I am trying to write a basic Twitter scraper in Python and while I have it so that it can scrape for hard coded terms, I'm trying to set it to take the search term from user input.
While my if/else statement accepts input when asked, it then fails to run stating that rawinput is not defined. The rawinput is within the if statement I've included my code below
I should mention I'm fairly new to Python.
I've tried removing the rawinput from the if/else and kept it separate but the same issue happens.
userinp = input("Select search type. 1 = tweets. 2 = people")
if userinp == 1:
entry = rawinput
query = u'q='
elif userinp == 2:
entry = rawinput
query = u'f=users&vertical=default&q='
searchurl = baseurl + query + entry
The expected result is that the user selects option 1 or 2 then is asked to enter their search term.
Results are:
Select search type. 1 = tweets. 2 = people1
Traceback (most recent call last):
File "Scrape.py", line 22, in <module>
userentry = rawinput('enter search term')
NameError: name 'rawinput' is not defined
Thanks in advance for any help given.
Use input() instead and don't forget the parentheses!
Additionally, input() converts your input to a string, so your if condition will never be met if it uses integers. Consider replacing with if userinp == '1': and elif userinp == '2':.
It should be raw_input() not rawinput. so your code should be like this...
userinp = input("Select search type. 1 = tweets. 2 = people")
if userinp == 1:
entry = raw_input()
query = u'q='
elif userinp == 2:
entry = raw_input()
query = u'f=users&vertical=default&q='
searchurl = baseurl + query + entry
Related
I am writing Oregon trail game and this is the code I have that is causing issues, I don't know why it is having issues. What I want to do is if they enter a name that contains a word that is in a list it will set the variable easter_mode to 1 if they don't then it will set easter_mode to 0. The words that need to be in the list are: (Sturtz, sturtz, Nate, nate)
Thank you
#asking name
player_name = input('What is your name:')
while len(player_name) >= 0:
if len(player_name) > 1:
print("Weclome" + str(player_name))
print('Which mode do you want to play?')
mode_choice = input('(easy) More modes comming soon:')
break
if len(player_name) == 1:
player_name_choice = input(str(player_name)+"? Are you kidding me? Only one letter? You might regreat it (Y/N):")
if player_name_choice == "y" or player_name_choice == "Y":
print("Ok Your Choice!!...")
mode_choice = 'easter'
break
if player_name_choice == "n" or player_name_choice == "N":
player_name = input('What is your name:')
else:
print("You do not type anything, try again")
player_name = input('What is your name:')
#Check Easter Egg Names
easter_names = ["nate sturtz", "Nate Sturtz", "Nate", "nate", "Sturtz", "sturtz"]
if player_name in easter_names:
easter_mode = 1
else:
easter_mode = 0
#easter eggs for name
if easter_mode == 1:
year_set = 2005
mode_choice = 'easter'
else:
year_set = input('Enter a year whatever you like:')
if year_set.isdigit():
return_num = 0
else:
return_num = 1
while return_num == 1:
print('Error,please try again!')
year_set = input('Enter a year whatever you like:')
if year_set.isdigit():
return_num = 0
else:
return_num = 1
year_set = int(year_set)
When I run the full file I get
Traceback (most recent call last):
File "Oregon.py", line 64, in <module>
player_name = input('What is your name:')
File "<string>", line 1, in <module>
NameError: name 'nate' is not defined
You can view the full code on Github
https://raw.githubusercontent.com/nsturtz/Oregon-Trail/master/Oregon.py
You'll get this error in Python 2. In Python 2, input() uses the exact value as your enter it.
In your example, you're typing nate and not 'nate'. The former value is a variable name (which is undefined in your code, hence the NameError), whereas the latter is a string.
In Python 3, input() behaves as you assume, and passes a string to your code.
If you are sure that you want to use Python 2, you can replace input() with raw_input() and it will interpret your input as a string rather than a variable name.
Under Python 2, you can use raw_input instead of input to prevent Python from interpreting the user input as Python code.
However, since Python 2 is deprecated, I strongly recommend against using it1. Use Python 3 instead, where input works as expected.
1 Except of course to maintain legacy products. But that doesn’t seem relevant here.
I'm trying to run the program in the following article:
https://blockgeeks.com/guides/python-blockchain-2/
I've copied all of the code into my Spyder IDE. When i run it there's a while loop which starts up asking the user to choose a number from the list of options it prints.
After selecting a number the program should perform the requested action. When i select it though it just loops back to the start of the while loop.
