Bash Variable Substitution – Substitution of Variable Followed by Underscore

bashshell-scriptvariable substitution

The variable BUILDNUMBER is set to value 230. I expect 230_ to be printed for the command echo $BUILDNUMBER_ but the output is empty as shown below.

# echo $BUILDNUMBER_

# echo $BUILDNUMBER
230

Best Answer

The command echo $BUILDNUMBER_ is going to print the value of variable $BUILDNUMBER_ which is not set (underscore is a valid character for a variable name as explicitly noted by Jeff Schaller)

You just need to apply braces (curly brackets) around the variable name or use the most rigid printf tool:

echo "${BUILDNUMBER}_"
printf '%s_\n' "$BUILDNUMBER"

PS: Always quote your variables.

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Shell – Assigning the output of a SQL query to variable

The terminating here-doc word must be the only characters on the line: no indenting allowed. Also, use $() instead of backticks -- they are nestable.

count=$(sqlplus -s $configuser/$configpass@$ORACLE_SID <<END
       set pagesize 0 feedback off verify off heading off echo off;
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       exit;
END
)
echo $count

http://www.gnu.org/software/bash/manual/bashref.html#Here-Documents

Bash – Double variable substitution in bash

You can do this with indirection

for progs in ${list[@]}; do
     a="${progs}_MIN_REQ"
     b="${progs}_CURR"
     echo "$progs: ${!a}:${!b}"
done

Best Answer

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Bash – Double variable substitution in bash

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