regex if (text contain this text) match this

(?(?=odp)-\d+\.\d+) won't work because (?=odp) is a positive lookahead that imposes a constraint on the pattern on the right, -\d+\.\d+. Namely, it requires odp string to occur exactly at the same location where - and a number are expected.

Use

(?<=ODP:)-\d+\.\d+
ODP:(-\d+\.\d+)

If lookbehinds are supported, the first variant is more viable. Otherwise, another option with capturing groups is good to use.

And if odp can appear anywhere, even after the number:

(?i)^(?=.*odp).*(-\d+\.\d+)

This will capture the value into a group.

EXPLANATION

--------------------------------------------------------------------------------
  (?i)                     set flags for this block (case-
                           insensitive) (with ^ and $ matching
                           normally) (with . not matching \n)
                           (matching whitespace and # normally)
--------------------------------------------------------------------------------
  ^                        the beginning of the string
--------------------------------------------------------------------------------
  (?=                      look ahead to see if there is:
--------------------------------------------------------------------------------
    .*                       any character except \n (0 or more times
                             (matching the most amount possible))
--------------------------------------------------------------------------------
    odp                      'odp'
--------------------------------------------------------------------------------
  )                        end of look-ahead
--------------------------------------------------------------------------------
  .*                       any character except \n (0 or more times
                           (matching the most amount possible))
--------------------------------------------------------------------------------
  (                        group and capture to \1:
--------------------------------------------------------------------------------
    -                        '-'
--------------------------------------------------------------------------------
    \d+                      digits (0-9) (1 or more times (matching
                             the most amount possible))
--------------------------------------------------------------------------------
    \.                       '.'
--------------------------------------------------------------------------------
    \d+                      digits (0-9) (1 or more times (matching
                             the most amount possible))
--------------------------------------------------------------------------------
  )                        end of \1

You can use the regex, (?i)(?<=odp:)[^,]*.

Explanation:

• (?i): Case-insenstitive flag
• (?<=odp:): Positive lookbehind for odp:
• [^,]*: Anything but ,

If you want the match to be restricted to numbers only, you can use the regex, (?i)(?<=odp:)(?:-\d+.\d+)

Explanation:

• (?i): Case-insenstitive flag
• (?<=odp:): Positive lookbehind for odp:
• (?:: Start non capturing group
  • -: Literal, -
  • \d+: 1+ digit(s)
  • .\d+: . followed by 1+ digit(s)
• ): End non capturing group

If the sign can be either + or -, you can use the regex, (?i)(?<=odp:)(?:[+-]\d+.\d+)

The pattern (?(?=odp)\-\d+\.\d+) is using a conditional (? stating in the if clause:

If what is directly to the right from the current position is odp, then match -\d+.\d+

That can not match.

What you also could do is match odp followed by any char other than a digit using \D* and capture the digit part in a group.

\bodp\b\D*(-\d+\.\d+)\b

The pattern matches:

• \bodp\b match odp between word boundaries to prevent a partial match
• \D* Optionally match any char other than a digit
• (-\d+\.\d+) Capture - and 1+ digits with a decimal part in group 1
• \b A word boundary

Regex demo

(?<=ODP:)(-\d+.\d+)

You can try using the negative look behind. This should solve for the code you ve provided.