获取具有名称的项目列表或基于其他表值的null

Question

我有这样的架构：

现在我想编写一个查询，只有当书借用时才返回带有借款人姓名的项目清单，否则借款人名称应为空。 我觉得它应该很容易;我一直在尝试多个连接和NULL条件，但我仍然无法得到我想要的。也许我应该改变我的架构？

Answer 1

SELECT items.*, borrowers.first_name, borrowers.last_name
FROM items
LEFT JOIN borrows ON borrows.item_id = items.id AND return_date IS NULL
LEFT JOIN borrowers ON borrowers.id = borrows.borrower_id

Answer 2

这样的事情可以解决问题：

select Items.title, case when b1.return_date is null then b2.first_name || ' ' || b2.last_name else null end as 'Borrower name' from Items i join Borrows b1 on b1.item_id = i.id join Borrowers b2 on b2.id = b1.borrower_id

这总是选择项目的标题，并连接借用者的first_name和last_name，但前提是return_date为null。否则，选择null作为'借用者名称'

Answer 3

您可以使用t = [[[0], [0], [0], [0], [0], [0], [0], [0], [0], [0], [0], [0], [0], [0], [0], [0]], [((3,), (2,), (2, 2), (2,)), ((2,), (1, 2), (2,), (0,))]] board_size = len(t[1][0]) cells = t[0] hnums = t[1][0] # These are the horizontal numbers vnums = t[1][1] # These are the vertical numbers # Here I'm building new lists using list comprehension. # I've noticed that some of the numbers are missing the second element. # For example here ((2,), (1, 2), (2,), (0,)), only the second element (1, 2) has two members. So I had to make sure this condition is taken care of by adding this x[1] if len(x) > 1 which means only take the second element if the length of the total tuple is more than 1. # If you haven't seen list comperhension before, you need to look them up because they are a huge topic. # at the end of these lines we will have something like this # h1 = [' ', ' ', 2, ' '] # h2 = [3, 2, 2, 2] h1 = [x[1] if len(x) > 1 else " " for x in hnums] h2 = [x[0] for x in hnums] v1 = [x[1] if len(x) > 1 else " " for x in vnums] v2 = [x[0] for x in vnums] # Here I'm building the two horizontal lines # heading_string will look like this ' {} {} {} {} ' because the board size is 4. Notice that this will work with other sizes if you changed the t variable. heading_string = " {} " * board_size # Here I'm just calling the format function and passing in the h1 variable. # The * will unpack the list. # The end result will be ' {} {} {} {} '.format(' ', ' ', 2, ' ') # So each parameter of the format function will be replaced with the {} placeholder print(heading_string.format(*h1)) print(heading_string.format(*h2)) # Here I'm just defining a function to take care of the logic of the cell text def cell_text(c): if c == [0]: return "?" elif c == [1]: return "." else: return "X" row = [] # a row list that will represent the each row of cells # for loop using enumerate and starting from 1. So i will equal to 1 in the first loop for i, cell in enumerate(cells, 1): row.append(cell_text(cell)) # appending cells to the row list # if i in the current loop is 4, 8, 12, 16 this condition will be true and we need to break the row to create a new one for the next row if i % board_size == 0: row_string = "[ {} ]" * board_size # Same thing as heading_string # Here I'm building the vertical numbers string # The numbers are stored in v1 and v2 # and they both have 4 members indexed 0, 1, 2, 3 # That's why I'm subtracting 1 from i because i starts from 1 not from 0 because of enumerate(cells, 1). # This will work fine for the first four elements # but in the fifth loop, i = 4 and we actually need v1[0] # and that's why I'm dividing by the board size to get the correct index vnum_string = " " + str(v2[(i-1) // board_size]) + " " \ + str(v1[(i-1) // board_size]) + " |" print(row_string.format(*row) + vnum_string) # Same as before except that I'm adding the vertical numbers string row = [] # Reset the row list 在LEFT OUTER JOIN表格上执行此操作，条件为Borrowers return_date。

这应该适合你：

IS NULL