UNIX for Dummies Questions & Answers

Im trying to set a filename to a variable and then read the file in using the variable but im getting a syntax error. any ideas?

#!/bin/bash



function scanFile()
{
while read $1
do
echo $filename
done
}

file1=report.log


scanFile() $file1

Leave off the () when calling the function, its only meant for when you're declaring it, and is optional even then.

read does not work that way. read NAME does not read from the file NAME, it reads into the variable $NAME. It usually reads from standard input, redirection is used if you want to change that. You'd want to do this:

Code:

function scanFile()
{
        # Local variable, incase you have FILENAME as a global variable somewhere else
        local FILENAME
        while read FILENAME
        do
                echo "$FILENAME"
        done < "$1"

        # Because BASH breaks the loop before processing the last input in some circumstances
        [ ! -z "$FILENAME" ] && echo "$FILENAME"        
}

do i need to set file1=report.log outside the variable still.

report.log is in the same directory i will be running my script from.

can i not set report.log to a variable and then use that variable in the loop?

What does "outside the variable" mean? You mean, just have the function get a name from a global variable instead of a parameter? Yes, you could, just replace "$1" with "$file1", but its generally a bad idea to use or modify global variables inside a function; If you ever forget which things it modifies or uses, you could be in for big headaches later when it starts doing things you didn't expect for no apparent reason.

I do sometimes make exceptions for log files myself, but I'm not certain how you're going to be using this.