how can I use sed to delete second line? - arrays

Question

I have a text file 3.txt, content like this
123.ooo
aaaa - bbbb ccccc, dddd dddd, eeee fff (hhhhh ii jjj) 890a fff rrr 98.jjjj.1234
444.kkk
read 3.txt into array VAR
#!/bin/bash
VAR=$(cat /tmp/3.txt)
LEN=${#VAR[#]}
for (( i=0; i<${LEN}; i++ ));
do
echo ${VAR[$i]}
sed -i -e "#"${VAR[$i]}"#d" /tmp/3.txt
cat /tmp/3.txt
done
I want to echo 123.ooo then delete the 123.ooo line in /tmp/3.txt
then echo aaaa - bbbb ccccc, dddd dddd, eeee fff (hhhhh ii jjj) 890a fff rrr 98.jjjj.1234 and delete this line in /tmp/3.txt
but second line have space, how can I use sed to delete second line?
Sorry about my poor English.
I want to echo first line, then delete first line in file.
echo second line, then delete second line in file.
echo third line, then delete third line in file.

Answer 1

Taking this as an exercise in using arrays and sed:
The issue with your code is that file is read as a single value into the variable VAR whereas you want to create an array with each line as its own array entry.
To assign to an array use
VAR=( ... )
To split the input on newlines, set the IFS variable to the value of a single newline.
You also have another issue with you sed editing script. You seem to want to use # as the delimiter for the regular expression (which is actually a "sed address"). If you want to do this, the first # has to be escaped (\#). It is only the arguments to the s and y sed editing commands that takes any unescaped character as delimiters.
#!/bin/bash
IFS=$'\n'
VAR=( $(cat /tmp/3.txt) )
LEN=${#VAR[#]}
for (( i=0; i<${LEN}; i++ ));
do
echo ${VAR[$i]}
sed -i -e "\#"${VAR[$i]}"#d" /tmp/3.txt
cat /tmp/3.txt
done
Or, with a bit of cleanup:
#!/bin/bash
IFS=$'\n'
VAR=( $(</tmp/3.txt) )
LEN=${#VAR[#]}
for (( i = 0; i < LEN; i++ )); do
echo ${VAR[$i]}
sed -i -e "\#${VAR[$i]}#d" /tmp/3.txt
cat /tmp/3.txt
done