shell - how to properly use comparison operators within an if-then statement -

what best way perform check/test output comes query called within shell script?

for example, have variable performs query:

export chk=`psql -h localhost -p 5432 -u foo foobar --tuples-only -c "select 1 result pg_database datname='foo_bar_1234'";` 

it return either 1=exist or empty string=not exist.

i tried code, doesn't work expected keeps going db not exist everytime if value empty string. tried "1" still same results.

if  [ "$chck" = 1 ];         echo "db exists."         exit  1      else         echo "db not exist." fi 

how correct and/or better way check this?

from comments sounds output of 1 may surrounded whitespace, 1 solution trim leading , trailing whitespace first, using read:

#!/usr/bin/env bash  export chck=$(psql -h localhost -p 5432 -u foo foobar --tuples-only -c "select 1 result pg_database datname='foo_bar_1234'")  # remove leading , trailing whitespace value. read -r chck <<<"$chck"  # whitespace trimmed, comparison string '1' should work intended. if  [[ $chck == '1' ]];   echo "db exists."   exit  1 else   echo "db not exist." fi 

• note unless need add $chck environment child processes can see it, there's no need use export.
• read -r chck <<<"$chck" takes advantage of fact read default removes leading , trailing whitespace input when reading single variable:
  • -r ensures \ instances in input left untouched - not strictly necessary here, practice in general.
  • <<< so-called bash here-string, sends argument via stdin command @ hand - read reads from. <<< single-line alternative posix-compliant here-documents.

---

another option use bash's regex-matching operator, =~, bypasses need trim whitespace first:

if [[ $chck =~ ^[[:space:]]*1[[:space:]]*$ ]]; # ... 

---

less stringently, if presence of 1 anywhere in string sufficient, using bash's pattern matching:

if [[ $chck == *1* ]]; # ... 

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even less stringently, if string empty (as opposed blank, meaning composed of whitespace characters) in non-existence case:

if [[ -n $chck ]]; # ... matches *any nonempty* string 

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since question generically tagged shell, here's posix-compliant equivalent of first solution:

export chck=$(psql -h localhost -p 5432 -u foo foobar --tuples-only -c "select 1 result pg_database datname='foo_bar_1234'")  # remove leading , trailing whitespace value. read -r chck <<eof $chck eof  # whitespace trimmed, comparison string '1' should work intended. if [ "$chck" = '1' ];   echo "db exists."   exit  1 else   echo "db not exist." fi 

• note use of here-document (<<) instead of bash-specific here-string (<<<), , [ ... ] (single-bracket) conditional form double-quoted variable reference.