What is the best way to break long lines having system commands which are being executed by " ` "
example:
my #data = `cat data.txt | perl -ne '/CTX:|ed-as-of time:\\s+(\\w+)\\s+[Feb|Mar|April|May]/ && print' | sed '\$!N;/Sync/P;D'|sed 'N;s/\\n/ /'`;
I tried to break it using " `." like below, but am getting errors (sh: 1: Syntax error: end of file unexpected:).
my #data = `cat data.txt | `.
`perl -ne '/CTX:|ed-as-of time:\\s+(\\w+)\\s+[Feb|Mar|April|May]/ && print' | sed '\$!N;/Sync/P;D'|sed 'N;s/\\n/ /'`;
The best ways are to assemble the command in a scalar variable
$cmd = "command1 --foo --bar | command2 2>log.err | "
. " command3 --are-we-done-yet | ...";
#data = `$cmd`; # or #data = qx($cmd)
or to assemble it as an argument to readpipe
#data = readpipe("command1 --foo --bar | command2 >log.err | "
. "command3 --are-we-done-yet | ...");
When I am using below command directly its working fine but when I am trying to put this in perl script its giving lots of error.
my $calculate = `echo "$value" | awk -F "SP=" '{print $2}' | awk -F ";" '{print $1}' | awk -F ":" '{print $2}' | sed 's/\,/\//g'`;
Data is like:
HM=Happy:man,1;SP=SAD:person,3;AP=Arrogant:personality,5;MD=Mad:Dynamic,9;PR=Power:raper,10;
I want output as
person/3
You can do that using a Perl regex pattern
my $calculate;
($calculate = $1) =~ tr~,~/~ if $value =~ /SP=[^:]*:([^;]*)/;
Something like this, perhaps:
#!/usr/bin/env perl
use strict;
use warnings;
use v5.20;
my $data
= 'HM=Happy:man,1;SP=SAD:person,3;AP=Arrogant:personality,5;MD=Mad:Dynamic,9;PR=Power:raper,10;';
my %hash = map {m/(.*?)=(.*)$/} split( ';', $data );
if ( $hash{'SP'} =~ m/:(\w+),(\d+)$/ ) {
my $result = $1 . '/' . $2;
say $result;
}
person/3
I have the following line in a Perl script:
my $temp = `sed 's/ /\n/g' /sys/bus/w1/devices/w1_bus_master1/10-000802415bef/w1_slave | grep t= | sed 's/t=//'`;
Which throws up the error:
"sed: -e expression #1, char 2: unterminated `s' command"
If I run a shell script as below it works fine:
temp1=`sed 's/ /\n/g' /sys/bus/w1/devices/w1_bus_master1/10-000802415bef/w1_slave | grep t= | sed 's/t=//'`
echo $temp1
Anyone got any ideas?
Perl interpretes your \n as a literal newline character. Your command line will therefore look something like this from sed's perspective:
sed s/ /
/g ...
which sed doesn't like. The shell does not interpret it that way.
The proper solution is not to use sed/grep in such a situation at all. Perl is, after all, very, very good at handling text. For example (untested):
use File::Slurp;
my #lines = split m/\n/, map { s/ /\n/g; $_ } scalar(read_file("/sys/bus...));
#lines = map { s/t=//; $_ } grep { m/t=/ } #lines;
Alternatively escape the \n once, e.g. sed 's/ /\\n/g'....
You need to escape the \n in our first regular expression. The backtick-operator in perl thinks it is a control-character and inserts a newline instead of the string \n.
|
V
my $temp = `sed 's/ /\\n/g' /sys/bus/ # ...
I want to replace path in
(setq myFile "/some/path")
in a file. I tried to do it with sed:
find ./_build/html -type f -name '*.html' | while read myFile; do
MyFile=`readlink -f "$myFile"`
sed -i "s/setq myFile [)]*/setq myFile \"$MyFile\"/" sphinx_nowrap.el
# and then some actions on file
done
and with perl:
find ./_build/html -type f -name '*.html' | while read myFile; do
MyFile=`readlink -f "$myFile"`
perl -ne "s/setq myFile .+/setq myFile \"$MyFile\")/" sphinx_nowrap.el
# and then some actions on file
done
but both give errors.
