Suppose I have a string "hello12 54 world23 43"
What I want is the first standalone number (having space before and after) and not the one attached to a word. So, for the above string it would be 54, and not 12.
I tried
sed -r 's|^([^.]+).*$|\1|; s|^[^0-9]*([0-9]+).*$|\1|'
but that gives me 12 (the first number in the string). Can anyone help?
Note : Can only use sed
Having GNU awk you can use the following command:
awk 'NF {print $1}' FPAT='\\y[[:digit:]]+\\y' file
The key here is the use of the FPAT variable which is a GNU extension. FPAT stands for field pattern and describes what is a field. In our case we want a field to be a number "enclosed" within word boundaries (\y, needs to get doubly escaped because it appears in a shell string).
NF {print $1} checks if the first field (number) exists; in that case the number of fields (NF) is greater than zero. If that's the case the first field will get printed.
Btw, probably your sed is able to do this?
echo "hello12 54 world23 43 " \
| sed 's/\(\b\|^\)\([0-9]\{1,\}\)\(\b\|$\)/\n\2\n/' \
| sed '/^[0-9]\{1,\}$/!d' \
| sed '1!d'
Sorry I can only guess if you can't say the exact sed version.
The first sed command extracts numbers that stands alone on separate lines. The second sed command deletes all lines which do not consist of a number only and the last one deletes everything except of the first line - if it exists.
It is probably cleaner to use grep with \b:
$ grep -Eo '\b[0-9]+\b' <<< "hello12 54 world23 43"
54
43
Note this shows all the matches, so you may want to pipe to head -1 to get just the first one.
From GNU grep → 3.3 The Backslash Character and Special Expressions:
‘\b’
Match the empty string at the edge of a word.
If you really need sed:
$ sed -r 's/.*?\b([0-9]+)\b.*/\1/' <<< "hello12 54 world23 43"
54
$ sed -r 's/.*?\b([0-9]+)\b.*/\1/' <<< "54 world23 43"
54
This catches the first block of [0-9]+ that occurs in a given line that constitutes a word itself. Then, it prints it back.
Removed since sed does not recognize the .*? non greedy regex matching.
Your regex has redundant parts that you could remove them. E.g s|^([^.]+).*$|\1| that does replace a line with itself. If you are sure there is only one number as such in your string below regex is enough otherwise check the other solutions to capture the first one:
sed -r "s/^.* ([0-9]+) .*/\1/"
Simulating lazy version (preferred way):
-r option)This works like greedy version except it is a must if your string may have more than one occurrence of such numbers.
Regex:
([0-9]+) .*|.
Usage:
$ sed -r "s/ ([0-9]+) .*|./\1/g" <<< " 54 foo 43 "
54
If you want to go with the oldest regex flavor still in use (POSIX BRE) then this is your choice. This works the same as above regex but written in BRE.
Regex:
\(\( \([0-9]*\) .*\)*.\)*
Usage:
$ sed "s/\(\( \([0-9]*\) .*\)*.\)*/\3/g" <<< " 54 foo 43 "
54
In lazy versions, global g modifier should be set.
