Mysql 如何根据条件中的值仅显示行？

我如何只向库尔特所在部门的人员展示

SELECT * FROM sample_table 
WHERE (name= 'kurt' OR department IN (select department from sample_table
where name = 'kurt'
group by department 
having count(*) > 1));
    

从这一点：

名称 部门 库尔特 2. 厕所 1. 做记号 1. 玫瑰 2. 安娜 2.

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一种方法使用现有逻辑：

选择s1.name，s1.department 来自样本_表s1 如果存在，则从样本表格s2中选择1 其中s2.department=s1.department，s2.name='kurt'；

---

一种方法使用现有逻辑：

选择s1.name，s1.department 来自样本_表s1 如果存在，则从样本表格s2中选择1 其中s2.department=s1.department，s2.name='kurt'； 试试这个：

SELECT *
FROM
    sample_table
WHERE
    department = (
        SELECT
            department
        FROM
            sample_table
        WHERE
            name = 'kurt'
        LIMIT 1
    )

试试这个：

SELECT *
FROM
    sample_table
WHERE
    department = (
        SELECT
            department
        FROM
            sample_table
        WHERE
            name = 'kurt'
        LIMIT 1
    )

答复如下

SELECT
    name, department
FROM
    sample_table
where
    department in (
        select distinct department
    from
        sample_table
    where
        name = 'kurt')

即使kurt在许多部门，这种情况也会出现

回答如下

SELECT
    name, department
FROM
    sample_table
where
    department in (
        select distinct department
    from
        sample_table
    where
        name = 'kurt')

---

即使kurt在许多部门，这也会返回

我甚至不知道EXISTS子句。谢谢你让我知道！我甚至不知道EXISTS子句。谢谢你让我知道！