Oracle, regexp_substr - how to match 'string'+regex - oracle

Question

I have a column that gives me lines like this:
<?xml version="1.0" encoding="UTF-8"?>
<xSettings>
<systemPropertyName>BLALBLA</systemPropertyName>
<minimumAmount>198.00</minimumAmount>
<closingAmount>198.00</closingAmount>
<useThisSetting>true</useThisSetting>
<SystemStep dayAfterPrevious="0">
<System SystemService="1" minimumAmount="450.00" />
</SystemStep>
<SystemStep dayAfterPrevious="8">
<message />
</SystemStep>
<SystemStep dayAfterPrevious="3">
<block />
</SystemStep>
<SystemStep dayAfterPrevious="1">
<message />
</SystemStep>
<SystemStep dayAfterPrevious="7">
<message />
</SystemStep>
</xSettings>
All numbers in it are variable and the BLA BLA is variable too. What I want is a select that only gives me
<minimumAmount>198.00</minimumAmount>
though this would be eve better:
198.00
basically, I can't figure out how to use regexp_substr to find a specific string , then return a number just after that may be from 1 to 4 digits and has two decimals after.

Answer 1

Do not use regular expressions to parse XML data - use a proper XML parser:
SELECT x.minimumAmount
FROM your_table t,
XMLTable(
'/xSettings',
PASSING XMLType( t.your_column )
COLUMNS minimumAmount NUMBER(5,2) PATH './minimumAmount'
) x
Or
SELECT TO_NUMBER(
EXTRACTVALUE( XMLType( your_column ), '/xSettings/minimumAmount' )
) AS minimumAmount
FROM your_table

Answer 2

XML data should not be queried using regular expressions, but here:
select
regexp_substr(col, '<minimumAmount>(.*)</minimumAmount>',1,1,null,1) minimum_amount
from your_table;
One other way is to use regexp_replace like this:
select
regexp_replace(col, '.*<minimumAmount>(.*)</minimumAmount>.*','\1') minimum_amount
from your_table;