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I have a file (file_name) which contains exactly one occurance of the string Result:, at the start of a line. I want to print all the characters after the string Result: in that line until I encounter a space. Which shell command should I use?
grep "Result: " file_name | tail -c +9
is not working.
The simplest way would be to use awk.
awk '/^Result: / {print $2}' file_name
That matches lines that begin with Result:, and prints the second field in the file, as defined by the default $IFS, which is whitespace.
: and afaik, the word before the next colon is searched.
Nov 7, 2011 at 4:39
Result: and the next space are what is desired. Awk will by default break up the line on white space. $1 = "Result:", $2 = the next word. Which is what the OP requested to extract.
Nov 7, 2011 at 5:10
blank'. Your command results for the line Result:two bla nk in bla.
Nov 7, 2011 at 5:29
Result: . This will now ignore lines like that in your example, and more closely match the initial request of the OP.
Nov 7, 2011 at 5:40
Sounds like what you want is:
sed -ne 's/^Result: \([^ ]*\).*/\1/p' file_name
Matches against the line with "Result: " and prints the first word after it. Other lines are not displayed.
^, but is unnecessary.
tillblank=$(egrep -o "^Result: [^ ]* " file)
echo ${tillblank/*:/}
egrep -o only outputs the pattern you searched for. The next line removes everything up to the colon.
Result: 2 apples, then your's would get 2 apples, not 2 as the OP suggested.
Result: 2 - the second line removes the Result:. Did you test it? You should!
Nov 8, 2011 at 1:45