CODEWITHSUNDEEP
bashparameter-passing
Yacob
Yacob

Reputation: 525

Accessing command line arguments beyond the 9th in bash script

I am trying to read command line arguments in bash but I have problems to read on the 10th column. Here is my sample script:

#-------------------------------------------------------
#!/bin/bash
an=$2  mn=$4  dy=$6  der=$8  new=$10  sec=(${12} ${13})
echo $an $mn  $dy   $der  $new  $sec 
#--------------------------------------------------------

I have run the above script "test.sh" as

./test.sh -yr cat  -mn Jan  -dy  tuesday  -der tt  -new car -sec 001 001

The output is:

cat Jan tuesday tt -yr0 001

But for variable $new ($10) the answer should have been car but I get -yr0

Any idea why?

Upvotes: 2

Views: 783

Answers (2)

Hai Vu
Hai Vu

Reputation: 40688

Here is the fix:

#!/bin/bash
an=$2  mn=$4  dy=$6  der=$8  new=${10}  sec="${12} ${13}"
echo $an $mn  $dy   $der  $new  $sec
  1. Instead of $10, which is "$1" with 0 appended, use ${10}
  2. sec="..." instead of using parentheses, unless you meant to use array

Upvotes: 3

jam
jam

Reputation: 3688

new=$10 is giving you $1 (i.e 'yr') appended with 0. You need some braces : ${10}

Upvotes: 4

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