Perl misses my DWIM (ternary ?:) and forces regex.
perl -e "print $bool ?'T' :'F'"
Use of ?PATTERN? without explicit operator is deprecated at -e line 1.
Search pattern not terminated or ternary operator parsed as search pattern at -e
line 1.
Why is that so?
Following two examples have correctly parsed ternary operator,
perl -e "print $bool ? 'T' :'F'"
perl -e "print [] ?'T' :'F'"
This is on windows, but almost same thing on *nix
perl -e 'print $bool ?"T" :"F"'
so it doesn't look like shell related.
The ?...? is an operator, as seen here, and the problem here is one of ambiguity. The compiler does not know if it is seeing a pattern or a ternary operator (as the subsequent warning says), so it guesses, and guesses wrong.
I think it has something to do with print() setting it up for a fall, in that print() has the syntax:
print FILEHANDLE LIST
So, it pre-determines $bool as a file handle, and ?'.... as a statement. Then it notices that $bool is not a file handle, but the ternary is already cast in its faulty role. You will notice that the following works as intended:
perl -e"print STDOUT $bool ?'t' :'f'"
Perl syntax has ambiguities.
print $bool ?...
can be parsed as
print FILEHANDLE LIST
+----------------- print
| +----------- FILEHANDLE
| | +------ LIST (A regex match ?...?)
| | |
_____ _____ ____
print $bool ?...
or as
print LIST
+----------------- print
| +--------- LIST (A conditional operator ...?...:...)
| |
_____ __________
print $bool ?...
Without looking too far ahead, perl incorrectly guesses that you meant the first one, but lets you know that it guessed.
Search pattern not terminated or ternary operator parsed as search pattern
Workarounds:
Adding a space after the ? (print $bool ? 'T' :'F';) makes it guess correctly (since it no longer looks as much like ?...?).
Adding a + before the file handle (print +$bool ?'T' :'F';) makes it guess correctly (since it's no longer a valid file handle expression).
Remove the space before the ? (print $bool?'T' :'F';) makes it guess correctly (since it's no longer a valid file handle expression).
Adding a file handle (print STDOUT $bool ?'T' :'F';) makes it guess correctly (since $bool must now be the start of the argument list).
Are you on Windows or on *nix?
On linux I get no warnings when I switch " and ' ( perl -e 'print $bool ? "T" :"F"'). The way you wrote it the $bool is interpolated by the shell, to an empty string, and your code becomes perl -e "print ?'T' :'F'" which correctly triggers the warning.
Related
It works as intended:
perl -ne "print uc" /etc/passwd
But following isn't (it just prints in original case":
perl -pe uc /etc/passwd
I don't understand what's wrong with it.
thanks.
You're doing different things. So it's not surprising that you get different results.
In the first example, you take the value of $_, pass it to uc and print the results (which is an upper case version of the original text).
In the second example, you take the value of $_, pass it to uc and print the value in $_. But you've done nothing to update $_ so you get the unaltered value. The fix (as you've already noted in a comment) is to update $_ with the value that is returned by uc.
perl -pe '$_ = uc' /etc/passwd
I need to pass system variable in upper case to perl subroutine.
For example, if the variable with name VARNAME (value 'super'), i need to pass "SUPER_MAN".
In general, if we use 'uc' option like in the example below, we can convert to upper case
perl -e 'print uc"$ENV{VARNAME}\n"'
But when we try to pass it in subroutine, we need to include uc function in the syntax and evaluate during runtime. To emulate that I was trying the below but not working, Where am I going wrong?
perl -e 'print ".uc($ENV{VARNAME})_MAN\n"'
.uc(super)_MAN
Alternate methods/approach is also welcome.
Take the uc out of the quotes "", since perl thinks you want the literal letters uc:
FOO=abc perl -e 'print "." . uc($ENV{FOO}) . "_MAN\n"'
.ABC_MAN
perldoc perlop - Quote and Quote like Operators
What's the use of <> in Perl. How to use it ?
If we simply write
<>;
and
while(<>)
what is that the program doing in both cases?
The answers above are all correct, but it might come across more plainly if you understand general UNIX command line usage. It is very common to want a command to work on multiple files. E.g.
ls -l *.c
The command line shell (bash et al) turns this into:
ls -l a.c b.c c.c ...
in other words, ls never see '*.c' unless the pattern doesn't match. Try this at a command prompt (not perl):
echo *
you'll notice that you do not get an *.
So, if the shell is handing you a bunch of file names, and you'd like to go through each one's data in turn, perl's <> operator gives you a nice way of doing that...it puts the next line of the next file (or stdin if no files are named) into $_ (the default scalar).
Here is a poor man's grep:
while(<>) {
print if m/pattern/;
}
Running this script:
./t.pl *
would print out all of the lines of all of the files that match the given pattern.
cat /etc/passwd | ./t.pl
would use cat to generate some lines of text that would then be checked for the pattern by the loop in perl.
