Do commands depending on the comparision

mr_bold · #1 (**permalink**) 04-12-2007

Dear all,

I am writing a shell, that do commands if the file count reaches 96. I wrote a shell, but doesn't work

Code:


#!/bin/sh
MIN_FILE_COUNT=96;
GET_PATH=/home/collection/data1; export GET_PATH

FILES_COUNT=`ls $dirname | wc -l`

if [ $FILES_COUNT > $MIN_FILE_COUNT ] then
echo "commands listed here"
fi


test.sh: line 9: syntax error near unexpected token `fi'
test.sh: line 9: `fi'

Thank you very much

msridh · #2 (**permalink**) 04-12-2007

Try this way.

#!/bin/ksh

MIN_FILE_COUNT=96
GET_PATH=/home/tcsmosr/tcsids; export GET_PATH

FILES_COUNT=`ls $dirname | wc -l`

if [ $FILES_COUNT -gt $MIN_FILE_COUNT ]
then
echo "commands listed here"
fi

Thanks and Regards,
msridh

matrixmadhan · #3 (**permalink**) 04-12-2007

Code:

if [ $FILES_COUNT > $MIN_FILE_COUNT ] then

should be

Code:

if [ $FILES_COUNT > $MIN_FILE_COUNT ]; then

mr_bold · #4 (**permalink**) 04-12-2007

Thanks to all.

Also this code works.

Code:

if awk 'BEGIN{if ('$FILES_COUNT' >= '$MIN_FILE_COUNT'); else exit 1}'
then
    echo true
else
    echo false
fi

anbu23 · #5 (**permalink**) 04-12-2007

Code:

if (( FILES_COUNT > MIN_FILE_COUNT )); then

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