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terdon
terdon
11:00 PM
Am I missing something here? See the comments:
7
A: Bash variable substitution of variable followed by underscore

terdonAs George Vassiliou already explained, that's because you're printing the variable $BUILDNUMBER_ instead of $BUILDNUMBER. The best way to get what you want is to use ${BUILDNUMBER}_ as George explained. Here are some more options: $ echo "$BUILDNUMBER"_ 230_ $ echo $BUILDNUMBER"_" 230_ $ printf ...

I see no reason to quote an underscore. And yes, variables should indeed be quoted, but the whole point of the one unquoted one was to demonstrate that "$var"_ and $var"_" both work when appending a string to a variable (which I know to be nothing but a number).
 

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