Regex Replace A Variable Within Brackets, Including Brackets -

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how rid of variable within square brackets, including brackets themselves? e.g. [152] or [153] or [154]. using yahoo pipes.

you can escape brackets (like other character special meaning) \.

s/\[\d+\]/replacement/

in yahoo pipes should work like: replace \[.+\] with (leave blank). maybe have check g flag.


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