sagittarius | A set of javascript most used utils | REST library
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QUESTION
I have made a program using python language which tells your Horoscope. I made that on Jupyter-notebook, Anaconda. I opened it on word its code is: {
...ANSWER
Answered 2021-Mar-18 at 05:38Colab is quite similar to jupyter notebook and is free to use. Just send the link to other once you add your code. https://colab.research.google.com/ is the link. A quick guide is: https://colab.research.google.com/github/jckantor/CBE30338/blob/master/docs/01.01-Getting-Started-with-Python-and-Jupyter-Notebooks.ipynb
QUESTION
I was hoping and looking for some other ideas or ways to do this. As you can see, I asked the user for the input which is mmddyyyy (05022001 = May 2, 2001) and used substring in order for me to to take the certain input in the index then parse it. Now, What I want to accomplish is to find another way to do this for example is using the formal date formatter which in my case I didn't use. I think date formatter is alot better than this, but I don't get the idea out of it.
Here's what I've got so far. Thanks in advance.
...ANSWER
Answered 2021-Jan-21 at 17:11You should inform the user of what format you expect.
Take the input string and make a parse attempt. Trap for exception in case of faulty input. No need for you to check the ranges of month and day. LocalDate.parse
makes those data entry validation checks for you.
QUESTION
When you type the zodiac sign in the parameter, an example is "Leo" or "LEO", the output should appear as "This user is a LEO" but when I do it only shows me the word "INVALID".
Can someone help me so that when I try to type a specific word it will appear as what the conditions states?
This is my code:
...ANSWER
Answered 2020-Nov-08 at 17:25Instead:
if zodiac = 'Capricorn''CAPRICORN'
You should have:
if zodiac = 'Capricorn' or zodiac = 'CAPRICORN'
QUESTION
I have a date of someone's birth as "12/02/1980 00:00:00" but I want to convert it into a zodiac name. So I want 12/02/1980 to be displayed as Sagittarius. Is that possible to do?
...ANSWER
Answered 2020-Oct-10 at 12:34Can do something like this
QUESTION
I'm making a console personality test and I want to print a quote for the inputed zodiac sign of the player. I don't want to use tons of if
statements, can I do this with dictionaries and loops?
Where the keys are the zodiac signs and the values are the specific quotes, then using a loop and a short if statement to loop the keys with the given input and print the corresponding value which is the quote.
Here's my code:
...ANSWER
Answered 2020-Aug-27 at 16:35You could just get the zodiac sign from the dictionary with
QUESTION
Encounter a problem, how to rename existing xticklabels in lineplot using dictionary ? Part of code:
...ANSWER
Answered 2020-Jul-22 at 09:24I assume your data are in a dataframe like this df
:
QUESTION
I have a problem to find very efficient way to search birthday in many list.
Here is my horoscope_dates function
...ANSWER
Answered 2020-May-17 at 22:38from datetime import date
horoscope_dates = {"aries": (date(2020, 3, 21), date(2020, 4, 20)),
"taurus": (date(2020, 4, 20), date(2020, 5, 21)),
"gemini": (date(2020, 5, 21), date(2020, 6, 22)),
}
month = int(input("Enter your month of birth: "))
day = int(input("Enter your day of birth: "))
year = 2020
birthday = date(year, month, day)
for horoscope, (start, end) in horoscope_dates.items():
if start <= birthday < end:
print(horoscope)
break
QUESTION
The code is all about taking the zodiac sign as the input in the 1st activity and when clicked OK navigates to the next activity where it displays the details as the web view of that particular zodiac sign. When ever I try to run the code in the emulator or in my mobile I get a warning project is continuously stopping and the application closes.
...ANSWER
Answered 2020-Apr-24 at 17:04You are calling startActivity
twice, also .setClass
is not needed:
QUESTION
Plese tell me sql query to select zodiac sign by birth date.
Here is the table structure.
...ANSWER
Answered 2020-Apr-12 at 02:23Note that varchar(5)
is not going to get you very far with most zodiac signs.
You can use a join
. Most databases have functions such as month()
and day()
, although they might be named a little differently.
The idea is:
QUESTION
I'm doing a task in Python, which is to create a zodiac calendar judging by input month and day of the month.
I've made a successful calendar, however, I have no idea how to solve the problem, if I input an incorrect month name or nonexistent date.
I have tried tons of things: to create a different hierarchy, or under each zodiac sign I wrote else: print ('Entering an incorrect date')
. I even tried to register with if day <0 and day> 31 or month! = 'March' or month! = 'April'
, etc... It turned out to be a huge useless function in the end, and nothing worked either.
I feel that there is a one-line solution here to detect days < 0 and > 31 and incorrect month name, but I just can't understand it. Could somebody help, please?
...ANSWER
Answered 2020-Mar-26 at 10:59You might want to look into the datetime module. With it, you can do comparisons between dates and times easily without the need for extraneous if statements.
For example, if one were to do something like:
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