How to print string between two character using Linux/bash command

Question

I am trying to find a Linux command which could give me the text between [ and ]. I tried standard grep and sed but none helped for far.Say the text is in LOGS. like -

    Starting mongo client..
Connecting to truststore.pki.rds.amazonaws.com (99.84.66.9:443)
global-bundle.pem    100% |********************************|  183k  0:00:00 ETA
MongoDB shell version v4.0.6
connecting to: mongodb://XXXx.amazonaws.com:27017/?authSource=XXX&gssapiServiceName=mongodb
Implicit session: session { "id" : UUID("XXXXXXX") }
MongoDB server version: 5.0.0
WARNING: shell and server versions do not match
----------------------------------------------------------------------------------------------------
        Listing Existing set of collections
----------------------------------------------------------------------------------------------------
[
    "10w",
    "11w",
    "Test1",
    "Test2",
    "QA_Testing",
    "QueuesTest",
    "TestPagination",
]

I wanted the text only this collection names inside []

Tried so far

1. echo "$LOGS" | grep -o "\[.*\]"
2. echo "$LOGS" | sed -n 's/.*\[\([^]]*\)\].*/\1/p'
3. echo "$LOGS" | grep -o '\[\K[^\]]+'
4. echo "$LOGS" | awk -F'[][]' '{for(i=2;i<=NF;i+=2) print $i}'

None of them helped so far. They are not showing any error but not also any output. Can anyone please help with the correct command which can help me

Answer 1

burcu's answer is perfect. As I prefer using the if statement in awk , here is my answer:

awk '{
    if ($0 == "[") {
        in_bracket = 1
        next
    }else if ($0 == "]"){
        in_bracket = 0
    }
    if (in_bracket) {
        print $0
    }
}' a.txt