Python expected an indented block

Question

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

User · Answer

There are several issues    elif option    2  and the subsequent elif-else should be aligned with the second if option    1  not with the for  The for x in range x  1  1   is missing a body  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  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

User · Answer

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    gt https   teamtreehouse com community what-is-a-indentationerror-expected-an-indented-block

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

This one is wrong at least               for x in range x  1  1           elif option    0

[python] Python "expected an indented block"

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