Question

I have a bash script that has this function in it:

function start_vi()
{
   echo "Please enter a file name with complete path to open in vi:"
   read input_file

   if [ -d "$input_file" ]
   then
      echo "You entered a directory."
      echo "Please try again and enter a readable/writable file."
   fi

   grep_var="file $input_file | grep -c data"

   if [ $? -eq 0 ]
   then
      vi $input_file
   else
      echo "File not found or invalid file type. Please try again."
   fi
}

For the most part the function works fine. My problem is that the two if statements work fine independently, eg, if I comment out one of them, the test works and it does what I want. But together, as written, for example, when I type in a directory at the prompt, vi opens it as a file, where the test should return an echo saying that it’s a directory, as it does when functioning alone.

Any ideas on why this is? I’m still relatively new at bash scripting so it is probably easy for the pros, but I have been banging my head against the wall for a while now.

Thanks in advance.

Answer 1

Add a return statement in the first if/then:

function start_vi()
{
   echo "Please enter a file name with complete path to open in vi:"
   read input_file

   if [ -d "$input_file" ]
   then
      echo "You entered a directory."
      echo "Please try again and enter a readable/writable file."
      return
   fi

   grep_var="file $input_file | grep -c data"

   if [ $? -eq 0 ]
   then
      vi $input_file
   else
      echo "File not found or invalid file type. Please try again."
   fi
}

Otherwise, it will print and then open the file anyway, as your second test should be like this:

   file $input_file | grep -c data

   if [ $? -eq 0 ]

The $? is the exit code of the last run command. Assigning to a variable (i.e. grep_var="...") sets $? to 0. What you seem to have wanted is the exit code of grep -c data – in that case, use backticks ` instead of quotes ” to run commands, like below. Or you can write it like this:

   grep_var=`file $input_file | grep -c data`

   if [ $grep_var != 0 ]

to compare the string value (i.e. what grep -c data returns – the count of data lines).

Doing some of the above should solve the problem.