使用bash或sed或awk从文件的字符串（以特定格式）中提取子字符串

Question

I have an input file, an example of which is shown below :
(?U0 ?U2 ?U9 ?U11 ?U21)
I want to extract all the numbers after ?U to an output file as:
0 2 9 11 21
Please help me in this regard, I am new to it.

Thanks

Answer 1

You could use grep but it produces output in each per line.

grep -oP '\?U\K\d+' file

or

$ echo '(?U0 ?U2 ?U9 ?U11 ?U21)' | grep -oP '\?U\K\d+' | paste -s -d " " -
0 2 9 11 21

Answer 2

Using sed you can do:

s='(?U0 ?U2 ?U9 ?U11 ?U21)'
sed 's/?U\([0-9]\+\)/\1/g; s/[()]//g' <<< "$s"
0 2 9 11 21

Answer 3

simple sed

echo "(?U0 ?U2 ?U9 ?U11 ?U21)" | sed 's/[()?U]//g'  

output
0 2 9 11 21

deleting all unneeded characters, you can put in set [...] if needed another characters

or more universal

echo "(?U0 ?U2 ?U9 ?U11 ?U21)" | sed 's/[^0-9 ]*//g'   

deleting all non-digit characters (and not space)

Answer 4

With grep:

out="$(grep -oP '(?<=\?U)\d+' filepath |tr -s '\n' ' ')"
out="${out% }"
echo "$out" >outfilepath