mysql - 如何在 sql 查询的空字段中显示文本？

Question

我必须从数据库中提取客户列表，但如果第二个地址字段为空，我想显示文本无提供。这是我的查询：

    select concat(first_name, " ", last_name) as CustomerName, address, 
    address2, postal_code 
    from customer
    inner join address 
    using (address_id);

如果 address2 为空白，我想显示“未提供”。我怎么做？我确信答案非常简单，但我脑子里放了个屁，我在任何地方都找不到答案。

好的，这就是我最终的结果：

    case when address2=" " or address2 is null then 'None Provided' 
    else address2 end as address2,

有没有更好的方法来实现这一目标？

Answer 1

您可以使用CASE：

select concat(first_name, " ", last_name) as CustomerName, 
   address, 
   case when address2 is null then 'None Provided' else address2 end as address2, 
   postal_code 
from customer
inner join address 
using (address_id);

甚至更好COALESCE：

select concat(first_name, " ", last_name) as CustomerName, 
   address, 
   COALESCE(address2,'None Provided') as address2, 
   postal_code 
from customer
inner join address 
using (address_id);

如果数据包含空字符串或 null，那么我会考虑使用：

select concat(first_name, " ", last_name) as CustomerName, 
   address, 
   case 
     when address2 is null or address2 = '' 
     then 'None Provided' 
     else address2 end as address2, 
   postal_code 
from customer
inner join address 
using (address_id);

Answer 2

您可以使用内置COALESCE函数来实现这一点。

返回列表中的第一个非 NULL 值，如果没有非 NULL 值，则返回 NULL。

select concat(first_name, " ", last_name) as CustomerName, address, 
COALESCE(address2, 'None provided') as address2, postal_code 
from customer
inner join address 
using (address_id);

Answer 3

select concat(first_name, " ", last_name) as CustomerName, address, 
    COALESCE(address2,'none provided') as address2, postal_code 
    from customer
    inner join address 
    using (address_id);

Answer 4

使用 COALESCE...

select concat(first_name, " ", last_name) as CustomerName, address, COALESCE(address2, 'None provided') as address2, postal_code from customer inner join address using (address_id);

COALESCE 将返回列表中的第一个非 NULL 值