I've read this and this and also this -- but can't make it work.
Edit:
Here's a perl error:
Having no space between pattern and following word is deprecated at -e line 1.
Bareword found where operator expected at -e line 1, near "s/setq myFile .+/setq myFile "/home"
String found where operator expected at -e line 1, at end of line
(Missing semicolon on previous line?)
syntax error at -e line 1, near "s/setq myFile .+/setq myFile "/home"
Can't find string terminator '"' anywhere before EOF at -e line 1.
and here's sed error:
sed: -e expression #1, char 34: unknown option to `s'
Edit 2:
So the solution is to change the delimeter char. And also sed expression should be changed:
sed -i "s!setq myFile .*!setq myFile \"$MyFile\")!" sphinx_nowrap.el
Looks like perl (and sed) recognizes the slash in the file path as the regex delimiter. You can use a different delimiter:
find ./_build/html -type f -name '*.html' | while read myFile; do
MyFile=`readlink -f "$myFile"`
perl -ne "s!setq myFile .+!setq myFile \"$MyFile\")!" sphinx_nowrap.el
# and then some actions on file
done
or for sed:
find ./_build/html -type f -name '*.html' | while read myFile; do
MyFile=`readlink -f "$myFile"`
sed -i "s!setq myFile [)]*!setq myFile \"$MyFile\"!" sphinx_nowrap.el
# and then some actions on file
done
Lets assume your $MyPath hold /foo/bar/baz. Then the Perl code reads as:
perl -ne "s/setq myFile .+/setq myFile \"/foo/bar/baz\")/" sphinx_nowrap.el
Your Regex is terminated with the third / character. To work around this, we can use another delimiter like s{}{}:
perl -ine "s{setq myFile .+}{setq myFile \"/foo/bar/baz\")}; print" sphinx_nowrap.el
I also added the -i Option (inplace editing) and a print statement so that something actually gets print out.
But probably it would be more elegant to pass the value aof $MyPath as a command line argument:
perl -ne 's{setq myFile .+}{setq myFile "$ARGV[0]")}; print' $MyPath <sphinx_nowrap.el >sphinx_nowrap.el
Let me try to explain this as clearly as I can...
I have a script that at some point does this:
grep -vf ignore.txt input.txt
This ignore.txt has a bunch of lines with things I want my grep to ignore, hence the -v (meaning I don't want to see them in the output of grep).
Now, what I want to do is I want to be able to know how many lines of input.txt have been ignored by each line of ignore.txt.
For example, if ignore.txt had these lines:
line1
line2
line3
I would like to know how many lines of input.txt were ignored by ignoring line1, how many by ignoring line2, and so on.
Any ideas on how can I do this?
I hope that made sense... Thanks!
Note that the sum of the ignored lines plus the shown lines may NOT add up to the total number of lines... "line1 and line2 are here" will be counted twice.
#!/usr/bin/perl
use warnings;
use strict;
local #ARGV = 'ignore.txt';
chomp(my #pats = <>);
foreach my $pat (#pats) {
print "$pat: ", qx/grep -c $pat input.txt/;
}
According to unix.stackexchange
grep -o pattern file | wc -l
counts the total number of a given pattern in the file. A solution, given this and the information, that you already use a script, is to use several grep instances to filter and count the patterns, which you want to ignore.
However, I'd try to build a more comfortable solution involving a scripting language like e.g. python.
This script will count the matched lines by hash lookup and save the lines to be printed in #result, where you may process them as you will. To emulate grep, just print them.
I made the script so it can print out an example. To use with the files, uncomment the code in the script, and comment the ones marked # example line.
