Oracle SQL-选择显式定义值的列

假设我有一张这样的桌子：

我的桌子

Name  |  Age  
Jack     20
Joe      33
John     4
Bob      0

假设我想从mytable中选择所有行以及一个自定义值列：

random:'rabbit'、'wild'、'nuts'

Name  |  Age  |  random
Jack     20      rabbit
Joe      33      wild 
John     4       nuts
Bob      0       NULL

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这怎么办呢？

你想要

case

表达式吗

select t.*, (case when age = 20 then 'rabbit' 
                  when age = 33 then 'wild' 
                  when age = 4 then 'nuts' 
             end) as random
from table t;

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您可以使用添加带有随机字符串的列

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[类似答案][2[

这有点灵活，但也更复杂

WITH random AS (
  SELECT row_number() OVER (ORDER BY (1)) rank, val FROM (
    SELECT 'rabbit' AS val FROM DUAL
    UNION ALL
    SELECT 'wild' AS val FROM DUAL
    UNION ALL
    SELECT 'nuts' AS val FROM DUAL
    UNION ALL
    SELECT NULL AS val FROM DUAL
  )
  ORDER BY dbms_random.value
),
myrows AS (
  SELECT row_number() OVER (ORDER BY (1)) rank, mytable.*
  FROM mytable
)
SELECT myrows.Name, myrows.Age, random.val
FROM myrows
LEFT JOIN random ON myrows.rank = random.rank;

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不，根本没有条件，我只是希望能够将一个值列表作为另一列传递给

select

，这是可能的吗？从何处获取值？最好将手动键控作为查询语句的一部分“手动键控”是什么意思？是否有这些值的表和函数？是否要为每一行提取一个随机值？

row\u number（）。
WITH random AS (
  SELECT row_number() OVER (ORDER BY (1)) rank, val FROM (
    SELECT 'rabbit' AS val FROM DUAL
    UNION ALL
    SELECT 'wild' AS val FROM DUAL
    UNION ALL
    SELECT 'nuts' AS val FROM DUAL
    UNION ALL
    SELECT NULL AS val FROM DUAL
  )
  ORDER BY dbms_random.value
),
myrows AS (
  SELECT row_number() OVER (ORDER BY (1)) rank, mytable.*
  FROM mytable
)
SELECT myrows.Name, myrows.Age, random.val
FROM myrows
LEFT JOIN random ON myrows.rank = random.rank;