编写MySQL查询：判断患者30天内是否再入院（标记1/0）

API 探路者

2026-8-8

MySQL查询：判断患者30天内是否再次入院

假设你的表名为admissions，以下两种方法均可实现需求：

方法一：使用EXISTS子查询

通过子查询检查同一患者是否存在满足时间条件的后续入院记录：

SELECT DISTINCT
    `Patient ID`,
    CASE
        WHEN EXISTS (
            SELECT 1
            FROM admissions b
            WHERE b.`Patient ID` = a.`Patient ID`
            AND b.`admission start` > a.`admission stop`
            AND DATEDIFF(b.`admission start`, a.`admission stop`) <= 30
        ) THEN 1
        ELSE 0
    END AS `30天内是否再入院`
FROM admissions a;

方法二：使用窗口函数LEAD

先获取每条入院记录的下一次入院时间，再分组判断是否存在30天内再入院的情况：

SELECT
    `Patient ID`,
    MAX(CASE
        WHEN next_admission_start IS NOT NULL AND DATEDIFF(next_admission_start, `admission stop`) <= 30 THEN 1
        ELSE 0
    END) AS `30天内是否再入院`
FROM (
    SELECT
        `Patient ID`,
        `admission start`,
        `admission stop`,
        LEAD(`admission start`) OVER (PARTITION BY `Patient ID` ORDER BY `admission start`) AS next_admission_start
    FROM admissions
) t
GROUP BY `Patient ID`;