I am using the PyQt library to take a screenshot of a webpage, then reading through a CSV file of different URLs. I am keeping a variable feed that incremements everytime a URL is processed and therefore should increment to the number of URLs.
Here's code:
webpage = QWebPage()
fo = open("C:/Users/Romi/Desktop/result1.txt", "w")
feed = 0
def onLoadFinished(result):
#fo.write( column1[feed])#, column2[feed], urls[feed])
#feed = 0
if not result:
print "Request failed"
fo.write(column1[feed])
fo.write(',')
fo.write(column2[feed])
fo.write(',')
#fo.write(urls[feed])
fo.write(',')
fo.write('404,image not created\n')
feed = feed + 1
sys.exit(1)
save_page(webpage, outputs.pop(0)) # pop output name from list and save
if urls:
url = urls.pop(0) # pop next url to fetch from list
webpage.mainFrame().load(QUrl(url))
fo.write(column1[feed])#,column2[feed],urls[feed],'200','image created','/n')
fo.write(',')
fo.write(column2[feed])
fo.write(',')
#fo.write(urls[feed])
fo.write(',')
fo.write('200,image created\n')
feed = feed + 1
else:
app.quit() # exit after last url
webpage.connect(webpage, SIGNAL("loadFinished(bool)"), onLoadFinished)
webpage.mainFrame().load(QUrl(urls.pop(0)))
#fo.close()
sys.exit(app.exec_())
It gives me the error:
local variable feed referenced before the assignment at fo.write(column1[feed])#,column2[feed],urls[feed],'200','image created','/n')
Any idea why?
This question is related to
python
python-2.7
python-3.x
variable-assignment
When Python parses the body of a function definition and encounters an assignment such as
feed = ...
Python interprets feed as a local variable by default. If you do not wish for it to be a local variable, you must put
global feed
in the function definition. The global statement does not have to be at the beginning of the function definition, but that is where it is usually placed. Wherever it is placed, the global declaration makes feed a global variable everywhere in the function.
Without the global statement, since feed is taken to be a local variable, when Python executes
feed = feed + 1,
Python evaluates the right-hand side first and tries to look up the value of feed. The first time through it finds feed is undefined. Hence the error.
The shortest way to patch up the code is to add global feed to the beginning of onLoadFinished. The nicer way is to use a class:
class Page(object):
def __init__(self):
self.feed = 0
def onLoadFinished(self, result):
...
self.feed += 1
The problem with having functions which mutate global variables is that it makes it harder to grok your code. Functions are no longer isolated units. Their interaction extends to everything that affects or is affected by the global variable. Thus it makes larger programs harder to understand.
By avoiding mutating globals, in the long run your code will be easier to understand, test and maintain.
Best solution: Don't use globals
>>> test1 = 0
>>> def test_func(x):
return x + 1
>>> test1 = test_func(test1)
>>> test1
1
As the Python interpreter reads the definition of a function (or, I think, even a block of indented code), all variables that are assigned to inside the function are added to the locals for that function. If a local does not have a definition before an assignment, the Python interpreter does not know what to do, so it throws this error.
The solution here is to add
global feed
to your function (usually near the top) to indicate to the interpreter that the feed variable is not local to this function.
You have to specify that test1 is global:
test1 = 0
def testFunc():
global test1
test1 += 1
testFunc()
Put a global statement at the top of your function and you should be good:
def onLoadFinished(result):
global feed
...
To demonstrate what I mean, look at this little test:
x = 0
def t():
x += 1
t()
this blows up with your exact same error where as:
x = 0
def t():
global x
x += 1
t()
does not.
The reason for this is that, inside t, Python thinks that x is a local variable. Furthermore, unless you explicitly tell it that x is global, it will try to use a local variable named x in x += 1. But, since there is no x defined in the local scope of t, it throws an error.
In order for you to modify test1 while inside a function you will need to do define test1 as a global variable, for example:
test1 = 0
def testFunc():
global test1
test1 += 1
testFunc()
However, if you only need to read the global variable you can print it without using the keyword global, like so:
test1 = 0
def testFunc():
print test1
testFunc()
But whenever you need to modify a global variable you must use the keyword global.
Source: Stackoverflow.com