Question

我在数据库中有三列代表一个人的名字，中间名和姓：

            People
|------------------------------------------|
| First_Name  |  Last_name    | Middle_name|
|------------------------------------------|
|  John       |  Hansen       |     T      |
|   NULL      |  Smith        |    NULL    |
|  Jacob      |    NULL       |     J      |
|  Michael    |  Johnson      |    NULL    |
|------------------------------------------|

获得null安全的最佳方法是Last，First Name + Middle Name。所以从上面的列表我会得到：

Hansen, John T
Smith
Jacob J
Johnson, Michael

到目前为止，我已经：

select concat_ws(', ', name_last, concat_ws(' ', name_first, name_middle)) as name from entity;

但它让我跟踪,，我不想要它们。

Answer 1

如果任何参数为CONCAT，您可以使用NULL将返回NULL的事实：

SELECT COALESCE(CONCAT(last_name, ', ', first_name), last_name, first_name)
FROM People

SQL Fiddle Demo

或者，既然您已编辑了问题以添加Middle_name列：

SELECT COALESCE(CONCAT(last_name, ', ', COALESCE(CONCAT(first_name, ' ', middle_name),first_name,middle_name)), 
                last_name, 
                COALESCE(CONCAT(first_name, ' ', middle_name),first_name,middle_name))
FROM People

SQL Fiddle Demo

Answer 2

反过来说：你有一个包含＆＃34; firstname middlename lastname＆＃34;的全名列。并希望按姓氏显示它

让它变得有效，至少它对荷兰名字很有用（＆＃34; van der Boogert＆＃34;还有单身（艺术家）名字。

此示例修剪了额外空格的anmes，然后计算全名中的空格数量，因此当它只有一个名称（艺术家名称）时，没有逗号是位置

SELECT
    TRIM(fullname) as originalfullname, 
    length(trim(fullname)) -length(replace(TRIM(fullname), ' ', '')) AS countedzeros,
    SUBSTRING_INDEX(TRIM(fullname), ' ', length(TRIM(fullname)) -length(replace(TRIM(fullname), ' ', '')))  AS first_name,
    SUBSTRING_INDEX(TRIM(fullname),  ' ', -1) AS last_name,
    CONCAT(
        SUBSTRING_INDEX(TRIM(fullname),  ' ', -1),
        IF(length(TRIM(fullname)) -length(replace(TRIM(fullname), ' ', '')),', ',''),
      SUBSTRING_INDEX(TRIM(fullname), ' ', length(TRIM(fullname)) -length(replace(TRIM(fullname), ' ', '')))
    ) as dispayname
    FROM catalog
    group by fullname


/*
Theo van den Boogaart => Boogaart, Theo van den
Hergé => Hergé
Jhonny Smith => Smith, Johnny

*/

sql姓氏逗号首先是null检查

2 个答案: