my $line ="Corner:Default,Output:fall_delay_slew_1,Mean=34.97p,Std- dev=1.767p,Min=30.02p,Max=39.71p"; #added semicolon
my $my_value="COND = Mean > 3"; #this has come from the parsed file.
$my_value =~ m/(\w+)\s*(.)\s*(\d+)/;
my $cond=$1;
my $sign=$2;
my $value=$3;
print "DEBUG:cond is $cond and sign $sign and value $value \n";
if ( $line =~ m/$cond=(.*?),/) {
if ( "$value $sign $1" ) {
print "$value is $sign than $1\n";
} else {
print "actual value is less\n";
}
}
If you see in the above if statement always evaluates to true.
How can I solve this kind of problem i.e $sign = "<" (could be any operator)
but when I want to compare it with $value I want it to function as an
operator and not as a string.
What you're willing to do (executing a string as code) can be done with
eval. That doesn't mean it is the most appropriate way of doing
it though. Do it only if you guarantee your input safety and check for
it.
A better way would be checking the operator your self and determining
how to proceed. If you use a recent Perl version, the given-when
feature can be handy to do this:
use feature 'switch'; # not needed if you already use 5.010 or greater
given ($sign) {
when ('<') { say "$cond less than $value" }
when ('>') { say "$cond greater than $value" }
default {
warn "unrecognized operator `$sign'\n";
# decide what to do
}
}
Related
I'm trying to compare a string to another.
If it's a JSON structure which contains things, I want to print "contains things".
If it's a JSON structure which doesn't contain thing, I print "empty"
If it something which is not between curly brackets "{}", i print that there's an error.
Here's what I've done :
if($content =~ m/{.+}/){
print "Contains things \n";
} elsif($content eq "{}"){
$job_status{$url}="";
print "empty \n";
} else {
print "Error \n";
}
When I pass "{}" to the variable $content, he does not enter the "elsif", but go to the "else", and throw an error.
I've tried to put "==" instead the "eq" in the if, even though I know it's for numbers. When so, he enters the "elsif", and print "empty", like he should do with the "eq", and throws :
Argument "{}" isn't numeric in numeric eq (==)".
I could use the JSON library but I prefer not.
Thanks for your help !
Bidy
It works for me. Does $content have a newline character? Try chomp $content;.
use warnings;
use strict;
my $content = '{}';
if($content =~ m/{.+}/){
print "Contains things \n";
} elsif($content eq "{}"){
print "empty \n";
} else {
print "Error \n";
}
__END__
empty
I can replicate the behaviour if I add a newline after the {}:
#!/usr/bin/perl
use strict;
use warnings;
my $content = "{}\n";
if($content =~ m/{.+}/){
print "Contains things \n";
} elsif($content eq "{}"){
print "empty \n";
} else {
print "Error \n";
}
It returns "Error", if I replace eq with ==, it returns empty, because both "{}" and "{}\n" are numerically 0. A warning is thrown as you mentioned.
You might try to chomp the $content before processing it.
A top-level JSON thingy can be an object ({...}) or an array ([...]), but you're only checking for one of those. If you merely want to see if it's empty, I'd check the length of the string:
chomp $possible_json;
if( $length $possible_json >= 3 ) { ... }
You might also consider Randal Schwartz's regex for JSON parsing. It doesn't handle everything, but it's often enough for simple things.
I'd probably end up breaking it up:
unless ($content) {print "Error\n"};
$content =~ /{(.*)}/
my $resp = $1;
if ($resp) {
print "Contains Things ($resp)\n";
} else {
print "Empty\n";
}
I am writing a program that takes numbers from the command line until the user enters a blank line.
Should the user enter something that is neither newline nor numeric, it notifies the user, and continues.
While everything works, I have use warnings turned on, and it doesn't seem to like the second if conditional if the enters something invalid.
Argument "foo" isn't numeric in numeric eq (==) at adder.pl line 25, <STDIN> line 4.
