I have a file of this type:
16:00 [61]Al-Najma - Al-Rifaa [62]5.06 [63]3.55 [64]1.57 4
and i want remove all the strings inside square parentheses in order to obtain
16:00 Al-Najma - Al-Rifaa 5.06 3.55 1.57 4
I am trying with sed in this manner:
sed 's/\[.*]//g' file1 > file2
but i obtain
16:00 1.57 4
and with
sed 's/\[.[1234567890]]//g' file1 > file2
does not work if the string contains more than 2 digit.
how can i do this?
your pattern allows only one character, adding a star behind the pattern widens it to all matching characters.
sed 's/\[.[1234567890]]*//g' file1 > file2
alternative:
sed 's/\[^\]*//g' file1 > file2
that means: after the starting "[" everything but the "]" is OK, and that for as many characters as there come (the "*")
for further reading on sed: http://www.grymoire.com/Unix/Sed.html
Your first regex does not work because the quantifier * is greedy, meaning it matches as many characters as possible. Since . also matches brackets, it continues to match until the last closing bracket ] it can find.
So you basically have two options: Use a non-greedy quantifier or restrict the types of characters you can match. You have tried the second solution. I would go with using a negated character class instead:
sed 's/\[[^]]*\]//g'
I'm not sure if sed has non-greedy quantifiers, but perl does:
perl -lpwe 's/\[.*?\]//g'
You already got the sed answer, so I will add other one using awk:
awk '
BEGIN {
FS = "\\[[^]]*\\]";
OFS = " "
}
{
for (i=1; i<=NF; i++)
printf "%s", $i
}
END {
printf "\n"
}
' <<<"16:00 [61]Al-Najma - Al-Rifaa [62]5.06 [63]3.55 [64]1.57 4"
Output:
16:00 Al-Najma - Al-Rifaa 5.06 3.55 1.57 4
awk:$ echo '16:00 [61]Al-Najma - Al-Rifaa [62]5.06 [63]3.55 [64]1.57 4' | awk -F '\[[0-9]*\]' '$1=$1'
16:00 Al-Najma - Al-Rifaa 5.06 3.55 1.57 4
This might work for you:
echo "16:00 [61]Al-Najma - Al-Rifaa [62]5.06 [63]3.55 [64]1.57 4" |
sed 's/\[[^]]*\]//g'
16:00 Al-Najma - Al-Rifaa 5.06 3.55 1.57 4
