how to grep a variable in the shell program? [duplicate] - bash

Question

This question already has answers here:
grep for expression containing variable
(2 answers)
Closed 9 years ago.
#!/bin/bash
for ((var=0; var<20; var++))
do
echo " Number is: $(grep 'Multiple_Frame = echo **$var**' 20mrf.txt | wc -l)" >>statisic.txt
done
This shell program cannot produce correct result which maybe the reason of wrong variable returning in the second grep command.
How can I grep a variable within the second echo sentence? to grep different things according to the var changing?
Many thanks!

Answer 1

As others have stated, the problem is that the single quotes prevent expansion of the variable. However, using $() allows you to use double quotes:
echo " Number is: $(grep "Multiple_Frame = echo **$var**" 20mrf.txt | wc -l)" >>statisic.txt
although I suspect something like this is what you meant:
echo " Number is: $(grep "Multiple_Frame = $var" 20mrf.txt | wc -l)" >>statisic.txt
You should also be aware that grep has an option to output the count so you can omit wc:
echo " Number is: $(grep -c "Multiple_Frame = $var" 20mrf.txt)" >>statisic.txt

Answer 2

#OP, doing what you do that way is rather inefficient. You are calling grep and wc 20 times on the same file. Open the file just once, and get all the things you want in 1 iteration of the file contents.
Example in bash 4.0
declare -A arr
while read -r line
do
case "$line" in
*"Multiple_Frame ="*)
line=${line#*Multiple_Frame = }
num=${line%% *}
if [ -z ${number_num[$num]} ] ;then
number_num[$num]=1
else
number_num[$num]=$(( number_num[$num]+1 ))
fi
;;
esac
done <"file"
for i in "${!number_num[#]}"
do
echo "Multiple_Frame = $i has ${number_num[$i]} counts"
done
similarly, you can use associative arrays in gawk to help you do this task.
gawk '/Multiple_Frame =/{
sub(/.*Multiple_Frame = /,"")
sub(/ .*/,"")
arr["Multiple_Frame = "$0]=arr["Multiple_Frame = "$0]+1
}END{
for(i in arr) print i,arr[i]
}' file

Answer 3

You have to store each substitution in a variable. Like this:
#!/bin/bash
for ((var=0; var < 20; var++))
do
count=`grep "Multiple_Frame = $var" 20mrf.txt | wc -l`
echo " Number is: $count" >> statisic.txt
done

Answer 4

Ok, the second [sic] problem is with your quoting on line 5. The reference to $var will never be expanded because it's contained within single quotes. You can fix that by replacing the single quotes (') with escaped double quotes (\").
The first [sic] problem is that you're trying to do too much in a single line, which causes your nesting problem with quotes. Break the line up into multiple commands, storing intermediary results as necessary. Yeah, it might run a tad slower, but you'll save a LOT of time debugging and maintaining it.
Trader's Second Law: If you have to choose between optimizing for performance, and optimizing for maintainability, ALWAYS choose to make your code more maintainable. Computers get faster all the time; Programmers don't.

Answer 5

The string inside $(...) is quoted with single quotes ('...') this quoting prevents the variable expansion. Use Double quotes instead. In your case:
#!/bin/bash
for ((var=0; var<20; var++))
do
echo " Number is: $(grep 'Multiple_Frame = echo **'"$var"'**' 20mrf.txt | wc -l)" >>statisic.txt
done
Note that in this example, it seems like the double quotes for the echo command are already opened, but you should note that the $(...) is evaluated first, and there is no double quotes inside it. So the change here is to close the single quote of the grep open double quote instead, and close the double quotes and reopen single quote later.
This lengthy explanation illustrates the benefit of breaking the expression apart, as suggested by other answers.