It appears to be ignoring the rest of the code in the while loop (the if statement part).
Confusingly if i take the parts of the code from the program which are used in the while loop and run them separately they work i.e if i run the below code and select the number 1 for my choice it will run the code in the if statement.
Why would the if statement run here but not in the main program?
#function 1:
def get_user_choice():
user_input = input("enter a number: ")
return user_input
#function 2:
def get_transaction_value():
tx_recipient = input('Enter the recipient of the transaction: ')
tx_amount = float(input('Enter your transaction amount '))
return tx_recipient, tx_amount
while True:
print("Choose an option")
print('Choose 1 for adding a new transaction')
print('Choose 2 for mining a new block')
print('Choose 3 for printing the blockchain')
print('Choose anything else if you want to quit')
user_choice = get_user_choice()
if user_choice == '1':
tx_data = get_transaction_value()
print(tx_data)
Update:
Sorry i realise i may not have been very clear what the problem is.
The above code is part of the code from the entire program and runs as expected in isolation from the main program.
The below code is the entire program from the article in the link. It includes all of the code in the program. If i run this main program the while loop doesn't use the if statement. It appears to just be breaking straight out of the loop after i select 1, 2 or 3 (any other number should break out of the loop anyway).
Here's a link for a screen shot showing what the console looks like after i have selected the number 1 for the option.
https://ibb.co/RNy2r0m
# Section 1
import hashlib
import json
reward = 10.0
genesis_block = {
'previous_hash': '',
'index': 0,
'transaction': [],
'nonce': 23
}
blockchain = [genesis_block]
open_transactions = []
owner = 'Blockgeeks'
def hash_block(block):
return hashlib.sha256(json.dumps(block).encode()).hexdigest()
# Section 2
def valid_proof(transactions, last_hash, nonce):
guess = (str(transactions) + str(last_hash) + str(nonce)).encode()
guess_hash = hashlib.sha256(guess).hexdigest()
print(guess_hash)
return guess_hash[0:2] == '00'
def pow():
last_block = blockchain[-1]
last_hash = hash_block(last_block)
nonce = 0
while not valid_proof(open_transactions, last_hash, nonce):
nonce += 1
return nonce
# Section 3
def get_last_value():
""" extracting the last element of the blockchain list """
return(blockchain[-1])
def add_value(recipient, sender=owner, amount=1.0):
transaction = {'sender': sender,
'recipient': recipient,
'amount': amount}
open_transactions.append(transaction)
# Section 4
def mine_block():
last_block = blockchain[-1]
hashed_block = hash_block(last_block)
nonce = pow()
reward_transaction = {
'sender': 'MINING',
'recipient': owner,
'amount': reward
}
open_transactions.append(reward_transaction)
block = {
'previous_hash': hashed_block,
'index': len(blockchain),
'transaction': open_transactions,
'nonce': nonce
}
blockchain.append(block)
# Section 5
def get_transaction_value():
tx_recipient = input('Enter the recipient of the transaction: ')
tx_amount = float(input('Enter your transaction amount '))
return tx_recipient, tx_amount
def get_user_choice():
user_input = input("Please give your choice here: ")
return user_input
# Section 6
def print_block():
for block in blockchain:
print("Here is your block")
print(block)
# Section 7
while True:
print("Choose an option")
print('Choose 1 for adding a new transaction')
print('Choose 2 for mining a new block')
print('Choose 3 for printing the blockchain')
print('Choose anything else if you want to quit')
user_choice = get_user_choice()
if user_choice == 1:
tx_data = get_transaction_value()
recipient, amount = tx_data
add_value(recipient, amount=amount)
print(open_transactions)
elif user_choice == 2:
mine_block()
elif user_choice == 3:
print_block()
else:
break
[1]: https://i.stack.imgur.com/FIrn7.png
When comparing values, Python takes a stronger route regarding data types than some other languages. That means no string in Python will equal a number.
Or in other terms "1" == 1 will be False.
That means you have to consider that in Python 3 you will receive a string from input() (not necessarily so in Python 2).
You can either compare this directly to another string:
user_choice = input()
if user_choice == "1":
print("You chose item 1")
Or you can convert it into a number first and compare it to a number:
user_choice = int(input())
if user_choice == 1:
print("You chose item 1")
Note that in the former case it might not be robust if the user enters extra spaces and in the latter case it will fail very loudly with an exception if the user doesn't enter an integer (or even nothing at all).
Both ways can be handled with extra code if necessary. In the former case, you can strip whitespace with user_input = input().strip() and in the latter case you can catch the exception with a try ... except ... block.
You have only handled the case for user_choice == '1'. If you enter anything other than 1, the program will return control to the beginning of the while loop.
I'll suggest you use a debugger to see what user_choice is before the if condition. If not, just use prints.
print("user_choice: {}, type: {}".format(user_choice, type(user_choice))
names=["aaa","bbb","ccc","ddd","eee"]
itMarks=[90,98,87,98,78]
def printMainMenu():
print(" Main Menu")
print(" =========")
print(" (1)Add Student")
print(" (2)Search Student")
print(" (3)Delete Student")
print(" (4)List Student")
print(" (5)Exit")
choice = int(input("Enter Your choice[1-5]:"))
return choice
def searchStudent(names,itMarks):
name = input("Enter Name")
i = names.index(names)
print("Index is" + i)
def deleteStudent(student,itMarks):
name = input("Enter Name to remove")
student.remove(names)
print("Successfully Deleted" + names)
def removeStudent(names):
name = input("Enter name to remove")
name.remove(name)
print("Successfully deleted" + names)
def addStudent(names, itMarkas):
name = input("Enter Name")
names.append(names)
itMarks = input("Enter IT Marks")
itMarks.append(itMarks)
def listStudent(names, itMarks):
for i in range(0, len(names)):
print(names[1], "", itMarks[i])
names = []
itMarks = []
choice = 1
while choice >= 1 and choice <= 4:
choice = printMainMenu()
if choice == 1:
addStudent(names, itMarks)
elif choice == 2:
searchStudent(names, itMarks)
elif choice == 3:
deleteStudent(names, itMarks)
elif choice == 4:
listStudent(names, itMarks)
elif choice == 5:
print("Exit from the program")
else:
print("invalid choice!")
choice = 1
I am new to the programming in Python. The following Python code is written to do some tasks with the array. There are two array named names and itMarks. And there are some functions :
addStudent() - To add students to the array
searchStudent() - To search a student with in the list.
deleteStudent() - To delete the given student from the list.
listStudent() - To list out the all the names of the students in the list.
When the program runs, it asks to select a choice. Then it do the task according to their choice. But when I run this coding it shows the errors.
Please help me. Thanks in advance.
ERROR :
When I select the choice 1 (Add student) and input name after the error is yield.
Traceback (most recent call last):
File "C:\Users\BAALANPC\Desktop\new 3.py", line 59, in <module>
addStudent(names, itMarks)
File "C:\Users\BAALANPC\Desktop\new 3.py", line 42, in addStudent
name = input("Enter Name")
File "<string>", line 1, in <module>
NameError: name 'rtrt' is not defined
Their so many mistakes in naming
In addStudent
def addStudent(names, itMarkas):
name = input("Enter Name")
names.append(name) # names cant appent it should be name
itMark = input("Enter IT Marks") # here itmark not itMarks
itMarks.append(itMark)
In searchStudent
def searchStudent(names,itMarks):
name = input("Enter Name")
i = names.index(name) # try to find index of name not names
print("Index is" + i)
In deleteStudent
def deleteStudent(student,itMarks):
name = input("Enter Name to remove")
student.remove(name) # try to remove name not names
print("Successfully Deleted" + name)
after change above I run its running you have to also change the naming of the variable for all methods
Output
Main Menu
=========
(1)Add Student
(2)Search Student
(3)Delete Student
(4)List Student
(5)Exit
Enter Your choice[1-5]:1
add student
Enter Name"aaa"
Enter IT Marks111
Main Menu
=========
(1)Add Student
(2)Search Student
(3)Delete Student
(4)List Student
(5)Exit
Enter Your choice[1-5]:
I'm assuming this is the correct form:
def searchStudent(names,itMarks):
name = input("Enter Name")
i = names.index(name)
print("Index is" + i)
note that I changed names to name.
also the same mistake again
def deleteStudent(student,itMarks):
name = input("Enter Name to remove")
student.remove(name)
print("Successfully Deleted" + names)
tl;dr revise your code
searchStudent(): You shouldn't need the itMarks argument if you're not using it inside your function at all. names refers to the list of names, but you are really trying to search name. i is an integer that is attempting to be concatenated with a string. Not allowed. It should be str(i).
deleteStudent(): Better to keep your arguments consistent and use names rather than student. Again, same problem as above, should be .remove(name) and you shouldn't need the itMarks argument. print statement should refer to name not names.
removeStudent(): This is the same code as deleteStudent(), but not used, so not sure why it's there.
addStudent(): Typo in the argument, .append(name). You have a global variable and a local variable named the same thing, which are conflicting to the program. Change the input set to itMark and .append(itMark).
listStudent(): print statement has a typo, 1 should be i. Not sure why the empty string is included as well.
