How to print all the elements of an array into file using bash

Question

I have an array of element in one variable. I have to print all the values of the array into a file. Initially, I read the values of an array from many files using sed and cut. I have print those values(array) into a file. When I use print statement within loop, the values are overwritten. Is there any other way to print all the elements into a file using commands like awk or cat???
Here is the code which I tried

for ((GPCount=0; GPCount< $GPCnt; GPCount++))  
do
G=${GPArr[$GPCount]}
G="$(sed -n '2p' $file.txt|cut -d" " -f7)"
printf "%d\n" "$G" >$out.txt
done

I know if I use this in loop it will overwrite, so I tried another block of code

for ((GPCount=0; GPCount< $GPCnt; GPCount++))  
do
$GPArr[$GPCount]="$(sed -n '2p' $file.txt|cut -d" " -f7)"
done
printf "%d\n" "${GPArr[@]}" >$out2.txt

there is no output and I am getting error( its giving the value but it couldnt direct the sed o/p to array variable

./test.sh: line 33: =33: command not found
./test.sh: line 33: =39: command not found
./test.sh: line 33: =45: command not found

If I modify the code like this

 for ((GPCount=0; GPCount< $GPCnt; GPCount++))  
 do
 G="$(sed -n '2p' $file.txt|cut -d" " -f7)"
 done
 printf "%d\n" "${GPArr[@]}" >$out2.txt

I am getting 0 as a output in the text file. I need the o/p like this in my text file

33
39
45

Kindly help me in this regard. Thanks in advance

Jahid · Accepted Answer · 2015-05-16 02:45:39Z

3

printf "%d\n" "$G" >>$out.txt

or

echo "$G" >>$out.txt

> will overwrite your whole file, while >> will append to the file

Usage:

for ((GPCount=0; GPCount< $GPCnt; GPCount++))  
do
G=${GPArr[$GPCount]}
G="$(sed -n '2p' $file.txt|cut -d" " -f7)"
printf "%d\n" "$G" >>$out.txt
done

edited May 16, 2015 at 2:45

answered May 16, 2015 at 2:36

Jahid

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Comments

alvits · Accepted Answer · 2015-05-16 02:25:59Z

0

The easiest way to accomplish your task is to use the bash builtin printf command. Note: this is bash builtin printf, not /usr/bin/printf.

for ((GPCount=0; GPCount< $GPCnt; GPCount++))  
do
printf -v GPArr[$GPCount] "%d" $(sed -n '2p' $file.txt|cut -d" " -f7)
done
printf "%d\n" "${GPArr[@]}" >$out2.txt

Give it a try.

answered May 16, 2015 at 2:25

alvits

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Comments

chepner · Accepted Answer · 2015-05-16 03:50:39Z

Your first loop fails because you seem to be trying to use G as an alias for ${GPArr[$GPCount]}. bash variables don't work that way (although in bash 4.3, you can declare a nameref which would work:

for ((GPCount=0; GPCount< $GPCnt; GPCount++))  
do
    declare -n G=${GPArr[$GPCount]}
    G="$(sed -n '2p' $file.txt|cut -d" " -f7)"
    printf "%d\n" "$G"
done > "$out.txt"

Also, make sure you aren't overwriting the file $out.txt each time through the loop. )

Your second loop just has a typo; you don't prefix a variable assignment with a dollar sign.

for ((GPCount=0; GPCount< $GPCnt; GPCount++))  
do
    # Not $GPArr[$GPCOUNT]=...
    GPArr[$GPCount]="$(sed -n '2p' $file.txt|cut -d" " -f7)"
done
printf "%d\n" "${GPArr[@]}" >$out2.txt

Incidentally, you don't need a loop counter at all; you can append to an array with the += operator.

for ((GPCount=0; GPCount< $GPCnt; GPCount++))  
do
    # Not $GPArr[$GPCOUNT]=...
    GPArr+=("$(sed -n '2p' $file.txt|cut -d" " -f7)")
done
printf "%d\n" "${GPArr[@]}" >$out2.txt

(In any case, it's not clear how the sed call is producing different values each time through the loop.)

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