How to fix "IndentationError: expected an indented block" in Python?

Question

I am trying to do a API test on a URL with Python, I have the following code block

       def simple_get(url):
        try:
            page_response = requests.get(page_link, timeout=5)
            if page_response.status_code == 200:
            # extract
            else:
                print(page_response.status_code)
                # notify, try again
        except requests.Timeout as e:
            print("It is time to timeout")
            print(str(e))
        except # other exception

When I run it give me the following error

File "<ipython-input-16-6291efcb97a0>", line 11
else:
   ^
IndentationError: expected an indented block

I dont understand why is the notebook still asking for indentation when I already have the "else" statement indented

make sure that you are not mixing tabs with spaces and use 4-spaces indentation, as per python standard — Francesco Montesano, Sep 02 '18 at 19:13
There needs to be a statement after the `if ...:` You can just put in a `pass` — jodag, Sep 02 '18 at 19:13
It's an ok question. Maybe try to interpret the error message more carefully. I interpret it as "Line 11 else is excepted to be indented.. why? ... ~thinking , thinking..~ oh, line 10 if doesn't have any task to perform " — kevinkayaks, Sep 02 '18 at 19:19
https://stackoverflow.com/questions/21764912/python-expected-an-indented-block — SherylHohman, Sep 03 '18 at 10:25
Possible duplicate of [Python: Expected an indented block](https://stackoverflow.com/questions/21764912/python-expected-an-indented-block) — SherylHohman, Sep 03 '18 at 10:25

score 1 · Accepted Answer · edited Sep 02 '18 at 19:39

1

Problem is that you did not tell the program what to do when the first condition is satisfied (if statement). If you are not sure about what to do in if, you can use python build in 'pass'.

if page_response.status_code == 200:
    pass
else:
    print(page_response.status_code)

edited Sep 02 '18 at 19:39

petezurich

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answered Sep 02 '18 at 19:14

shahidammer

1,026
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1

Thanks, this worked, moving on to the next exception – Jin Sep 02 '18 at 19:19
what's the next one @Jin? – kevinkayaks Sep 02 '18 at 19:20
1

I think this answer would be more valuable if you would write one or two sentence why this fixes the problem. – syntonym Sep 02 '18 at 19:20
@syntonym Thanks – shahidammer Sep 02 '18 at 19:22
@syntonym, this fixed the problem because pass I did not set a condition for the if statement, therefore the "else" statement has indentation problem. After setting the pass condition, the "else" statement no longer has indentation problem – Jin Aug 19 '19 at 17:31

score 1 · Answer 2 · answered Sep 02 '18 at 19:48

This is a very basic concept in Python to start coding:
Lines which start with # are ignored within a code block.

The code here

    if page_response.status_code == 200:
    # extract
    else:
        print(page_response.status_code)

literally translated as

    if page_response.status_code == 200:
    else:
        print(page_response.status_code)

and therefore produces IndentationError.

You can solve it by placing at least pass command or any working line into to the if statement.

Similar question is already asked before:
Python: Expected an indented block

score 0 · Answer 3 · answered Sep 02 '18 at 20:26

import requests
from bs4 import BeautifulSoup
def simple_get(url):
    try:
        page_response = requests.get(url, timeout=5)
        if page_response.status_code == 200:
            print(page_response.status_code)
            pass
            # extract
        else:
            print(page_response.status_code)
            # notify, try again
    except requests.Timeout as e:

        print("It is time to timeout")
        print(str(e))
simple_get("https://www.nytimes.com/")

How to fix "IndentationError: expected an indented block" in Python?

3 Answers3