So you see, while(<>) gets you a very standard UNIX command line behavior...process all of the files I give you, or process the thing I piped to you.
<>;
is a short way of writing
readline();
or if you add in the default argument,
readline(*ARGV);
readline is an operator that reads a line from the specified file handle. Reading from the special file handle ARGV will read from STDIN if #ARGV is empty or from the concatenation of the files named by #ARGV if it's not.
As for
while (<>)
It's a syntax error. If you had
while (<>) { ... }
it get rewritten to
while (defined($_ = <>)) { ... }
And as previously explained, that means the same as
while (defined($_ = readline(*ARGV))) { ... }
That means it will read lines from (previously explained) ARGV until there are no more lines to read.
It is called the diamond operator and feeds data from either stdin if ARGV is empty or each line from the files named in ARGV. This webpage http://docstore.mik.ua/orelly/perl/learn/ch06_02.htm explains it very well.
In many cases of programming with syntactical sugar like this, Deparse of O is helpful to find out what's happening:
$ perl -MO=Deparse -e 'while(<>){print 42}'
while (defined($_ = <ARGV>)) {
print 42;
}
-e syntax OK
Quoting perldoc perlop:
The null filehandle <> is special: it can be used to emulate the
behavior of sed and awk, and any other Unix filter program that takes
a list of filenames, doing the same to each line of input from all of
them. Input from <> comes either from standard input, or from each
file listed on the command line.
it takes the STDIN standard input:
> cat temp.pl
#!/usr/bin/perl
use strict;
use warnings;
my $count=<>;
print "$count"."\n";
>
below is the execution:
> temp.pl
3
3
>
so as soon as you execute the script it will wait till the user gives some input.
after 3 is given as input,it stores that value in $count and it prints the value in the next statement.
I've got a file with various wildcards in it that I want to be able to substitute from a (Bash) shell script. I've got the following which works great until one of the variables contains characters that are special to regexes:
VERSION="1.0"
perl -i -pe "s/VERSION/${VERSION}/g" txtfile.txt # No problems here
APP_NAME="../../path/to/myapp"
perl -i -pe "s/APP_NAME/${APP_NAME}/g" txtfile.txt # Error!
So instead I want something that just performs a literal text replacement rather than a regex. Are there any simple one-line invocations with Perl or another tool that will do this?
The 'proper' way to do this is to escape the contents of the shell variables so that they aren't seen as special regex characters. You can do this in Perl with \Q, as in
s/APP_NAME/\Q${APP_NAME}/g
but when called from a shell script the backslash must be doubled to avoid it being lost, like so
perl -i -pe "s/APP_NAME/\\Q${APP_NAME}/g" txtfile.txt
But I suggest that it would be far easier to write the entire script in Perl
Use the following:
perl -i -pe "s|APP_NAME|\\Q${APP_NAME}|g" txtfile.txt
Since a vertical bar is not a legal character as part of a path, you are good to go.
I don't particularly like this answer because there should be a better way to do a literal replace in Perl. \Q is cryptic. Using quotemeta adds extra lines of code.
But... You can use substr to replace a portion of a string.
#!/usr/bin/perl
my $name = "Jess.*";
my $sentence = "Hi, my name is Jess.*, dude.\n";
my $new_name = "Prince//";
my $name_idx = index $sentence, $name;
if ($name_idx >= 0) {
substr($sentence, $name_idx, length($name), $new_name);
}
print $sentence;
Output:
Hi, my name is Prince//, dude.
You don't have to use a regular expression for this (using substr(), index(), and length()):
perl -pe '
foreach $var ("VERSION", "APP_NAME") {
while (($i = index($_, $var)) != -1) {
substr($_, $i, length($var)) = $ENV{$var};
}
}
'
Make sure you export your variables.
You can use a regex but escape any special characters.
Something like this may work.
APP_NAME="../../path/to/myapp"
APP_NAME=`echo "$APP_NAME" | sed -e '{s:/:\/:}'`
perl -i -pe "s/APP_NAME/${APP_NAME}/g" txtfile.txt
Use:
perl -i -pe "\$r = qq/\Q${APP_NAME}\E/; s/APP_NAME/\$r/go"
Rationale: Escape sequences
I managed to get a working solution, partly based on bits and pieces from other peoples' answers:
app_name='../../path/to/myapp'
perl -pe "\$r = q/${app_name//\//\\/}/; s/APP_NAME/\$r/g" <<<'APP_NAME'
This creates a Perl variable, $r, from the result of the shell parameter expansion:
${app_name//\//\\/}
${ # Open parameter expansion
app_name # Variable name
// # Start global substitution
\/ # Match / (backslash-escaped to avoid being interpreted as delimiter)
/ # Delimiter
\\/ # Replace with \/ (literal backslash needs to be escaped)
} # Close parameter expansion
All that work is needed to prevent forward slashes inside the variable from being treated as Perl syntax, which would otherwise close the q// quotes around the string.
