平时做技术实践时，很多问题不是概念不会，而是细节没串起来。拿“Hive实现方法连续N天登陆语法实例代码”来说，它看着像小点，放到项目里常会牵出环境、配置、兼容性和维护成本。下面按实际采用顺序，把思路、关键写法和容易踩坑的地方讲清楚，便于大家直接对照操作。

目录

• Sql方式实现连续N天登陆
• 1.采用lag&lead+datediff窗口函数
• 2.采用date_add函数
• 代码实现思路
• 总结

Sql方式实现连续N天登陆

构造测试数据

create table dwd.login_log as
select 1 as user_id, "2020-01-01" as login_date
union all
select 1 as user_id, "2020-01-02" as login_date
union all
select 1 as user_id, "2020-01-07" as login_date
union all
select 1 as user_id, "2020-01-08" as login_date
union all
select 1 as user_id, "2020-01-09" as login_date
union all
select 1 as user_id, "2020-01-10" as login_date
union all
select 2 as user_id, "2020-01-01" as login_date
union all
select 2 as user_id, "2020-01-02" as login_date
union all
select 2 as user_id, "2020-01-04" as login_date

若日期格式不规范，能够将其转换为标准格式

create table dwd.login_log as
select user_id,to_date(from_unixtime(UNIX_TIMESTAMP(login_date,'yyyy-MM-dd'))) as login_date
from tmp.login_log; -- tmp库为原始数据

1.采用lag&lead+datediff窗口函数

• 理解这一步时，比如求连续三天登陆，能够将当天上一条数据和下一条数据都拿到，随后保证now-lag=lead-now=1即可;
• 从实现思路看，若是连续多天，能够取更多的数据，或者将数据全部更改为lag或者lead函数;
• datediff(date1, date2) - Returns the number of days between date1 and date2

select user_id 
from 
  (select user_id
  from
      (select user_id,
            lag(login_date,1) over(partition by user_id order by login_date) as lag_login_date,
            login_date,
            lead(login_date,1) over(partition by user_id order by login_date) as lead_login_date
      from dwd.login_log)t1
  where datediff(login_date,lag_login_date)=1 and datediff(lead_login_date,login_date)=1)t2
group by user_id;

2.采用date_add函数

• 结合项目来看，通用的，先对user_id分区排序，随后将日期减去rank天，查看有多少条数据即可;
• 优点在于能够统计具体连续登陆多少天,以及连续登陆的实际情况；
• date_add(start_date, num_days) - Returns the date that is num_days after start_date

select user_id,con_login_date,count(*) nums
from
    (select user_id,login_date,rk,date_add(login_date,1 - rk) as con_login_date
    from 
        (select user_id,login_date,rank() over(partition by user_id order by login_date) rk
        from dwd.login_log)t1
    )t2
group by user_id,con_login_date
having count(*) >= 3;

• t1表的查询结果

• 结合项目来看，t2表的查询结果，归一化的日期(也就是上述取前1 - rk)能够自己定义

• 在这个场景下，group by后的查询结果，第三列能够按照session内统计来理解，就是这批连续登陆内连续登陆的天数

代码实现思路

• 在这个场景下，采用代码来实现连续N天登陆，核心逻辑就是按照日期排序，新日期如果和旧日期相差1天就保留在HashMap里面，Size超过N即可输出user_id,否则清空

package cn.lang.spark_core
import java.text.{ParseException, SimpleDateFormat}
import java.util.Calendar
import org.apache.spark.sql.SparkSession
object ContinuousLoginDays {
  def main(args: Array[String]): Unit = {
    // env
    val spark: SparkSession = SparkSession
      .builder()
      .appName("ContinuousLoginDays")
      .master("local[*]")
      .getOrCreate()
    val sc = spark.sparkContext
    // source,可以是load hive(开启hive支持)或者parquet列式文件(定义好schema)
    val source = sc.textFile("/user/hive/warehouse/dwd/login_log")
    case class Login(uid: Int, loginTime: String) // 可以kryo序列化
    /** get date last `abs(n)` days defore or after biz_date   *
     * example biz_date = 20200101 ,last_n = 1,return 20191231 */
    def getLastNDate(biz_date: String,
                     date_format: String = "yyyyMMdd",
                     last_n: Int = 1): String = {
      val calendar: Calendar = Calendar.getInstance()
      val sdf = new SimpleDateFormat(date_format)
      try
        calendar.setTime(sdf.parse(biz_date))
      catch {
        case e: ParseException => // omit
      }
      calendar.set(Calendar.DATE, calendar.get(Calendar.DATE) - last_n)
      sdf.format(calendar.getTime)
    }
    // transform
    val result = source
      .map(_.split("t"))
      .map(iterm => Login(iterm(0).toInt, iterm(1)))
      .groupBy(_.uid) // RDD[(Int, Iterable[Login])]
      .map(iterm => {
        // 用于给此uid标记是否符合要求
        var CONTINUOUS_LOGIN_N = false
        val logins = iterm._2
          .toSeq
          .sortWith((v1, v2) => v1.loginTime.compareTo(v2.loginTime) > 0)
        var lastLoginTime: String = ""
        var loginDays: Int = 0
        logins
          .foreach(iterm => {
            if (lastLoginTime == "") {
              lastLoginTime = iterm.loginTime
              loginDays = 1
            } else if (getLastNDate(iterm.loginTime) == lastLoginTime) {
              lastLoginTime = iterm.loginTime
              loginDays = 2
            } else {
              lastLoginTime = iterm.loginTime
              loginDays = 1
            }
          })
        if (loginDays > 3) CONTINUOUS_LOGIN_N = true
        /** 此处可以使用集合将连续登陆的情况保留，
         * 也可以直接按照是否连续登陆N天进行标记
         */
        (iterm._1, CONTINUOUS_LOGIN_N)
      })
      .filter(_._2)
      .map(_._1)
    // sink
    result.foreach(println(_))
  }
}

总结

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