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How to find out if a character in a string is an integer


How to find out if a character in a string is an integer

By : user3849553
Date : October 22 2020, 11:12 PM
To fix the issue you can do Use isdigit
code :


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How to find integer representing code point of special character? TypeError: ord() expected a character, but string of l

How to find integer representing code point of special character? TypeError: ord() expected a character, but string of l


By : Juan Andrey Valverde
Date : March 29 2020, 07:55 AM
Hope that helps I found classmethod int.from_bytes(bytes, byteorder, *, signed=False) which does the job instead of ord(). Code:
code :
characters = ['Č', 'č', 'Š', 'š', 'Ž', 'ž']
codecs = ['cp852', 'iso8859_2', 'cp1250', 'mac_latin2', 'utf-8', 'utf_16_le', 'utf_16_be']

for letter in characters:
    for codec in codecs:
        decCodePoint = int.from_bytes(letter.encode(codec), byteorder='big') #code point integer
        print(letter + ' ' + codec + ' ' + str(decCodePoint) + ' ' + str(hex(decCodePoint)) + ' ' + str(oct(decCodePoint))) #i also convert decimal integer to hexadecimal and octal
Č cp852 172 0xac 0o254
Č iso8859_2 200 0xc8 0o310
Č cp1250 200 0xc8 0o310
Č mac_latin2 137 0x89 0o211
Č utf-8 50316 0xc48c 0o142214
Č utf_16_le 3073 0xc01 0o6001
Č utf_16_be 268 0x10c 0o414
T-SQL string concatenation `'string' + str(integer)` introduces extra space character

T-SQL string concatenation `'string' + str(integer)` introduces extra space character


By : brick
Date : March 29 2020, 07:55 AM
this one helps. The problem is STR:
code :
select REPLACE(str(Id), ' ', '-')
from Users
where DisplayName = 'Jon Skeet';

OUTPUT:
-----22656
select CONCAT('https://stackoverflow.com/users/', id) AS link
from Users
where DisplayName = 'Jon Skeet';
C string problem. Entering integer N and string, I want to print out character on Nth position. (If N>strlen(str))

C string problem. Entering integer N and string, I want to print out character on Nth position. (If N>strlen(str))


By : user2324724
Date : March 29 2020, 07:55 AM
This might help you Okay, I will try to explain it! if you have N and L (let say the length of the String). Then the position of the character at the Nth place will be calculated by a formula like: N = (N%L)-1; (subtracted one due to the array starts with 0) This formula works like this: Suppose the I have string "avlgh" and my N = 6, then at then after counting to 5 there will be a remainder that will be the index of the Character at the position you are looking for.
I hope this helps, and if don't others will do the better job at explaining this to you
Efficiently find character in a string formed by repeated substring by integer

Efficiently find character in a string formed by repeated substring by integer


By : Mary
Date : March 29 2020, 07:55 AM
hope this fix your issue If you know that some string s is another string t repeated n times then the character with index k in string s is equal to the character with index k2 = k mod t.length in the string t. We can use that to solve this task:
code :
len = 0
for each character ch in s
    if ch is digit
        len = len * digit
    else
        len = len + 1
 reverseS = reverse(s)
 curLen = len
 for each character ch in reverseS
     if ch is digit
         curLen = curLen / digit
         k = k mod curLen 
     else 
         if k == (curLen-1) then return ch as answer
         curLen = curLen - 1
Find the last 5 digits even when there is a character and convert them to integer

Find the last 5 digits even when there is a character and convert them to integer


By : Fifty Down
Date : March 29 2020, 07:55 AM
may help you . If you are running MySQL 8.0 / MariaDB 10.0.5, you can use regexp_substr() for this. This should be as simple as:
code :
select cast(regexp_substr(col_1, '[0-9]{1,5}$') as unsigned) from table_A
col_1            | col_1_new
:--------------- | :--------
978568-16258     | 16258    
ERGF99252697     | 52697    
SP-988824-189241 | 89241    
SP-456790-568723 | 68723    
SP-456790-568    | 568
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