Find regular expression in a file matching a given value - regex

Question

I have some basic knowledge on using regular expressions with grep (bash).
But I want to use regular expressions the other way around.
For example I have a file containing the following entries:
line_one=[0-3]
line_two=[4-6]
line_three=[7-9]
Now I want to use bash to figure out to which line a particular number matches.
For example:
grep 8 file
should return:
line_three=[7-9]
Note: I am aware that the example of "grep 8 file" doesn't make sense, but I hope it helps to understand what I am trying to achieve.
Thanks for you help,
Marcel

Answer 1

As others haven pointed out, awk is the right tool for this:
awk -F'=' '8~$2{print $0;}' file
... and if you want this tool to feel more like grep, a quick bash wrapper:
#!/bin/bash
awk -F'=' -v seek_value="$1" 'seek_value~$2{print $0;}' "$2"
Which would run like:
./not_exactly_grep.sh 8 file
line_three=[7-9]

Answer 2

My first impression is that this is not a task for grep, maybe for awk.
Trying to do things with grep I only see this:
for line in $(cat file); do echo 8 | grep "${line#*=}" && echo "${line%=*}" ; done
Using while for file reading (following comments):
while IFS= read -r line; do echo 8 | grep "${line#*=}" && echo "${line%=*}" ; done < file

Answer 3

This can be done in native bash using the syntax [[ $value =~ $regex ]] to test:
find_regex_matching() {
local value=$1
while IFS= read -r line; do # read from input line-by-line
[[ $line = *=* ]] || continue # skip lines not containing an =
regex=${line#*=} # prune everything before the = for the regex
if [[ $value =~ $regex ]]; then # test whether we match...
printf '%s\n' "$line" # ...and print if we do.
fi
done
}
...used as:
find_regex_matching 8 <file
...or, to test it with your sample input inline:
find_regex_matching 8 <<'EOF'
line_one=[0-3]
line_two=[4-6]
line_three=[7-9]
EOF
...which properly emits:
line_three=[7-9]
You could replace printf '%s\n' "$line" with printf '%s\n' "${line%%=*}" to print only the key (contents before the =), if so inclined. See the bash-hackers page on parameter expansion for a rundown on the syntax involved.

Answer 4

This is not built-in functionality of grep, but it's easy to do with awk, with a change in syntax:
/[0-3]/ { print "line one" }
/[4-6]/ { print "line two" }
/[7-9]/ { print "line three" }
If you really need to, you could programmatically change your input file to this syntax, if it doesn't contain any characters that need escaping (mainly / in the regex or " in the string):
sed -e 's#\(.*\)=\(.*\)#/\2/ { print "\1" }#'

Answer 5

As I understand it, you are looking for a range that includes some value.
You can do this in gawk:
$ cat /tmp/file
line_one=[0-3]
line_two=[4-6]
line_three=[7-9]
$ awk -v n=8 'match($0, /([0-9]+)-([0-9]+)/, a){ if (a[1]<n && a[2]>n) print $0 }' /tmp/file
line_three=[7-9]
Since the digits are being treated as numbers (vs a regex) it supports larger ranges:
$ cat /tmp/file
line_one=[0-3]
line_two=[4-6]
line_three=[75-95]
line_four=[55-105]
$ awk -v n=92 'match($0, /([0-9]+)-([0-9]+)/, a){ if (a[1]<n && a[2]>n) print $0 }' /tmp/file
line_three=[75-95]
line_four=[55-105]
If you are just looking to interpret the right hand side of the = as a regex, you can do:
$ awk -F= -v tgt=8 'tgt~$2' /tmp/file

Answer 6

You would like to do something like
grep -Ef <(cut -d= -f2 file) <(echo 8)
This wil grep what you want but will not display where.
With grep you can show some message:
echo "8" | sed -n '/[7-9]/ s/.*/Found it in line_three/p'
Now you would like to transfer your regexp file into such commands:
sed 's#\(.*\)=\(.*\)#/\2/ s/.*/Found at \1/p#' file
Store these commands in a virtual command file and you will have
echo "8" | sed -nf <(sed 's#\(.*\)=\(.*\)#/\2/ s/.*/Found at \1/p#' file)