使用 spark.sql parse_url() 从包含大括号或管道的 URL 中提取主机

Question

我需要从数百万 URL 中提取主机。一些 URL 格式不正确并且 return NULL。在许多情况下，我看到大括号 ({}) 或管道 (|) 导致问题，其他时候我看到多个散列 (#) 字符导致问题。

这是我的代码，其中包含我需要解析的 URL：

val b = Seq(
    ("https://example.com/test.aspx?doc={1A23B4C5-67D8-9012-E3F4-A5B67890CD12}"),
    ("https://example.com/test.aspx?names=John|Peter"),
    ("https://example.com/#/test.aspx?help=John#top"),
    ("https://example.com/test.aspx?doc=1A23B4C5-67D8-9012-E3F4-A5B67890CD12"),
    ).toDF("url_col")

b.createOrReplaceTempView("temp")
spark.sql("SELECT parse_url(`url_col`, 'HOST') as HOST, url_col from temp").show(false)

预期输出：

+-----------+------------------------------------------------------------------------+
|HOST       |url_col                                                                 |
+-----------+------------------------------------------------------------------------+
|example.com|https://example.com/test.aspx?doc={1A23B4C5-67D8-9012-E3F4-A5B67890CD12}|
|example.com|https://example.com/test.aspx?names=John|Peter                          |
|example.com|https://example.com/#/test.aspx?help=John#top                           |
|example.com|https://example.com/test.aspx?doc=1A23B4C5-67D8-9012-E3F4-A5B67890CD12  |
+-----------+------------------------------------------------------------------------+

当前输出：

+-----------+------------------------------------------------------------------------+
|HOST       |url_col                                                                 |
+-----------+------------------------------------------------------------------------+
|null       |https://example.com/test.aspx?doc={1A23B4C5-67D8-9012-E3F4-A5B67890CD12}|
|null       |https://example.com/test.aspx?names=John|Peter                          |
|null       |https://example.com/#/test.aspx?help=John#top                           |
|example.com|https://example.com/test.aspx?doc=1A23B4C5-67D8-9012-E3F4-A5B67890CD12  |
+-----------+------------------------------------------------------------------------+

当 URL 包含无效字符或格式错误时，是否有办法强制 parse_url 到 return 主机？或者有更好的方法吗？

Answer 1

您可以使用 regexp_extract 函数提取域（regex 的示例）：

spark.sql("""
    SELECT  regexp_extract(url_col, "^(?:https?:\/\/)?(?:[^@\n]+@)?(?:www.)?([^:\/\n?]+)", 1) as HOST, 
            url_col 
    FROM  temp
""").show(false)

//+-----------+------------------------------------------------------------------------+
//|HOST       |url_col                                                                 |
//+-----------+------------------------------------------------------------------------+
//|example.com|https://example.com/test.aspx?doc={1A23B4C5-67D8-9012-E3F4-A5B67890CD12}|
//|example.com|https://example.com/test.aspx?names=John|Peter                          |
//|example.com|https://example.com/#/test.aspx?help=John#top                           |
//|example.com|https://example.com/test.aspx?doc=1A23B4C5-67D8-9012-E3F4-A5B67890CD12  |
//+-----------+------------------------------------------------------------------------+

使用 spark.sql parse_url() 从包含大括号或管道的 URL 中提取主机

Extract HOST from URL containing braces or pipes using spark.sql parse_url()

scala

apache-spark

apache-spark-sql