Underneath your function def's, you restate your variables as empty lists. This can lead to ValueErrors from a lot of your functions as you're trying to look something up or modify something in an empty list. Simply delete this code.
Additionally, any error will break your while loop. I suggest adding more booleans or using a try except clause to catch these errors.
Good luck!
I'm new in Python and in Telegram bot, I hope you can help me understand this with a simple example.
What I need is to define a method that returns me a string to complete a url.
In Python what I need is:
user = input("Insert a username to see the graph:")
graphUrl = "https://www.graphsss123.com/ser/graph/" + user + "-123.jpg"
print(graphUrl)
how could I get the same result using Telepot?
Thank you
There is no exclusive function for receiving messages in telepot(I guess!!),so you will have to maintain the states (here i have done that using step).The snippet below is an example for adding 2 numbers given by user through telegram
ref:https://github.com/nickoala/telepot/issues/209
global step, no1
if step == 1:
if msg['text'] == 'add':
bot.sendMessage(chat_id, "input no1")
step = 2
else:
bot.sendMessage(chat_id, "please provide no")
elif step == 2:
no1 = msg['text']
bot.sendMessage(chat_id, "input no2")
step = 3
elif step == 3:
no2 = msg['text']
no3=no1+no2
bot.sendMessage(chat_id,no3)
step = 1
My code does not write to a file, what am I doing wrong? I am trying to program to continue to ask for products until the user does not enter a product code. I want all products to be saved in the file.
store_file = open("Database.txt", "w")
NewProduct = ""
while NewProduct != False:
contine = input("Press 1 to enter a new product press 2 to leave: ")
if contine == "1":
print("Enter your product information")
information = []
product = input("What's the product code: ")
information.append(product)
description = input("Give a description of the product: ")
information.append(description)
price = input("Enter price of product: ")
information.append(price)
information = str(information)
clean = information.replace("]","").replace("[","").replace(",","").replace("'","")
store_file.write(clean)
elif contine == "2":
NewProduct = False
else:
print("Your input is invalid")
store_file.close
I got the program working with the following adjustments. See comments for explanations:
store_file = open("Database.txt", "w")
NewProduct = ""
while NewProduct != False:
continue = raw_input("Press 1 to enter a new product press 2 to leave: ")
#Changed to raw_input because input was reading in an integer for 1 rather than a
#string like you have set up. This could be specific to my IDE
if continue == "1":
print("Enter your product information")
information = []
product = raw_input("What's the product code: ")
information.append(product)
description = raw_input("Give a description of the product: ")
information.append(description)
price = raw_input("Enter price of product: ")
information.append(price)
information = str(information)
clean = information.replace("]","").replace("[","").replace(",","").replace("'","")
store_file.write(clean + "\n")
#Added a line break at the end of each file write
elif contine == "2":
NewProduct = False
else:
print("Your input is invalid")
store_file.close() #Added parentheses to call the close function
I'm assuming the problem here is that you're using Python 2, and input isn't doing what you think it does. In Python 2, input evals the input as if it were Python source code, so if someone enters 2, it's going to return the int value 2, not "2". In Python 2, you want to use raw_input, always (eval-ing random user input not being secure/reliable).
Also, while on CPython (the reference interpreter) files tend to naturally close themselves when they go out of scope, you made an effort to close, but forgot to actually call the close method; store_file.close looks up the method without calling it, store_file.close() would actually close it. Of course, explicit close is usually the wrong approach; you should use a with statement to avoid the possibility of forgetting to close (or of an exception skipping the close). You can replace:
store_file = open("Database.txt", "w")
...
store_file.close()
with:
with open("Database.txt", "w") as store_file:
... do all your work that writes to the file indented within the with block ...
... When you dedent from the with block, the file is guaranteed to be closed ...
There are other issues though. What you're doing with:
information = str(information)
information = information.replace("]","").replace("[","").replace(",","").replace("'","")
is terrible. I'm 99% sure what you really wanted was to just join the inputs with spaces. If you switch all your input calls to raw_input (only on Python 2, on Python 3, input is like raw_input on Python 2), then your list is a list of str, and you can just join them together instead of trying to stringify the list itself, then remove all the list-y bits. You can replace both lines above with just:
information = ' '.join(information)