Failed to echo values of variable inside "for loop"

Question

I was writing up a bash script to collect hardware information of each servers. But at some portion of the script I am unable to print values of variable that is given multiple times in a single echo command

#!/bin/bash
NO=$(df -hT | grep ext | wc -l)
for i in $(seq 1 $NO);do export PART$i=$(df -hT | grep ext | head -$i | tail -1 | awk '{print $1}'); done
echo "Total $NO ext partitions found"
for i in $(seq 1 $NO);do echo "Ext Partition$i is ${PART$i} ";done

In my system, I have two ext partitions, so variable $NO would store value 2.

The output I expect and wanted is as follows:

Total 2 ext partitions found
Ext Partition1 is /dev/sda7 
Ext Partition2 is /dev/sda8

But I am having trouble printing "$PART$i" as expected. Right now I am getting error for "for loop" in last line.

# /root/disk.sh
Total 2 ext partitions found
/root/disk.sh: line 5: Ext Partition$i is ${PART$i} : bad substitution

Can you tell me how to correctly invoke and echo the value of PART1 and PART2 in last for loop?.

Kurt Stutsman · Accepted Answer · 2016-02-22 17:26:30Z

3

I would suggest using an array variable instead:

#!/bin/bash
NO=$(df -hT | grep ext | wc -l)
declare -a PART
for i in $(seq 1 $NO);do PART[$i]=$(df -hT | grep ext | head -$i | tail -1 | awk '{print $1}'); done
echo "Total $NO ext partitions found"
for i in $(seq 1 $NO);do echo "Ext Partion$i is ${PART[$i]} ";done

You could further simplify this code as:

PART=( $(df -hT | awk '/ext/ { print $1 }') )
NO=${#PART[@]}
echo "Total $NO ext partitions found"
for i in ${!PART[@]}; do echo "Ext Partion$i is ${PART[$i]} "; done

Note that the simplification does change the indexing to be 0-based indexing.

edited Feb 22, 2016 at 17:26

answered Feb 22, 2016 at 16:42

Kurt Stutsman

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3 Comments

vjwilson Over a year ago

thank you so much. First one is working fine for me. For second one, the last for loop "for i in ${!x[@]}; do echo "Ext Partion$i is ${PART[$i]} "; done" print nothing on screen.

Kurt Stutsman Over a year ago

@johnwilson Oops forgot to change my test variable x back to PART in the second example hehe :) Thanks for letting me know.

Kurt Stutsman Over a year ago

@johnwilson The ${!PART[@]} is not indirect expansion in this case. That is syntax to return the keys of the array i.e. the indexes.

rob mayoff · Accepted Answer · 2016-02-22 17:00:21Z

2

You need to use indirect expansion:

v=PART$i
echo "Ext Partition$i is ${!v}"

From the bash manual:

If the first character of parameter is an exclamation point (!), it introduces a level of variable indirection. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion. The exceptions to this are the expansions of ${!prefix*} and ${!name[@]} described below. The exclamation point must immediately follow the left brace in order to introduce indirection.

edited Feb 22, 2016 at 17:00

answered Feb 22, 2016 at 16:43

rob mayoff

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1 Comment

vjwilson Over a year ago

It works. But can you tell what does that exclamation mark mean and what is actually does >> ${!v} ?

IT_User · Accepted Answer · 2016-02-22 16:39:48Z

-1

You are trying to call ${PART$i} which is two separate variables. Try:

echo "Ext Partition$i is ${PART}${i}"

answered Feb 22, 2016 at 16:39

IT_User

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1 Comment

vjwilson Over a year ago

Yeah, I have tried that one already, but it prints only the value of counter i in each iteration like as follows: <code> # /root/disk.sh Total 2 ext partitions found Ext Partion1 is 1 Ext Partion2 is 2 </code>

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