Hi Sean,
Thanks for quick response!
I have tried with string literal 'r' as a prefix that also gave an empty
result..
spark.sql(r"select regexp_extract('[11] [22]
[33]','(^\[OrderID:\s)?(?(1).*\]\s\[UniqueID:\s([a-z0-9A-Z]*)\].*|\[.*\]\s\[([a-z0-9A-Z]*)\].*)',1)
as anyid").show()
and as I mentioned when I am using 2 backslashes it is giving an exception
as follows:
: java.util.regex.PatternSyntaxException: Unknown inline modifier near
index 21
(^\[OrderID:\s)?(?(1).*\]\s\[UniqueID:\s([a-z0-9A-Z]*)\].*|\[.*\]\s\[([a-z0-9A-Z]*)\].*)
Kind Regards,
Sachit Murarka
On Wed, Dec 2, 2020 at 9:07 PM Sean Owen <sro...@gmail.com> wrote:
> As in Java/Scala, in Python you'll need to escape the backslashes with \\.
> "\[" means just "[" in a string. I think you could also prefix the string
> literal with 'r' to disable Python's handling of escapes.
>
> On Wed, Dec 2, 2020 at 9:34 AM Sachit Murarka <connectsac...@gmail.com>
> wrote:
>
>> Hi All,
>>
>> I am using Pyspark to get the value from a column on basis of regex.
>>
>> Following is the regex which I am using:
>>
>> (^\[OrderID:\s)?(?(1).*\]\s\[UniqueID:\s([a-z0-9A-Z]*)\].*|\[.*\]\s\[([a-z0-9A-Z]*)\].*)
>>
>> df = spark.createDataFrame([("[1234] [3333] [4444] [66]",),
>> ("abcd",)],["stringValue"])
>>
>> result = df.withColumn('extracted value',
>> F.regexp_extract(F.col('stringValue'),
>> '(^\[OrderID:\s)?(?(1).*\]\s\[UniqueID:\s([a-z0-9A-Z]*)\].*|\[.*\]\s\[([a-z0-9A-Z]*)\].*)',
>> 1))
>>
>> I have tried with spark.sql as well. It is giving empty output.
>>
>> I have tested this regex , it is working fine on an online regextester .
>> But it is not working in spark . I know spark needs Java based regex ,
>> hence I tried escaping also , that gave exception:
>> : java.util.regex.PatternSyntaxException: Unknown inline modifier near
>> index 21
>>
>> (^\[OrderID:\s)?(?(1).*\]\s\[UniqueID:\s([a-z0-9A-Z]*)\].*|\[.*\]\s\[([a-z0-9A-Z]*)\].*)
>>
>>
>> Can you please help here?
>>
>> Kind Regards,
>> Sachit Murarka
>>
>
