Searching two patterns in files that appear in the same file but may not appear on the same line

From a directory, I need to find all files that contain a decimal numbers say 42.7 AND the keyword “foo” or “goo”. How could I achieve that?

Suppose I have a directory with three files

file1.txt
=======
double x = 2.7
foo();

file2.txt
===========
double u = 5.7

file3.txt
===========
goo(42.0); 

The search command should give file1.txt and file3.txt. What is a search command to achieve this?

i searched for the solutions but all I could find deal with having the patterns simultaneously appearing in the same line. Also I got difficulty in dealing with the decimal dot.

Answer

Use 2 grep -l commands to list the file names (not the matched lines) that contain the regex. Connect them by xargs, for example like so:

grep -Pl '(d+[.]?d*|d*[.]?d+)' file?.txt | xargs grep -Pl '(foo|goo)'

Example:

Create the input files. I use a few more examples in addition to the ones listed in the question to illustrate the patterns/files that are found:

cat > file1.txt <<EOF
double x = 2.7
foo();
EOF

cat > file2.txt <<EOF
double u = 5.7
EOF

cat > file3.txt <<EOF
goo(42.0);
EOF

cat > file4.txt <<EOF
foo(4);
EOF

cat > file5.txt <<EOF
goo(.42);
EOF

cat > file6.txt <<EOF
goo(.);
EOF

Run grep -l ... | xargs grep -l ... to find the matching files:

grep -Pl '(d+[.]?d*|d*[.]?d+)' file?.txt | xargs grep -Pl '(foo|goo)'

Prints:

file1.txt
file3.txt
file4.txt
file5.txt

Here, grep uses the following options:
-P : Use Perl regexes.
-l : list the file names only, not the matching lines.

The regex has the following parts:
d* : any digit 0-9, repeated 0 or more times.
d+ : same, repeated 1 or more times.
[.] : literal dot (.). Otherwise, without escaping, . means any character.

SEE ALSO:

-l
--files-with-matches

Suppress normal output; instead print the name of each input file from which output would normally have been printed. The scanning of each file stops on the first match. (-l is specified by POSIX.)

grep manual