Extracting string pattern that is not within square brackets using regex

Question

I'm trying to extract alpha numeric characters from strings that also have some info within square brackets.

Ex:

this is some sample text [first sentence]
[second sentence][important] some more sample text
[not important] this is sample as well

I want the output to be:

this is some sample text
some more sample text
this is sample as well

I tried using negative look ahead, extracting patterns before'[' (works only for a few cases).

@CaiusJard I'm querying a set of string values logged in a table using presto. It uses the same regex functionality as javascript. — starHopp3r
– starHopp3r, Commented Aug 21, 2019 at 13:32
Instead of trying to extract stuff not in brackets can you run a regex replace to remove everything that is in brackets? Eg replace " *\[.*?\] *" with ""? — Caius Jard
– Caius Jard, Commented Aug 21, 2019 at 13:39

Caius Jard · Accepted Answer · 2019-08-21 13:44:32Z

1

Per my comment, and after a quick look at the fine presto manual, could you:

SELECT regexp_replace('[second sentence][important] some more sample text', ' *\[.*?\] *');

Regex is any number of spaces, then sq bracket then any number of any char up to the next sq bracket, then sq bracket, then any number of spaces.

I dug the function out of the manual (no access to presto/never used), I presume by providing only two arguments it implicitly replaces matches with nothing

edited Aug 21, 2019 at 13:44

answered Aug 21, 2019 at 13:42

Caius Jard

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2 Comments

starHopp3r Over a year ago

Thank you, that worked. However I used REGEXP_REPLACE(string_field, '\[.*?\]', '')

Caius Jard Over a year ago

No problems - I had the extra space-asterisk in there to try and nuke the trailing and leading spaces around brackets too. You could also TRIM, or whatever presto does for removing white spaces from ends of strings

Sam L · Accepted Answer · 2019-08-21 13:45:49Z

0

You could try to use a Regex for this, but I think that making your own function for this would work well.

function getText(bracketedText) {
  let idx = 0
  let newIdx = 0
  let str = ''
  while (newIdx !== -1) {
    newIdx = bracketedText.indexOf('[', idx)
    if (newIdx < 0) {
      str += bracketedText.slice(idx, bracketedText.length)
    } else {
      str += bracketedText.slice(idx, newIdx)
    }
    idx = bracketedText.indexOf(']', newIdx + 1) + 1
  }
  return str
}

This should be fairly efficient at stripping out anything in brackets.

answered Aug 21, 2019 at 13:45

Sam L

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