I was writing a code solution for password validation question so what's wrong with my code and it is not giving the desired opt

Question

ps=input()
h=['!', '@', '#', '

Answer 1

You probably think that this condition is fulfilled like this:
if ( len(ps) >= 7 and x[0] and x[1] and h[0] and h[1] ) in ps:

In fact it is fulfilled like this:
if len(ps) >= 7 and x[0] and x[1] and h[0] and ( h[1] in ps ):

As a result, x[0] and x[1] and h[0] will always be True.

(h[1] in ps) will be True if the @ character is contained in ps.

As a result, you got the condition:
if len(ps) >= 7 and True and True and True and ( h[1] in ps ): ☺️

Answer 2

If you use some condition in exception handling, then you must raise a specific exception, otherwise except will not work.

try:
if len(h)==2 in ps: #bug
pass
if len(x)==2 in ps: #bug
pass
if len(ps)>=7:
print("Strong")
else:
raise ValueError()
except:
print("Weak")

You can read in more detail in the course "Intermediate Python", lesson 31.1 "Raise exceptions".

Answer 3

How will it work correctly if you haven't fixed any bugs?

Answer 4

You need to first count the number of characters h and the number of numbers x contained in ps.
From the received result, compose an output condition.

Answer 5

Thanks but it still not working:
ps=input()
h=['!', '@', '#', '#x27;, '%', '&', '*']
x=[1,2,3,4,5,6,7,8,9,0]
try:
if len(h)==2 in ps:
pass
if len(x)==2 in ps:
pass
if len(ps)>=7:
print("Strong")
else:
raise ValueError()
except ValueError:
print("Weak")

Answer 6

It is only printing strong

Answer 7

O yes I changed my codes totally

Answer 8

ps=input()
h=['!','@','&','#','#x27;,'*','%']
if h[0]and h[1]and x[0]and x[1] in ps:
print("Strong")
else:
print("Weak")

Answer 9

But it is ignoring the middle code

Answer 10

Like if I do it like this:
if len(ps)>=7 and x[0]and x[1]and h[0]and h[1] in ps:

Answer 11

It is ignoring the middle code

Answer 12

x[0]and x[1]

Answer 13

😀

Answer 14

😨😱😲🤣🤐

Answer 15

😁

Answer 16

I did it https://code.sololearn.com/cW2Sj8r0Iky1/?ref=app