How to access the variables after if-condition when the variable is defined inside the if-condition in python

问题

I need to access variable from outside of if-condition when the variable is created inside the if-condition in python. The variable types which are inside if-condition are test is <type, str> and vn is <type, instance>.

I have tried the below way but it has not worked for me.

In the below code I need to access vn and test variables

for DO in range(count) :
    atnnames = doc.getElementsByTagName("atnId")[DO]
    atn = atnnames.childNodes[0].nodeValue
    if atn == line[0]:
        vn = doc.getElementsByTagName("vn")[DO]
        vncontent = vn.childNodes[0].nodeValue
        y = vncontent.encode('utf-8')
       # print y
        if '-' in y:
            slt = (int(y.split('-')[0][-1]) + 1)
            test = y.replace(y.split('-')[0][-1], str(slt))
       #     print test
        else:
            slt = (int(y.split('.')[-1]) + 1)
            test = y.replace(y.split('.')[-1], str(slt))
       #     print test
    else:
        #print test
        vn.firstChild.nodeValue = test
print vn.firstChild.nodeValue

The error I'm getting when I run the above code is

UnboundLocalError: local variable 'test' referenced before assignment

I tried by defining the variables as None before for loop.

and It is throwing below error. AttributeError: 'NoneType' object has no attribute 'firstChild'

回答1:

Your problem appears to be the fact that you are referencing a variable outside of its scope. Essentially what is happening is in your if statement you are creating a variable exclusively for use within the if scope. Effectively when you have said print vn.firstChild.nodeValue you can also imagine it as being any other variable such as print undefinedVar. What is occuring is your are referencing (calling) upon the variable before it has even been defined.

However, no worries here since this is very easy to fix. What we can do is simply create your vn and test variables outside of the if scope, hence inside your actual method by doing the following:

vn = None
test = None

for DO in range(count) :
    atnnames = doc.getElementsByTagName("atnId")[DO]
    atn = atnnames.childNodes[0].nodeValue
    if atn == line[0]:
        vn = doc.getElementsByTagName("vn")[DO]
        vncontent = vn.childNodes[0].nodeValue
        y = vncontent.encode('utf-8')
       # print y
        if '-' in y:
            slt = (int(y.split('-')[0][-1]) + 1)
            test = y.replace(y.split('-')[0][-1], str(slt))
       #     print test
        else:
            slt = (int(y.split('.')[-1]) + 1)
            test = y.replace(y.split('.')[-1], str(slt))
       #     print test
    else:
        #print test
        vn.firstChild.nodeValue = test
print vn.firstChild.nodeValue

This basically just creates an empty variable in the outermost scope. I've set the values to None since they get defined once your for loop runs. So what happens now is you have a variable which has been declared outside, and is None at the start, but as you run your for loop you are not creating a temporary variable just inside the if statement, but you are actually changing the value of

回答2:

define the variable before the if block with None, then update it in the if block. Consider the following:

y = None
x = 1
print(y)
if x == 1:
    y = "d"
else:
    y = 12
print(y)

来源：https://stackoverflow.com/questions/54630146/how-to-access-the-variables-after-if-condition-when-the-variable-is-defined-insi

标签

python

variables

if-statement

scope

access