Code:
use strict;
use warnings;
use v5.10;
use Data::Dumper; # example line
# Example data.
my #ignore = ('line1' .. 'line9'); # example line
my #input = ('line2' .. 'line9', 'fo' .. 'fx', 'line2', 'line3'); # example line
#my $ignore = shift; # first argument is ignore.txt
#open my $fh, '<', $ignore or die $!;
#chomp(my #ignore = <$fh>);
#close $fh;
my #result;
my %lookup = map { $_ => 0 } #ignore;
my $rx = join '|', map quotemeta, #ignore;
#while (<>) { # This processes the remaining arguments, input.txt etc
for (#input) { # example line
chomp; # Required to avoid bugs due to missing newline at eof
if (/($rx)/) {
$lookup{$1}++;
} else {
push #result, $_;
}
}
#say for #result; # This will emulate grep
print Dumper \%lookup; # example line
Output:
$VAR1 = {
'line6' => 1,
'line1' => 0,
'line5' => 1,
'line2' => 2,
'line9' => 1,
'line3' => 2,
'line8' => 1,
'line4' => 1,
'line7' => 1
};
while IFS= read -r pattern ; do
printf '%s:' "$pattern"
grep -c -v "$pattern" input.txt
done < ignore.txt
grep with -c counts matching lines, but with -v added it counts non-matching lines. So, simply loop over the patterns and count once for each pattern.
This will print the number of ignored matches along with the matching pattern:
grep -of ignore.txt input.txt | sort | uniq -c
For example:
$ perl -le 'print "Coroline" . ++$s for 1 .. 21' > input.txt
$ perl -le 'print "line2\nline14"' > ignore.txt
$ grep -of ignore.txt input.txt | sort | uniq -c
1 line14
3 line2
I.e., A line matching "line14" was ignored once. A line matching "line2" was ignored 3 times.
If you just wanted to count the total ignored lines this would work:
grep -cof ignore.txt input.txt
Update: modified the example above to use strings so that the output is a little clearer.
This might work for you:
# seq 1 15 | sed '/^1/!d' | sed -n '$='
7
Explanation:
Delete all lines except those that match. Pipe these matching (ignored) lines to another sed command. Delete all these lines but show the line number only of the last line. So in this example 1 thru 15, lines 1,10 thru 15 are ignored - a total of 7 lines.
EDIT:
Sorry misread the question (still a little confused!):
sed 's,.*,sed "/&/!d;s/.*/matched &/" input.txt| uniq -c,' ignore.txt | sh
This shows the number of matches for each pattern in the the ignore.txt
sed 's,.*,sed "/&/d;s/.*/non-matched &/" input.txt | uniq -c,' ignore.txt | sh
This shows the number of non-matches for each pattern in the the ignore.txt
If using GNU sed, these should work too:
sed 's,.*,sed "/&/!d;s/.*/matched &/" input.txt | uniq -c,;e' ignore.txt
or
sed 's,.*,sed "/&/d;s/.*/non-matched &/" input.txt | uniq -c,;e' ignore.txt
N.B. Your success with patterns may vary i.e. check for meta characters beforehand.
On reflection I thought this can be improved to:
sed 's,.*,/&/i\\matched &,;$a\\d' ignore.txt | sed -f - input.txt | sort -k2n | uniq -c
or
sed 's,.*,/&/!i\\non-matched &,;$a\\d' ignore.txt | sed -f - input.txt | sort -k2n | uniq -c
But NO, on large files this is actually slower.
Are both ignore.txt and input.txt sorted?
If so, you can use the comm command!
$ comm -12 ignore.txt input.txt
How many lines are ignored?
$ comm -12 ignore.txt input.txt | wc -l
Or, if you want to do more processing, combine comm with awk.:
$ comm ignore.txt input.txt | awk '
END {print "Ignored lines = " igtotal " Lines not ignored = "commtotal " Lines unique to Ignore file = " uniqtotal}
{
if ($0 !~ /^\t/) {uniqtotal+=1}
if ($0 ~ /^\t[^\t]/) {commtotal+=1}
if ($0 ~ /^\t\t/) {igtotal+=1}
}'
Here I'm taking advantage with the tabs that are placed in the output by the comm command:
* If there are no tabs, the line is in ignore.txt only.
* If there is a single tab, it is in input.txt only
* If there are two tabs, the line is in both files.
By the way, not all the lines in ignore.txt are ignored. If the line isn't also in input.txt, the line can't really be said to be ignored.
With Dennis Williamson's Suggestion
comm ignore.txt input.txt | awk '
!/^\t/ {uniqtotal++}
/^\t[^\t]/ {commtotal++}
/^\t\t/ {igtotal++}
END {print "Ignored lines = " igtotal " Lines not ignored = "commtotal " Lines unique to Ignore file = " uniqtotal}'