[SOLVED] [scripting] try to passing arguments to for cycle (inside a function)

vomplete · 04-26-2011, 06:22 PM

Hi

The code:

Quote:

#!/bin/bash
# Dummy list
ls -1 / >> list.txt

# Function definition
ReadFileLines() {
for i in $(cat $1)
do
$2
done
}

# Run
ReadFileLines "list.txt" "echo $i"

Problem: I need a method to maintain the $i variable.
In fact, actually, this variable get lost when executed.
I think that an escape can preserve this variable and permit its execution inside the function, but I've no idea about.

Any ideas?

corp769 · 04-26-2011, 06:29 PM

Are you trying to run the [ReadFileLines "list.txt" "echo $i" ] only once, or do you want it ran every time the for loop sequences? From what I can see though, you can declare another variable and set it to the value of i, just make sure to declare it outside of the for-loop scope.

Kind regards,

Josh

crts · 04-26-2011, 06:35 PM

Hi,

try this

Code:

#!/bin/bash
# Dummy list
ls -1 / > list.txt
# Function definition
ReadFileLines() {
for i in $(cat $1)
do
eval $2
done
}
# Run
ReadFileLines "list.txt" 'echo $i'

While the above example "works" I still wonder why you would want to do this in such a peculiar way. Why don't you simply do
echo $i

inside the for-loop instead of expanding a variable that holds an instruction?

vomplete · 04-26-2011, 07:01 PM

Quote:

Originally Posted by crts

Hi,

try this

Code:

#!/bin/bash
# Dummy list
ls -1 / > list.txt
# Function definition
ReadFileLines() {
for i in $(cat $1)
do
eval $2
done
}
# Run
ReadFileLines "list.txt" 'echo $i'

While the above example "works" I still wonder why you would want to do this in such a peculiar way. Why don't you simply do
echo $i

inside the for-loop instead of expanding a variable that holds an instruction?

thanks 10000!!!

solved!