CODEWITHSUNDEEP
linuxbashshellunixawk
logan
logan

Reputation: 8346

Storing awk command matched pattern output lines into array indexes

I have a file where i am searching for 2 patterns if matched then storing each matched line into array value. But It is not storing values

My Shell command 

#Removing new line chars
a=`sed ':a;N;$!ba;s/\n/ /g' sample.txt`
#storing each matched pattern row by row 
while read v1; do
y[i]="$v1"
(( i++ ))
done < <(awk -F '<abc>|</abc>' '{for (i=2; i<=NF; i+=2) print $i}' <<< "$a")

Outputs empty values:

echo ${y[0]} is empty
echo ${y[1]} is empty
echo ${y[2]} is empty

it should be 

echo ${y[0]} = 1. I am here to show
echo ${y[1]} = 2. I am here to show
echo ${y[2]} = 3. I am here to show

My file is : sample.txt

<abc>
1. I am here to show
</abc>
<no>
</no>
<abc>
2. I am here to show
</abc>
<abc>
3. I am here to show
</abc>
<no>
</no>

Upvotes: 1

Views: 593

Answers (2)

Mark Setchell
Mark Setchell

Reputation: 207445

I think this is a bit easier on the eye maybe:

#!/bin/bash
declare -a y
while read x; do
   y[i]=$x
   ((i++))
done <  <(awk '/^<abc>/   {p=1;next}
               /^<\/abc>/ {p=0;next} p' sample.txt)
echo ${y[*]}

In the awk, the variable p determines if we are printing the current line. It is set when we find <abc> and cleared when we find </abc>.

Upvotes: 1

logan
logan

Reputation: 8346

I found the problem myself...

variable $i was already set to some value so when i do i++ it was storing the value in to some array index.

i did i=0 then re executed, it works

Upvotes: 0

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