Else 语句继续执行 PYTHON

Question

This functions searches for records in my database. Apart from if condition the else statement ( "Record not found") also keeps on executing even if the condition is true. OUTPUT screenshot

def displaySearchAcc():
try:
    command = "SELECT * FROM BANK"
    mycursor.execute(command)
    s = mycursor.fetchall()
    ch = input("Enter the account number to be searched : ")
    for i in s:
        i = list(i)
        if i[0] == ch:
            print("*" * 125)
            print("ACCNO", "NAME", "MOBILE", "EMAIL", "ADDRESS", "CITY", "COUNTRY", "BALANCE")
            print("=" * 125)
            for j in i:
                print("%14s" % j, end=' ')
                print()
        else:
            print("record not found")
except:
    print("Table not found")

Answer 1

def displaySearchAcc():
try:
    command = "SELECT * FROM BANK"
    mycursor.execute(command)
    s = mycursor.fetchall()
    ch = input("Enter the account number to be searched : ")
    for i in s:
        if i[0] == ch:
            print("*" * 125)
            print("ACCNO", "NAME", "MOBILE", "EMAIL", "ADDRESS", "CITY", "COUNTRY", "BALANCE")
            print("=" * 125)
            for j in i:
                print("%14s" % j, end=' ')
                print()
        else:
            print("record not found")
except:
    print("Table not found")

Answer 2

You don't even need the for loop. You can perform the search just by using SQL statements which is way faster than the iterative method you are using. Assuming the column name is account_number

def displaySearchAcc():
    account_no = input("Enter the account number to be searched : ")
    command = f"SELECT * FROM BANK WHERE account_number = '{account_no}'"
    row = mycursor.fetchone()
    if row is not None:
        print("*" * 125)
        print(
            "ACCNO", "NAME", "MOBILE", "EMAIL", "ADDRESS", "CITY", "COUNTRY", "BALANCE"
        )
        print("=" * 125)
        for col in row:
            print("s" % col, end=" ")
            print()
    else:
        print("Record not found")

If you really insist on using the iterative method, then the mistake you were making was the line i = list(i) . i is already a list and by putting i in list you are making list of lists (2D list) which messes up your iteration.

def displaySearchAcc():
    try:
        command = "SELECT * FROM BANK"
        mycursor.execute(command)
        all_rows = mycursor.fetchall()
        account_no = input("Enter the account number to be searched : ")
        for row in all_rows:
            if row[0] == account_no:
                print("*" * 125)
                print("ACCNO", "NAME", "MOBILE", "EMAIL", "ADDRESS", "CITY", "COUNTRY", "BALANCE")
                print("=" * 125)
                for col in row:
                    print("%14s" % col, end=' ')
                    print()
                break; # No need to iterate further
        else:
            print("record not found")
    except:
        print("Table not found")

Also, use better variable names.