In the replacement part, use the variable $r (the $ is escaped, to prevent it from being treated as a shell variable within double quotes).
Testing it out:
$ app_name='../../path/to/myapp'
$ perl -pe "\$r = q/${app_name//\//\\/}/; s/APP_NAME/\$r/g" <<<'APP_NAME'
../../path/to/myapp
I have a file, someFile, like this:
$cat someFile
hdisk1 active
hdisk2 active
I use this shell script to check:
$cat a.sh
#!/usr/bin/ksh
for d in 1 2
do
grep -q "hdisk$d" someFile && echo "$d : ok"
done
I am trying to convert it to Perl:
$cat b.sh
#!/usr/bin/ksh
export d
for d in 1 2
do
cat someFile | perl -lane 'BEGIN{$d=$ENV{'d'};} print "$d: OK" if /hdisk$d\s+/'
done
I export the variable d in the shell script and get the value using %ENV in Perl. Is there a better way of passing this value to the Perl one-liner?
You can enable rudimentary command line argument with the "s" switch. A variable gets defined for each argument starting with a dash. The -- tells where your command line arguments start.
for d in 1 2 ; do
cat someFile | perl -slane ' print "$someParameter: OK" if /hdisk$someParameter\s+/' -- -someParameter=$d;
done
See: perlrun
Sometimes breaking the Perl enclosure is a good trick for these one-liners:
for d in 1 2 ; do cat kk2 | perl -lne ' print "'"${d}"': OK" if /hdisk'"${d}"'\s+/';done
Pass it on the command line, and it will be available in #ARGV:
for d in 1 2
do
perl -lne 'BEGIN {$d=shift} print "$d: OK" if /hdisk$d\s+/' $d someFile
done
Note that the shift operator in this context removes the first element of #ARGV, which is $d in this case.
Combining some of the earlier suggestions and adding my own sugar to it, I'd do it this way:
perl -se '/hdisk([$d])/ && print "$1: ok\n" for <>' -- -d='[value]' [file]
[value] can be a number (i.e. 2), a range (i.e. 2-4), a list of different numbers (i.e. 2|3|4) (or almost anything else, that's a valid pattern) or even a bash variable containing one of those, example:
d='2-3'
perl -se '/hdisk([$d])/ && print "$1: ok\n" for <>' -- -d=$d someFile
and [file] is your filename (that is, someFile).
If you are having trouble writing a one-liner, maybe it is a bit hard for one line (just my opinion). I would agree with #FM's suggestion and do the whole thing in Perl. Read the whole file in and then test it:
use strict;
local $/ = '' ; # Read in the whole file
my $file = <> ;
for my $d ( 1 .. 2 )
{
print "$d: OK\n" if $file =~ /hdisk$d\s+/
}
You could do it looping, but that would be longer. Of course it somewhat depends on the size of the file.
Note that all the Perl examples so far will print a message for each match - can you be sure there are no duplicates?
My solution is a little different. I came to your question with a Google search the title of your question, but I'm trying to execute something different. Here it is in case it helps someone:
FYI, I was using tcsh on Solaris.
I had the following one-liner:
perl -e 'use POSIX qw(strftime); print strftime("%Y-%m-%d", localtime(time()-3600*24*2));'
which outputs the value:
2013-05-06
I was trying to place this into a shell script so I could create a file with a date in the filename, of X numbers of days in the past. I tried:
set dateVariable=`perl -e 'use POSIX qw(strftime); print strftime("%Y-%m-%d", localtime(time()-3600*24*$numberOfDaysPrior));'`
But this didn't work due to variable substitution. I had to mess around with the quoting, to get it to interpret it properly. I tried enclosing the whole lot in double quotes, but this made the Perl command not syntactically correct, as it messed with the double quotes around date format. I finished up with:
set dateVariable=`perl -e "use POSIX qw(strftime); print strftime('%Y-%m-%d', localtime(time()-3600*24*$numberOfDaysPrior));"`
Which worked great for me, without having to resort to any fancy variable exporting.
I realise this doesn't exactly answer your specific question, but it answered the title and might help someone else!
That looks good, but I'd use:
for d in $(seq 1 2); do perl -nle 'print "hdisk$ENV{d} OK" if $_ =~ /hdisk$ENV{d}/' someFile; done
It's already written on the top in one long paragraph but I am also writing for lazy developers who don't read those lines.
Double quotes and single quote has big different meaning for the bash.
So please take care
Doesn't WORK perl '$VAR' $FILEPATH
WORKS perl "$VAR" $FILEPATH