I would like to make a bash script that uses 2 arguments, file1 file2 that copies all lines from the file1 that contains the letter b into file2 . I have found the solution to determine if a string is contains the letter
if [[ $string == *"b"* ]]; then
echo "It's there!"
fi
I just can figure how to apply this code to my problem, and run through each line of a random file.
In the course description i have found that this problem can be solved with the usage of head -n tail -n cat echo wc -c wc -l wc -w if case test , but we don't have to limit ourselves to the usage of just these commands.
This is the reason why grep has been invented:
grep "b" file1.txt >>file2.txt
(This copies all lines from file1.txt, containing the character b, to file2.txt)
Related
The command that I'm making wants the first input to be a file and search how many times a certain pattern occurs within the file, using grep and sed.
Ex:
$ cat file1
oneonetwotwotwothreefourfive
Intended output:
$ ./command file1 one two three
one 2
two 3
three 1
The problem is the file does not have any lines and is just a long list of letters. I'm trying to use sed to replace the pattern I'm looking for with "FIND" and move the list to the next line and this continues until the end of file. Then, use $grep FIND to get the line that contains FIND. Finally, use wc -l to find a number of lines. However, I cannot find the option to move the list to the next line
Ex:
$cat file1
oneonetwosixone
Intended output:
FIND
FIND
twosixFIND
Another problem that I've been having is how to use the rest of the input, not including the file.
Failed attempt:
file=$1
for PATTERN in 2 3 4 5 ... N
do
variable=$(sed 's/$PATTERN/find/g' $file | grep FIND $file | wc -l)
echo $PATTERN $variable
exit
Another failed attempt:
file=$1
PATTERN=$($2,$3 ... $N)
for PATTERN in $*
do variable=$(sed 's/$PATTERN/FIND/g' $file | grep FIND $file | wc-1)
echo $PATTERN $variable
exit
Any suggestions and help will be greatly appreciated. Thank you in advance.
Non-portable solution with GNU grep:
file=$1
shift
for pattern in "$#"; do
echo "$pattern" $(grep -o -e "$pattern" <"$file" | wc -l)
done
If you want to use sed and your "patterns" are actually fixed strings (which don't contain characters that have special meaning to sed), you could do something like:
file=$1
shift
for pattern in "$#"; do
echo "$pattern" $(
sed "s/$pattern/\n&\n/g" "$file" |\
grep -e "$pattern" | wc -l
)
done
Your code has several issues:
you should quote use of variables where word splitting may happen
don't use ALLCAPS variable names - they are reserved for use by the shell
if you put a string in single-quotes, variable expansion does not happen
if you give grep a file, it won't read standard input
your for loop has no terminating done
This might work for you (GNU bash,sed and uniq):
f(){ local file=$1;
shift;
local args="$#";
sed -E 's/'${args// /|}'/\n&\n/g
s/(\n\S+)\n\S+/\1/g
s/\n+/\n/g
s/.(.*)/echo "\1"|uniq -c/e
s/ *(\S+) (\S+)/\2 \1/mg' $file; }
Separate arguments into file and remaining arguments.
Apply arguments as alternation within a sed substitution command which splits words into lines separated by a newline either side.
Remove unwanted words and unwanted newlines.
Evaluate the manufactured file within a sed substitution using the uniq command with the -c option.
Rearrange the output and print the result.
The problem is the file does not have any lines
Great! So the problem reduces to putting newlines.
func() {
file=$1
shift
rgx=$(printf "%s\\|" "$#" | sed 's#\\|$##');
# put the newline between words
sed 's/\('"$rgx"'\)/&\n/g' "$file" |
# it's just standard here
sort | uniq -c |
# filter only input - i.e. exclude fourfive
grep -xf <(printf " *[0-9]\+ %s\n" "$#")
};
func <(echo oneonetwotwotwothreefourfive) one two three
outputs:
2 one
1 three
3 two
I can do the following using a for loop
for f in *.txt; do grep 'RINEX' $f |wc -l; done
Is there any possibility to get an individual file report by running one liner?
Meaning that I want to grep & wc one file at the time in a similar fashion like
grep 'RINEX' *.txt
UPDATE:
grep -c 'RINEX' *.txt
returns the name of each file and its corresponding number of occurrences. Thx #Evert
grep is not the right tool for this task.
grep does line based match, e.g. line grep 'o' <<< "fooo" will return 1 line. however we have 3 os.
