python3/lib/python3.8/site-packages/wikipedia/wikipedia.py:389:
GuessedAtParserWarning: No parser was explicitly specified, so I'm using the best available HTML
parser for this system ("html5lib"). This usually isn't a problem, but if you run this code on
another system, or in a different virtual environment, it may use a different parser and behave
differently.
The code that caused this warning is on line 389 of the file
/opt/virtualenvs/python3/lib/python3.8/site-packages/wikipedia/wikipedia.py. To get rid of this
warning, pass the additional argument 'features="html5lib"' to the BeautifulSoup constructor.
lis = BeautifulSoup(html).find_all('li')
error
i am using repl.it to program this
if anyone has any idea about why it keeps displaying the error anyway please please let me know :D
First of all, thank you all for commenting :D it helped a lot
At the end the solution that worked for me was inspired from Askold Ilvento's comment
although
with warnings.catch_warnings(): warnings.simplefilter("ignore")
didn't work when i adapted it a little bit it did the job!
this is the code that solved the problem and allowed me to ignore what was actually a warning (not an exception) and stop displaying it
import warnings
warnings.catch_warnings()
warnings.simplefilter("ignore")
i just added this at the start of the script and it solved the problem :D
once again all the credit for this code goes to Askold Ilvento i just had to add a little bit to it to make it work
I'm working on a Python project and as of now, my code has over 400+ lines. At one point, I had to write a multi-line comment about a small bug which needs a work around, and the interpreter decided to throw a syntax error.
According to the interpreter, the syntax error is occuring at elif.
I re-checked my indentation, converted tabs to spaces etc. Nothing seems to work.
if some_condition_1 == True:
do_something()
"""
Sub stage (b):
Refer documentation [1.7A] for ...
....
....
....
"""
elif condition_1 == True:
if condition_2 == False:
list.append(item)
However, if I remove the multi-line comment, the code executes fine.
Any idea what's going wrong? Please note that the code sample I've show above, is at very top of the file, and there's no chance for anything to go wrong elsewhere.
This is an indentation error. Your "multi-line comment" (really multi-line string) must be indented under the if block just like anything else.
""" These kinds of things """ are not really comments in Python. You're just creating a string and then throwing the value away (not storing it anywhere). Since Python doesn't have true multi-line comments, many people use them this way. However, since they are not true comments (they aren't ignored by the interpreter), they must obey all normal syntax rules, including indentation rules.
(Do note that when I say "creating a string" I'm speaking loosely. CPython, at least, has an optimization not to create an object here.)
I'm returning 0 all over the place in a python script but would prefer something more semantic, something more readable. I don't like that magic number. Is there an idea in python similar to how in C you can return EXIT_SUCCESS instead of just 0?
I was unable to find it here:
https://docs.python.org/3.5/library/errno.html
I'm returning 0
return is not how you set your script's exit code in Python. If you want to exit with an exit code of 0, just let your script complete normally. The exit code will automatically be set to 0. If you want to exit with a different exit code, sys.exit is the tool to use.
If you're using return values of 0 or 1 within your code to indicate whether functions succeeded or failed, this is a bad idea. You should raise an appropriate exception if something goes wrong.
Since you discuss return here, it seems like you may be programming Python like C. Most Python functions should ideally be written to raise an exception if they fail, and the calling code can determine how to handle the exceptional conditions. For validation functions it's probably best to return True or False - not as literals, usually, but as the result of some expression like s.isdigit().
When talking about the return value of a process into its environment you caonnt use return because a module isn't a function, so a return statement at top level would be flagged as a syntax error. Instead you should use sys.exit.
Python might seem a little minimalist in this respect, but a call to sys.exit with no arguments defaults to success (i.e. return code zero). So the easiest way to simplify your program might be to stop coding it with an argument where you don't want to indicate failure!
As the documentation reminds us, integer arguments are passed back, and string arguments result in a return code of 1 and the string is printed to stderr.
The language doesn't, as far as I am aware, contain any constants (while it does have features specific to some environments, if it provided exit codes their values might need to be implementation- or platform-specific and the language developers prefer to avoid this where possible.
I am developing a tool which has to accept a file as an input, check syntax errors, compile it and do something after that.
For example,
I have a file run.py:
a=5
b=c
print b
This should clearly show a syntax error while compiling because 'c' is not defined
I tried to use
try:
py_compile.compile("source_program/run.py", doraise=True)
print "Compiled"
except:
print "Error while compiling"
I get the output "Compiled" instead of "Error while compiling"
If I modify the run.py file as:
a=5
b=c/ #Instead of b=c
print b
Then I get the output "Error while compiling"
What don't I get an error message in the first case?
It's not a syntax error. b=c is perfectly valid syntax, whether or not c exists. In fact, some other module could have done
import __builtin__
__builtin__.c = 3
in which case there would be a built-in c variable with value 3 available to all modules, and your code would run fine.
For a somewhat less pathological example, if the file contains a * import such as
from numpy import *
the import will dump a whole bunch of names into the module's global namespace, and there's no way to tell what those names are. Even without import *, though, Python can't be sure that a reference to an unknown name is an error at compile time.
If you want to detect semantic errors such as this, you'll need a more complex analysis of the program. Integrating with an existing linter like pylint, as suggested by NPE, is likely to be more productive than writing your own tool. If you really want to do it yourself, you can parse the code with ast.parse and examine the AST, going statement by statement to see what variables exist at what points. You'll still never catch all bugs, but you'll find quite a few.
It's a tricky one, for many reasons.
It might not be a bad idea to try and integrate with pylint instead of trying to come up with your own.
c not being defined actually isn't a compile-time error. Python, when run, only runs into problems with undefined variables during runtime. This is not something that would be caught by any Python compiler.
This is a bit of an odd question; it came up in the context of a tool that exposes a Python API, which we spend a lot of time querying interactively from the REPL. The particular idiom causing issues is something like this:
for var in slow_generator_of_giant_list():
stats = update(stats, var)
print stats
To enter this at the REPL, I can type this:
>>> for var in slow_generator_of_giant_list():
... stats = update(stats, var)
...
If I now attempt to type the print, I get a syntax error due to improper indentation. (Or else I put the print inside the loop and do it on every iteration.)
But if I hit enter to go to the next line, the loop runs immediately, and I have to wait for it to finish, or type the print command in the face of possible output coming at me, etc.
Obviously I can define a function containing the above, and it might be worth saving into a file anyway, but in the general case we're constructing these on the fly, and it would be nice to have a way to "schedule" a command to run after the end of a loop from the REPL. In a language with block delimiters, I could of course put it after the ending delimiter (and any necessary statement separator). But my coworkers and I were stumped trying to do something similar here.
Is there perhaps an ugly abuse of Pythonic syntax that will do the trick that my coworkers and I couldn't think of? Or a recommended way to avoid the problem while still making it easy to throw together ad hoc interactive queries?
Thanks for any pointers.
Not beautiful, but this should work:
>>> mygen = slow_generator_of_giant_list()
>>> try:
... while True: stats = update(stats, mygen.next())
... except StopIteration:
... print stats
...
I would just say that you would find it easier just to not use the interactive shell for this.
It's not much effort to save a file and run it. You only have to keep it around for as long as you use it.
I actually have found this answering on SO. I keep a file open in my text editor with a terminal in the right directory, and just use it as a scratchpad for mocking up answers in.