regex - ORACLE REGEXP_SUBSTR that match digits between square brackets -

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i have varchar column no have specific format.

these examples:

afdsbfgf, jbhttp://www.iabvdfbdos.com/view.php?p_id=170405, arcm, cocm, fbus, bv[123545]

i need match substring between brackets when string 'bv[123545]'. substring numeric.

i tried that:

regexp_substr (my_column, '[^bv\[]+[[:digit:]]+')

but matches 

'jbhttp://www.iabvdfbdos.com/view.php?p_id=170405' , returns "_id=170405"

thanks in advance

assumes brackets appear once in line. grabs first sub-group (the part in parens) of first occurrence of 'bv' followed left square bracket, followed 1 or more digits, followed right square bracket. square brackets escaped define character class in regex engine otherwise.

sql> select regexp_substr('afdsbfgf, jbhttp://www.iabvdfbdos.com/view.php?p_id=170405, arcm, cocm, fbus, bv[123545]', 'bv\[(\d+)\]', 1, 1, null, 1 ) dual;  regexp ------ 123545  sql>

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