[python] Python "expected an indented block"

Let me start off by saying that I am COMPLETELY new to programming. I have just recently picked up Python and it has consistently kicked me in the head with one recurring error -- "expected an indented block" Now, I know there are several other threads addressing this problem and I have looked over a good number of them, however, even checking my indentation has not given me better results. I have replaced all of my indents with 4 spaces and even rewritten the code several times. I'll post this counter assignment I got as an example.

option == 1
while option != 0:
    print "MENU"
    option = input()
    print "please make a selection"
    print "1. count"
    print "0. quit"
    if option == 1:
        while option != 0:
            print "1. count up"
            print "2. count down"
            print "0. go back"
            if option == 1:
                print "please enter a number"
                for x in range(1, x, 1):
                    print x
                elif option == 2:
                    print "please enter a number"
                    for x in range(x, 1, 1):
                elif option == 0:
                    break
                else:
                    print "invalid command"
    elif option == 0:
        break

Python is very picky about white space and indentation, more so than many languages. The reason is, rather than using curly braces and semi colons (like javascript or php) python looks for a return character (press enter/return on your keyboard) instead of the semicolon, and a colon with a tab after it for a opening curly brace. When the next piece of code is unindented, it expects that this is the same as a closing curly brace in Javascript or PHP.

From ==>https://teamtreehouse.com/community/what-is-a-indentationerror-expected-an-indented-block

Starting with elif option == 2:, you indented one time too many. In a decent text editor, you should be able to highlight these lines and press Shift+Tab to fix the issue.

Additionally, there is no statement after for x in range(x, 1, 1):. Insert an indented pass to do nothing in the for loop.

Also, in the first line, you wrote option == 1. == tests for equality, but you meant = ( a single equals sign), which assigns the right value to the left name, i.e.

option = 1

#option = 1
#while option != 0:

print ("MENU")
print("please make a selection")
print("1. count")
print("0. quit")
option = int(input("MAKE Your Selection  "))
if option == 1:
    print("1. count up")
    print("2. count down")
    print("0. go back")
    option = int(input("MAKE Your Selection  "))
    if option == 1:
        x = int(input("please enter a number   "))
        for x in range(1, x, 1):
            print (x)

    elif option == 2:
        x = int(input("please enter a number   "))
        for x in range(x, 0, -1):
            print (x) 
    elif option == 0:
        print("hi")
    else:
        print("invalid command")
else:
    print ("H!111")
_________________________________________________________________________


You can try this code! It works.

The body of the loop is indented: indentation is Python’s way of grouping statements. At the interactive prompt, you have to type a tab or space(s) for each indented line. In practice you will prepare more complicated input for Python with a text editor; all decent text editors have an auto-indent facility. When a compound statement is entered interactively, it must be followed by a blank line to indicate completion (since the parser cannot guess when you have typed the last line). Note that each line within a basic block must be indented by the same amount.

src: ##

##https://docs.python.org/3/tutorial/introduction.html#using-python-as-a-calculator

Your for loop has no loop body:

elif option == 2:
    print "please enter a number"
    for x in range(x, 1, 1):
elif option == 0:

Actually, the whole if option == 1: block has indentation problems. elif option == 2: should be at the same level as the if statement.

in python .....intendation matters, e.g.:

if a==1:
    print("hey")

if a==2:
   print("bye")

print("all the best")

In this case "all the best" will be printed if either of the two conditions executes, but if it would have been like this

if a==2:
   print("bye")
   print("all the best")

then "all the best" will be printed only if a==2

This one is wrong at least:

            for x in range(x, 1, 1):
        elif option == 0:

There are several issues:

1. elif option == 2: and the subsequent elif-else should be aligned with the second if option == 1, not with the for.

2. The for x in range(x, 1, 1): is missing a body.

3. Since "option 1 (count)" requires a second input, you need to call input() for the second time. However, for sanity's sake I urge you to store the result in a second variable rather than repurposing option.

4. The comparison in the first line of your code is probably meant to be an assignment.

You'll discover more issues once you're able to run your code (you'll need a couple more input() calls, one of the range() calls will need attention etc).

Lastly, please don't use the same variable as the loop variable and as part of the initial/terminal condition, as in:

            for x in range(1, x, 1):
                print x

It may work, but it is very confusing to read. Give the loop variable a different name:

            for i in range(1, x, 1):
                print i

Your last for statement is missing a body.

Python expects an indented block to follow the line with the for, or to have content after the colon.

The first style is more common, so it says it expects some indented code to follow it. You have an elif at the same indent level.