[SOLVED] Confusion on get the first parameter on Shell Script

MarcosPauloBR · 03-24-2011, 09:55 AM

Hi people!

I'm getting a error message on this script

Code:

if [ $1 -e '1' ]
then envia_arquivo
else 
       if [ $1 -e 2 ]
        then echo 'Implementar!'
        else echo 'Os parametros devem ser 1(para envio de arquivos) ou 2(para receber arquivos)'
             exit 1
       fi
fi

when o call this file with 1 as the parameter, I get this message:

./envio.sh: line 25: [: -e: binary operator expected
./envio.sh: line 28: [: -e: binary operator expected

OBS: There are more lines, because of this the line number are higher then the script.

I'm doing anything wrong?

Thanks!!

sycamorex · 03-24-2011, 10:01 AM

The -e flag refers to FILES. You could do it as follows:

Code:

if [ "$1" != "" ]; then
    echo "Positional paremeter exists"
else
    echo "Positional Parameter $1 does not exist"
fi

MarcosPauloBR · 03-24-2011, 10:05 AM

I need to compare the parameter with a number.

How can I do this??

Thanks for your time!!

sycamorex · 03-24-2011, 10:21 AM

What if you modify the script I posted above and include the number in " "?

MarcosPauloBR · 03-24-2011, 10:28 AM

I did it!

I modified to if [ $1 = '1'] and it works!

Thanks!

chrism01 · 03-25-2011, 01:25 AM

Actually, for numbers you should use

Code:

if [[ $1 -eq 1 ]]

strings use

Code:

if [[ $1 = "somestr" ]]

http://tldp.org/LDP/abs/html/comparison-ops.html
http://tldp.org/LDP/abs/html/fto.html
http://tldp.org/LDP/abs/html/testconstructs.html