I don't like running the program with this warning. How can I improve my code?
This is my program
#!/usr/bin/perl
use strict;
use warnings;
#declare variable
my $number = 0; #final answer
my $input;
#prompt the user
print "Input some integers, line by line. When you are done, press return to add them up." . "\n";
while (1) {
#get input from user
$input = <STDIN>;
#remove newlines
chomp($input);
#user pnches in newline
if ($input eq '') { #if the answer is new line
#quit the loop
last;
} #end of if statement
#user punches in bad input
elsif ($input == 0 && $input ne '0' && $input ne '') {
#tell the user what happened and how to rectify it
print "Input must be an integer." . "\n";
} # end of elsif statement
else {
chomp($input);
$number += $input;
} # end of else statement
} #end of while
print "Total is: $number\n";
Perl does DWIM very well. It is famous for it.
So, whatever language you have come from - it looks like C - forget about checking for both strings and numbers: a Perl scalar variable is whatever you ask it to be.
That means something like
elsif ($input == 0 && $input ne '0' && $input ne '') {
makes little sense. Anything read from the keyboard is initially a string, but it will be a number if you want. You are asking for $input to evaluate as zero but not to be the literal string 0. That applies to very few strings, for instance 00 or 0e0.
I think this is what you meant to write. Please take a look.
Isn't it clearer without comments?
use strict;
use warnings;
print "Input some integers line by line. When you are done, press return to add them up\n";
my $total = 0;
while (<>) {
chomp;
last unless /\S/;
if (/\D/) {
print "Input must be an integer\n";
next;
}
$total += $_;
}
print "Total is: $total\n";
Since Perl is untyped, and you are using $input as both a number and a string, you get that warning because "foo" isn't a number and "==" is used to compare equality of numbers.
You first need to check to see if $input is a number or not. One suggestion:
if ($input =~ /^\d+$/)
{
$number += $input;
}
else
{
print "Input must be an integer.\n";
}
I have an if clause in perl, where as condition I need to compare two variables if they match as strings. But my code doesnt work and the strings never match:
if(trim($file) eq trim($fields[0])) {
print "OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO";
}
For the definition of trim I have used:
sub trim($)
{
my $string = shift;
$string =~ s/^\s*(.*?)\s*$/$1/;
return $string;
}
Moreover I have used this before for the variables to compare.
my #fields= split(/\;/,$_);
Any help? Thanks!
Your code is correct, so your strings are different.
To find the differences, I recommend the following code since it will reveals differences that might not be noticeable by just printing the strings:
use Data::Dumper;
{
local $Data::Dumper::Useqq=1;
print Dumper($file, $fields[0]);
}
By the way, the following is more elegant and possibly faster:
sub trim {
my $string = shift;
$string =~ s/^\s+//;
$string =~ s/\s+\z//;
return $string;
}
And IIRC, the following is even faster (for a drop in readability):
sub trim {
my $string = shift;
$string =~ s/^\s+|\s++\z//g;
return $string;
}
It is my understanding that when writing a Unix shell program you can iterate through a string like a list with a for loop. Does this mean you can access elements of the string by their index as well?
For example:
foo="fruit vegetable bread"
How could I access the first word of this sentence? I've tried using brackets like the C-based languages to no avail, and solutions I've read online require regular expressions, which I would like to avoid for now.
Pass $foo as argument to a function. Than you can use $1, $2 and so on to access the corresponding word in the function.
function try {
echo $1
}
a="one two three"
try $a
EDIT: another better version is:
a="one two three"
b=( $a )
echo ${b[0]}
EDIT(2): have a look at this thread.
Using arrays is the best solution.
Here's a tricky way using indirect variables
get() { local idx=${!#}; echo "${!idx}"; }
foo="one two three"
get $foo 1 # one
get $foo 2 # two
get $foo 3 # three
Notes:
$# is the number of parameters given to the function (4 in all these cases)
${!#} is the value of the last parameter
${!idx} is the value of the idx'th parameter
You must not quote $foo so the shell can split the string into words.
With a bit of error checking:
get() {
local idx=${!#}
if (( $idx < 1 || $idx >= $# )); then
echo "index out of bounds" >&2
return 1
fi
echo "${!idx}"
}
Please don't actually use this function. Use an array.
I have written a code to get the url of a website and then search for a string and then compare that string(actually a number) with a hardcoded number
#!/usr/bin/perl
use LWP::Simple;
my $oldversion =36;
$pageURL="http://www.google.com/isos/preFCS5.3/LATESTGOODCVP/";
my $simplePage=get($pageURL);
my $newPage = "$simplePage";
my $str = (split("href=\"CVP-LATEST-5.3.0.",$newPage ))[1];
my $version = substr("$str",0,2);
print $version; // HERE IT PRINT 37 WHICH IS CORRECT
if($version =! $oldVersion )
{
print $version; // BUT HERE IT PRINTS 1 WHICH IS WRONG. HOW IS IT CHANGING ?
##-- fetch the zip and save it as perlhowto.zip
my $status = getstore("http://www.google.com/isos/preFCS5.3/LATESTGOODCVP/CVP-LATEST-5.3.0.$version.iso", "CVP-LATEST-5.3.0.$version.iso");
}
else
{
print("Currently new version\n");
}
Why is it changing the value ? its not able to download the file becuase of that.
You mean !=, not =!, which is an assignment of a negation.
Also, split always uses a regex (except for the very special case of a string that has a single space), so those .s in 5.3.0. will match any non-newline. You probably want to \-escape them.
You may be interested in the uscan script in the debian devtools package.
You have got your "not equals" operator backwards. It should be != rather than =!.
By using =! you are in effect saying "set $version to the negated value of $oldversion".
Here is the offending line
if($version =! $oldVersion ) # Should be if($version != $oldVersion )
Also notice that by using the != operator you are telling perl that $version and $oldversion contain numbers. For string comparisons you should use the ne operator, which assumes that these variables contain strings.
if($version ne $oldVersion ) # String inequality
Here is the documentation for equality operators -
http://perldoc.perl.org/perlop.html#Equality-Operators
It's because you are assigning to $version the value !$oldVersion in this "test":
if($version =! $oldVersion )
And $oldVersion is nothing--but $oldversion is 37. You are assigning $version the boolean negation of an undefined variable. Undefined is boolean false, and so the negation is boolean true or 1.
If you read very much on perl, you're bound to come across the advice to use strict and warnings. Had you done that, it would have told you, among other things:
Global symbol "$oldVersion" requires explicit package name at - line 21.
This means that you didn't declare $oldVersion as lexical (my) or package-level (our) in this package, so if you want to use it, please include the package where you're getting it. In a vast majority of cases, a seasoned Perl programmer will recognize this as "Ugh, I didn't declare $oldVersion!" and the reason is that you declared $oldversion.
Your use of split doesn't make a lot of sense here. What you really want are the two digits following the CVP-LATEST-5.3.0. string. You're also not really doing anything by assigning one variable to another with the addition of quotes ($newPage = "$simplePage").
And, of course, as others have pointed out, the comparison is != not =!.
I'd rewrite this as:
use strict;
use warnings;
use LWP::Simple;
my $oldVersion = 36;
my $url = 'http://www.google.com/isos/preFCS5.3/LATESTGOODCVP/';
my $newPage = get($url)
or die "Cannot retrieve contents from $url\n";
if ( $newPage =~ /href=\"CVP-LATEST-5\.3\.0\.(\d\d)/ ) {
my $version = $1;
if ( $version != $oldVersion ) {
my $status = getstore($url . "CVP-LATEST-5.3.0.$version.iso",
"CVP-LATEST-5.3.0.$version.iso");
} else {
print "Already at most recent version\n";
}
} else {
die "Cannot find version tag in contents from $url\n";
}