- variable
- user input
- if-else
- for(break,continue)
- while(break,continue)
- Comprehensive example
- Operation
1. Variables
1.1. To understand the pointing problem of variables, see the following example:
1 #Variables are used to store content 2 name= ' SYR ' #python is a dynamic language, no need to declare data types 3 name2= name#name2 points to 4 print(name,name2)#SYR 5 name= ' XYQ ' #name Pointing to XYQ, the point of name2 has not changed 6 print(name,name2)#XYQ SYR 7 #Rules of variable definition: only the combination of alphanumeric and underscore can be said, the first character cannot be a number 8 #Variable names are semantic 9 #How to represent constants in python, variable names are capitalized: PI
1.2. Rules for variable definition:
- Variable names can only be any combination of letters, numbers or underscores
- The first character of a variable name cannot be a number
- The following keywords cannot be used as variable names: ['and', 'as', 'assert', 'break', 'class', 'continue', 'def', 'del', 'elif', 'else', 'except', 'exec', 'finally', 'for', 'from', 'global', 'if', 'import', 'in', 'is', 'lambda', 'not', 'or ', 'pass', 'print', 'raise', 'return', 'try', 'while', 'with', 'yield']
2. User input
2.1 The difference between input and raw_input:
Remember to use input() directly in python3.x, and use raw_input() directly in python2.x. The difference between input() and raw_input() in python2.x can be referred to:
https://www.cnblogs.com/gengcx/p/6707024.html
2.2. The default output of input is str type. If you need int type, you need to add conversion int() by yourself.
username=input('username:') # age1=input('age:') age=int(input('age:')) # print(type(age))
In the above example, the type of age1 is str and the type of age is int
2.3. Several output methods of print:
① Be careful not to add a comma before %(username,username,age,job,salary), which is different from the c language
username=input('username:') age=input('age:') job=input('job:') salary=input('salary:') print(''' --------------info of %s----------- Name:%s Age:%s Job:%s Salary:%s '''%(username,username,age,job,salary))
②.format usage
print(''' --------------info1 of {_name}------------ Name:{_name} Age:{_age} Job:{_job} Salary:{_salary} '''.format(_name=username, _age=age, _job=job, _salary=salary))
③.Format usage
print(''' ---------------info2 of {0}-------------- Name:{0} Age:{1} Job:{2} Salary:{3} '''.format(username,age,job,salary))
3. if-esle
Relatively simple, not to mention
my_age = 28 user_input = int(input("input your guess num:")) if user_input == my_age: print("Congratulations, you got it !") elif user_input < my_age: print("Oops,think bigger!") else: print("think smaller!")
Fourth, for
4.1. Simple for statement
for i in range(10): print("loop:", i )
nested for statement
#nested loop for i in range(10): print('------',i) for j in range(10): print(j) if j>5: break #The large loop is executed 10 times, and the small loop is executed 6 times
4.2, for combined with break, continue
break is to end the current entire loop, the following output should be: 0, 1, 2, 3, 4, 5
for i in range(10): if i>5: break #Don't go down, just jump out of the whole loop print("loop:", i )
continue is to end the current loop and continue to the next loop. The following output should be 5, 6, 7, 8, 9
for i in range(10): if i<5: continue #Do not go down, go directly to the next loop print("loop:", i )
4.3、for-else
The following program will be executed when the guess is correct: print('you have tried too many times')?
The answer is: no.
Because break jumps out of the for loop, else and for are side by side, and the else is executed only when the three loops of i in for are executed.
for i in range(3): guess_age = int(input("guess age:")) if guess_age == age_of_sun: print('yes,you got it') break elif guess_age > age_of_sun: print('think smaller') else: print('think bigger') else: print('you have tried too many times')
五、while
5.1. The following is an infinite loop, which will run
count = 0 while True: print("You are the wind and I am the sand, lingering to the end of the world...", count) count +=1
5.2. Combining while with break and continue
break, continue is the same as the above for for.
while count<3: guess_age = int(input("guess age:")) if guess_age==age_of_sun: print('yes,you got it') break elif guess_age>age_of_sun: print('think smaller') else: print('think bigger')
5.3、while-else
In this example, only the conut in the while loop ends from 1-2, and print("It's wrong to guess so many times, you idiot.")
my_age = 28 count = 0 while count < 3: user_input = int(input("input your guess num:")) if user_input == my_age: print("Congratulations, you got it !") break elif user_input < my_age: print("Oops,think bigger!") else: print("think smaller!") count += 1 #each loop counter +1 else: print("It's wrong to guess so many times, you idiot.")
6. Comprehensive examples
age_of_sun=25 count=0 for i in range(0,3): guess_age=int(input('guess age:')) if guess_age==age_of_sun: print('you got it') break break is to end the current entire loop elif guess_age>age_of_sun: print('smaller') else: print('smaller') count+=1 if count==3: confirm=input('continue?') if confirm != 'n': count=0
7. Homework
Work requirements:
- Write the login interface
- Enter username and password
- Welcome after successful authentication
- Lock account after entering three times
Work ideas:
- Account locked with file storage more than three times
- Before the user logs in, determine whether the account is in the file, and if so, mention it
First edition of the work:
#Author:Yueru Sun #Write the login interface, enter the username and password, after the authentication is successful, the welcome message will be displayed, and it will be locked after entering it three times. _username='sunyueru' _password='sunny@0321' def lock(username): #Write username to lock file f=open('lockuser','r+') f.write(username) f.close() def check_lock(username): f=open('lockuser','r+') for name in f: if username in name: print("Your account is locked") f.close() def main(): count=0 while True: username=input('Please enter username:') # Determine if the account is locked check_lock(username) password=input('Please enter the password:') if username==_username and password==_password: print("Login successful") break else: count+=1 print('Login failed') if count>=3: #lock account lock(username) print('You have entered wrong three times and your username is locked') break main()
Improved version: