简单的打印语句不适用于python中的if语句的方法

Question

I started learning Python code recently and one simple print statement is giving me trouble since last 4 days.

Problem: the print statement is not working inside the validatePostcode(postcode) method for if-statement. The assigned value is 200 (status code) which is printing fine without the if-statement. Also, when I compare with the True (result value) for that API it works fine without if-statement, why it is not working after I apply the if and try to compare?

Error:

  File "./py_script3.py", line 32
    print ("Congrats")
        ^
IndentationError: expected an indented block

---

    #!/usr/bin/env python3


    import os,re,sys


    import urllib.request as req
    import json

    def loadJsonResponse(url):
        #return json.loads(req.urlopen(url).read().decode('utf-8'))['result']
        #return json.loads(req.urlopen(url).read().decode('utf-8'))['status']
        print ("I am in loadJsonResponse before returning string")
        string = json.loads(req.urlopen(url).read().decode('utf-8'))
        return string
        print ("I am in loadJsonResponse after returning string")

    def lookuppostcode(postcode):
        url = 'https://api.postcodes.io/postcodes/{}'.format(postcode)
        return loadJsonResponse(url)

    def validatePostcode(postcode):
        url = 'https://api.postcodes.io/postcodes/{}/validate'.format(postcode)
        #return loadJsonResponse(url)
        string = json.loads(req.urlopen(url).read().decode('utf-8'))
        Value = str(string['status'])
        print (Value)
        if Value == 200 :
        print ("Congrats")

    def randomPostcode():
        url = 'https://api.postcodes.io/random/postcodes'
        return loadJsonResponse(url)

    def queryPostcode(postcode):
        url = 'https://api.postcodes.io/postcodes?q={}'.format(postcode)
        return loadJsonResponse(url)

    def getAutoCompletePostcode(postcode):
        url = 'https://api.postcodes.io/postcodes/{}/autocomplete'.format(postcode)
        return loadJsonResponse(url)

    #Input = input("Enter the postcode : ")
    #print(lookuppostcode('CB3 0FA'))
    validatePostcode('CB3 0FA')
    #print(queryPostcode('HU88BT'))
    #print(randomPostcode(Input))

Answer 1

You should indent the print statement like so:

if Value == 200 :
   print ("Congrats")

You can read more about this here !

Answer 2

From https://docs.python.org/2.0/ref/indentation.html :

Leading whitespace (spaces and tabs) at the beginning of a logical line is used to compute the indentation level of the line, which in turn is used to determine the grouping of statements.

By doing

if Value == 200:
print ("Congrats")

Python interprets the two lines as two different groups of statements. What you should do is:

if Value == 200:
    print ("Congrats")

Answer 3

This piece of code (which is generating the error):

if Value == 200 : 
print ("Congrats")

Should be

if Value == 200 : 
    print ("Congrats")

Because python expects an indented block after the conditional, just like the message error is saying to you

Answer 4

Need to add an indent after the if statement. You can do so by pressing return after typing the colon