Question

我知道为什么我会得到以下结果？

scala> val b = to_timestamp($"DATETIME", "ddMMMYYYY:HH:mm:ss")
b: org.apache.spark.sql.Column = to_timestamp(`DATETIME`, 'ddMMMYYYY:HH:mm:ss')

scala> sourceRawData.withColumn("ts", b).show(6,false)
+------------------+-------------------+-----------+--------+----------------+---------+-------------------+
|DATETIME          |LOAD_DATETIME      |SOURCE_BANK|EMP_NAME|HEADER_ROW_COUNT|EMP_HOURS|ts                 |
+------------------+-------------------+-----------+--------+----------------+---------+-------------------+
|01JAN2017:01:02:03|01JAN2017:01:02:03 | RBS       | Naveen |100             |15.23    |2017-01-01 01:02:03|
|15MAR2017:01:02:03|15MAR2017:01:02:03 | RBS       | Naveen |100             |115.78   |2017-01-01 01:02:03|
|02APR2015:23:24:25|02APR2015:23:24:25 | RBS       |Arun    |200             |2.09     |2014-12-28 23:24:25|
|28MAY2010:12:13:14| 28MAY2010:12:13:14|RBS        |Arun    |100             |30.98    |2009-12-27 12:13:14|
|04JUN2018:10:11:12|04JUN2018:10:11:12 |XZX        | Arun   |400             |12.0     |2017-12-31 10:11:12|
+------------------+-------------------+-----------+--------+----------------+---------+-------------------+

我正在尝试将DATETIME（以ddMMMYY：HH：mm：ss格式）转换为Timestamp（显示在上面的最后一列），但它似乎没有转换为正确的值。我提到了以下帖子，但没有帮助：

Better way to convert a string field into timestamp in Spark

任何人都可以帮助我？

Answer 1

使用y（年）而不是Y（周年）：

spark.sql("SELECT to_timestamp('04JUN2018:10:11:12', 'ddMMMyyyy:HH:mm:ss')").show
// +--------------------------------------------------------+
// |to_timestamp('04JUN2018:10:11:12', 'ddMMMyyyy:HH:mm:ss')|
// +--------------------------------------------------------+
// |                                     2018-06-04 10:11:12|
// +--------------------------------------------------------+

Answer 2

试试这个UDF：

val changeDtFmt = udf{(cFormat: String,
                         rFormat: String,
                         date: String) => {
  val formatterOld = new SimpleDateFormat(cFormat)
  val formatterNew = new SimpleDateFormat(rFormat)
  formatterNew.format(formatterOld.parse(date))
}}

sourceRawData.
  withColumn("ts", 
    changeDtFmt(lit("ddMMMyyyy:HH:mm:ss"), lit("yyyy-MM-dd HH:mm:ss"), $"DATETIME")).
  show(6,false)

Answer 3

尝试以下代码

我为表格

创建了一个示例数据框“df”

+---+-------------------+
| id|               date|
+---+-------------------+
|  1| 01JAN2017:01:02:03|
|  2| 15MAR2017:01:02:03|
|  3|02APR2015:23:24:25 |
+---+-------------------+
val t_s= unix_timestamp($"date","ddMMMyyyy:HH:mm:ss").cast("timestamp")

df.withColumn("ts",t_s).show()

+---+-------------------+--------------------+
| id|               date|                  ts|
+---+-------------------+--------------------+
|  1| 01JAN2017:01:02:03|2017-01-01 01:02:...|
|  2| 15MAR2017:01:02:03|2017-03-15 01:02:...|
|  3|02APR2015:23:24:25 |2015-04-02 23:24:...|
+---+-------------------+--------------------+

由于

如何将自定义日期时间格式转换为时间戳？

3 个答案: