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bash script error message if else

bash script error message if else

Here is a snippet of my bash code ..can't seem to figure out the error I get; however, if I comment out the first 2 if statements runs fine ..the error I get is "syntax error: unexpected end of file", points the last line

if [[ $# -eq 0 || $# -gt 1 ]] && usage
if [[ $1 != "process1" || $1 != "process2" ]] && usage

if [ "$1" = "$DB" ]; then
echo "$1 process already running..."
exit 1

RE: bash script error message if else

Replace this:
if [[ $# -eq 0 || $# -gt 1 ]] && usage
if [[ $1 != "process1" || $1 != "process2" ]] && usage
with either this:
[[ $# -eq 0 || $# -gt 1 ]] && usage
[[ $1 != "process1" || $1 != "process2" ]] && usage
or this:
if [[ $# -eq 0 || $# -gt 1 ]]; then usage; fi
if [[ $1 != "process1" || $1 != "process2" ]]; then usage; fi

In fact, I'd use this:
[[ $# -eq 1 ]] || usage
[[ $1 != "process1" && $1 != "process2" ]] && usage

Hope This Helps, PH.
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RE: bash script error message if else

I'm calling my script like this: ./processName wws2

below is my syntax, how can I use the argument I'm passing to this command below..doesn't seem to like it being passed in as a parameter?

VAR=`ps -ef | grep apdel | awk '/$1/ {print $10}' | cut -d\/ -f5`

RE: bash script error message if else

Single quotes do not allow the variable substitution. It needs to be in double quotes. But then you need to protect the "$10" with a backslash.

Maybe this?


VAR=`ps -ef | grep apdel | awk "/$1/ {print \$10}" | cut -d\/ -f5` 

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