Just starting with python, encountering 1 problem...

[Russ356]

I have just started to learn python, and I am loving it, I have tried to learn other languages before, but this is really workin, but I am having one problem,

Any time I have an "If" statement, Such as:

if shape = 1:

or

if temperature > 50:

Anyone know the problem?

[Russ356]

Here is a complete Code I am using as I am just starting out,

# Area Calculation Program

print "Welcome to the Area Calculation program"
print "---------------------------------------"
print

# Print out the menu:
print "Please select a shape:"
print "1  Rectangle"
print "2  Circle"

# Get the user's choice:
shape = input("> ")

#calculate the area:
if shape = 1:
    height = input("Please enter the height:  ")
    width = input("Please enter the width:  ")
    area = height*width
    print "The area is", area
else:
    radius = input("Please enter the radius:  ")
    area = 3.14(radius**2)
    print "The area is", area

[bwkaz]

And, err, what's wrong with it? Error messages? Does it just not run right?

[Russ356]

Hah, I am a idiot, didn't post the error message..

  File "C:/Documents and Settings/Russell Johnston/My Documents/Python stuff/askcalc", line 16
    if shape = 1:
             ^
SyntaxError: invalid syntax

[Pierre Lambion]

if shape == 1: ??

[Russ356]

Thank you so much!

[Russ356]

New problem though --

Please select a shape:
1  Rectangle
2  Circle
> 2
Please enter the radius:  2
Traceback (most recent call last):
  File "C:\Documents and Settings\Russell Johnston\My Documents\Python stuff\askcalc", line 23, in ?
    area = 3.14(radius^2)
TypeError: 'float' object is not callable

I am assuming something is going wrong with the process where it cannot determine the answer, hence the float, I am guessing though.

[Russ356]

Meh, I fixed it, Just did radius*radius*3.14 instead.

[Russ356]

New Problem...

temperature = input("What is the temperature of the spam?")

    if temperature > 50:
        print "The salad is properly cooked."
    else:
        print "Cook the salad some more."

The problem is with the "if", Which I cannot find out the problem to.

[Strike]

Originally posted by Russ356
New problem though --

Please select a shape:
1  Rectangle
2  Circle
> 2
Please enter the radius:  2
Traceback (most recent call last):
  File "C:\Documents and Settings\Russell Johnston\My Documents\Python stuff\askcalc", line 23, in ?
    area = 3.14(radius^2)
TypeError: 'float' object is not callable

I am assuming something is going wrong with the process where it cannot determine the answer, hence the float, I am guessing though.

Text(stuff) is the syntax for a function call (where Text is the name of the function). That's what it was parsing it as. Also the exponentiation operator in python is **, not ^. You want:

    area = 3.14 * (radius**2)

[Strike]

Originally posted by Russ356
New Problem...

temperature = input("What is the temperature of the spam?")

    if temperature > 50:
        print "The salad is properly cooked."
    else:
        print "Cook the salad some more."

The problem is with the "if", Which I cannot find out the problem to.

Error message please.

[Strike]

Actually, I see an error

input (use raw_input, btw) returns a string, which you can't compare to an integer. Either cast the input to an integer or ... yeah, just do that.

[Russ356]

I am assuming the raw helped but the error message says

File "C:\Documents and Settings\Russell Johnston\My Documents\Python stuff\Spamsaladtemp.py", line 3
if temperature > 50:
^
SyntaxError: invalid syntax

[Russ356]

I got a new Error too!


Code and message:

if f == 1:
        # Do something...
    elif f == 2:
# Do something else...
    elif f == 3:
# Do something completely different...
    else:
# If all else fails...

  File "C:/Documents and Settings/Russell Johnston/My Documents/Python stuff/elseif.pyw", line 3
    elif foo == 2:
       ^
SyntaxError: invalid syntax

It seems it doesn't like when I use "if" any time.

[Strike]

Originally posted by Russ356
I got a new Error too!


Code and message:

if f == 1:
        # Do something...
    elif f == 2:
# Do something else...
    elif f == 3:
# Do something completely different...
    else:
# If all else fails...

  File "C:/Documents and Settings/Russell Johnston/My Documents/Python stuff/elseif.pyw", line 3
    elif foo == 2:
       ^
SyntaxError: invalid syntax

It seems it doesn't like when I use "if" any time.

Empty if/elif/else clauses are syntax errors. You can't have an if statement with no thing in it.