This one-liner should do what you want:
awk -F'RINEX' 'FILENAME!=f{if(f)print f,s;f=FILENAME;s=0}
{s+=(NF-1)}
END{print f,s}' /path/*.txt
Let's say I have a program foo that finds files with a certain specification and that the output of running foo is:
file1.txt
file2.txt
file3.txt
I want to print the contents of each of those files (preferably with the file name prepended). How would I do this? I would've thought piping it to cat like so:
foo | cat
would work but it doesn't.
EDIT:
My solution to this problem prints out each file and prepends the filename to each line of output is:
foo | xargs grep .
This gets output similar to:
file1.txt: Hello world
file2.txt: My name is foobar.
<your command> | xargs cat
You need xargs here:
foo | xargs cat
In order to allow for file names that have spaces in them, you'll need something like this:
#/bin/bash
while read -r file
do
# Check for existence of the file before using cat on it.
if [[ -f $file ]]; then
cat "$file"
# Don't bother with empty lines
elif [[ -n $file ]]; then
echo "There is no file named '$file'"
fi
done
Put this a script. Let's call it myscript.sh. Then, execute:
foo | myscript.sh
foo | xargs grep '^' /dev/null
why grep on ^ ? to display also empty lines (replace with "." if you want only non-empty lines)
why is there a /dev/null ? so that, in addition to any filename provided in "foo" output, there is at least 1 additionnal file (and a file NOT maching anything, such as /dev/null). That way there is AT LEAST 2 filenames given to grep, and thus grep will always show the matching filename.
This question already has answers here:
Find lines from a file which are not present in another file [duplicate]
(4 answers)
Closed 8 years ago.
I will fetch the file names from the file say: FILE_A, and will search these file names in another file say: File_B Using the script say: script.sh
I want to print those file names which are not present in a file say: FILE_B.
I use the code but it didn't work.
Code in the script->script.sh is as follows:
#!/bin/bash
while read line
do
grep -v "$line" FILE_B
done<FILE_A
please help me. why it is not working and what is the solution of it?
grep can read its input from a file; no need for a loop.
grep -Fxvf FILE_A FILE_B
The -F option specifies that the input is literal strings, not regular expressions. Otherwise an input which contains regex metacharacters would not match itself; or not only itself. For example, the regular expression a.c matches "aac", "abc", etc.
The -x option requires a full-line match. Otherwise, the input "bc" would match on any line containing it as a substring, such as "abcd".
The -v option says to print non-matching lines instead of matching.
Finally, the lowercase -f option specifies a file name as its argument to use as input for the patterns to match.
comm is good for this, but it requires the input files to be sorted. If that's not a problem:
# lines in FILE_A that are not in FILE_B
comm -23 <(sort FILE_A) <(sort FILE_B)
No extra linefeed between while and do
grep -v expr file will
print all lines of those files, not containing expr. What you want, is just the result whether it's found or not. You need to test the
exit state.
Try:
#!/bin/bash
while read line
do
grep -q "$line" FILE_B || echo "$line"
done<FILE_A
grep returns exit 0 if a line was found. The || concatenation with echo means: execute echo when exit state != 0- i.e. when $line was not found.
This script works but does not print what you want. For each filename in FILE_A it prints all the OTHER filenames in FILE_B. Instead you should print the filename yourself if grep does not find it:
while read line
do
grep "$line" FILE_B >/dev/null || echo "$line"
done <FILE_A
Use this instead
#!/bin/bash
while read line
do
if grep -qw $line "file_B"
then
echo $line
fi
done < file_A
I got wrong result from the wc -l command. After a long :( checking a found the core of the problem, here is the simulation:
$ echo "line with end" > file
$ echo -n "line without end" >>file
$ wc -l file
1 file
here are two lines, but missing the last "\n". Any easy solution?
For the wc line is what ends with the "\n" char. One of solutions is grep-ing the lines. The grep not looking for the ending NL.
e.g.
$ grep -c . file #count the occurrence of any character
2
the above will not count empty lines. If you want them, use the
$ grep -c '^' file #count the beginnings of the lines
2
from man page of wc
-l, --lines
print the newline counts
form man page of echo
-n do not output the trailing newline
so you have 1 newline in your file and thus wc -l shows 1.
You can use the following awk command to count lines
awk 'END{print NR}' file