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 def get_open_tickets():
    query = ("SELECT tickets.ticketid, status.status, solutions.solution, employee.name AS employee"  
            "tickets.customername, tickets.customeremail, tickets.submitteddate, tickets.issue"  
            "FROM employees INNER JOIN"  
            "tickets ON employees.employeeid = tickets.employeeid INNER JOIN"  
            "solutions ON tickets.solutionid = solutions.solutionid INNER JOIN"  
            "status ON tickets.statusid = status.statusid"  
            "WHERE tickets.statusid=1 OR tickets.statusid=2")
     with closing(conn.cursor()) as cur:
         cur.execute(query)
         results = cur.fetchall()
    
     tickets = []
     for row in results:
         tickets.append(make_ticket(row))
     return tickets
  

приведенный выше код в визуальном коде выдает следующую ошибку. пожалуйста, предложите выход.

Если бы мне пришлось догадываться, даже если вы не предоставили сообщение об ошибке:

 SELECT tickets.ticketid, status.status, solutions.solution, employee.name AS employeetickets.customername, tickets.customeremail, tickets.submitteddate, tickets.issueFROM employees INNER JOINtickets ON employees.employeeid = tickets.employeeid INNER JOINsolutions ON tickets.solutionid = solutions.solutionid INNER JOINstatus ON tickets.statusid = status.statusidWHERE tickets.statusid=1 OR tickets.statusid=2
  

вам не хватает пробелов.

Например:

 tickets.issueFROM employees