Perl and PostgreSQL escaping $ - perl

I have an SQL statement executed from perl.
$my_sql = qq (
SELECT 1 AS ....
);
Then I do
my $sm = $dbh->prepare($my_sql);
$sm->execute;
The issue is that my SQL has regex like this in many places.
value ~ '^[1-9][0-9]?/[1-9][0-9]?/[1-9][0-9]{3}$':
So I get errors when I execute the perl script because of these '$'.
Use of uninitialized value $' in concatenation (.) or string at
DBD::Pg::st execute failed: execute called with an unbound placeholder at
How can I avoid these errors and make the SQL statement work?

qq(something) is just another way to say "something" while q(something) means 'something'.
The difference between "this" and 'this' is interpolation, which is enabled only inside double quotes.
my $foo = 123;
print "<< $foo >>"; # prints << 123 >>
print '<< $foo >>'; # prints << $foo >>
So, since $ is not special symbol inside '', just change qq to q.

Related

why the syntax `&name arg1 arg2 ...` can't be used to call a Perl subroutine?

for a Perl subroutine, if passing 0 argument, I can use 4 forms to call it. But if passing 1 or more arguments, there is one form that I can't use, please see below:
sub name
{
print "hello\n";
}
# 4 forms to call
name;
&name;
name();
&name();
sub aname
{
print "#_\n";
}
aname "arg1", "arg2";
#&aname "arg1", "arg2"; # syntax error
aname("arg1", "arg2");
&aname("arg1", "arg2");
The error output is
String found where operator expected at tmp1.pl line 16, near "&aname "arg1""
(Missing operator before "arg1"?)
syntax error at tmp1.pl line 16, near "&aname "arg1""
Execution of tmp1.pl aborted due to compilation errors.
Can someone explain the error output from the compiler's point of view? I don't understand why it complains about missing operator.
Thanks
It's documented in perlsub:
To call subroutines:
NAME(LIST); # & is optional with parentheses.
NAME LIST; # Parentheses optional if predeclared/imported.
&NAME(LIST); # Circumvent prototypes.
&NAME; # Makes current #_ visible to called subroutine.
With &NAME "arg", perl sees &NAME() "ARG", so it thinks there's an operator missing between the sub call and "ARG".
In Perl 5, you don't need the & in most cases.

(4 + sub) not equals to (sub + 4)?

(edit) TL;DR: my problem was that I though the Win32 API defines were true integer constants (as in the platform SDK headers) while the Win32 Perl wrapper defines them as subs. Thus caused the one-liner parsing misunderstood.
While testing in a one-liner a call to Win32::MsgBox, I am puzzled by the following : giving that the possible arguments for MsgBox are the message, a sum of flags to chose the kind of buttons (value 0..5) and message box icon "constants" (MB_ICONSTOP, ...) and the title
calling perl -MWin32 -e"Win32::MsgBox world, 4+MB_ICONQUESTION, hello" gives the expected result
while the looking similar code perl -MWin32 -e"Win32::MsgBox world, MB_ICONQUESTION+4, hello" is wrong
I first though that it comes from my lack of parenthesis, but adding some perl -MWin32 -e"Win32::MsgBox (world, MB_ICONQUESTION+4, hello)" gives exactly the same wrong result.
I tried with a colleague to dig deeper and display the parameters that are passed to a function call (as the MB_xxx constants are actually subs) with the following code
>perl -Mstrict -w -e"sub T{print $/,'called T(#'.join(',',#_).'#)'; 42 }; print $/,'results:', join ' ,', T(1), T+1, 1+T"
that outputs
called T(#1#)
called T(##)
called T(#1,43#)
results:42 ,42
but I can't understand why in the list passed to join() the args T+1, 1+T are parsed as T(1, 43)...
B::Deparse to the rescue:
C:>perl -MO=Deparse -MWin32 -e"Win32::MsgBox world, MB_ICONQUETION+4, hello"
use Win32;
Win32::MsgBox('world', MB_ICONQUESTION(4, 'hello'));
-e syntax OK
C:>perl -MO=Deparse -MWin32 -e"Win32::MsgBox world, 4+MB_ICONQESTION, hello"
use Win32;
Win32::MsgBox('world', 4 + MB_ICONQUESTION(), 'hello');
-e syntax OK
The MB_ICONQUESTION call in the first case is considered a function call with the arguments +4, 'hello'. In the second case, it is considered as a function call with no arguments, and having 4 added to it. It is not a constant, it seems, but a function.
In the source code we get this verified:
sub MB_ICONQUESTION { 0x00000020 }
It is a function that returns 32 (00100000 in binary, indicating a bit being set). Also as Sobrique points out, this is a flag variable, so you should not use addition, but the bitwise logical and/or operators.
In your case, it just accepts any arguments and ignores them. This is a bit confusing if you are expecting a constant.
In your experiment case, the statement
print $/,'results:', join ' ,', T(1), T+1, 1+T
Is interpreted
print $/,'results:', join ' ,', T(1), T(+1, (1+T))
Because execution from right to left goes
1+T = 43
T +1, 43 = 42
T(1) = 42
Because plus + has higher precedence than comma ,, and unary + even higher.
To disambiguate, you need to do use parentheses to clarify precedence:
print $/,'results:', join ' ,', T(1), T()+1, 1+T
# ^^-- parentheses
As a general rule, one should always use parentheses with subroutine calls. In perldoc perlsub there are 4 calling notations:
NAME(LIST); # & is optional with parentheses.
NAME LIST; # Parentheses optional if predeclared/imported.
&NAME(LIST); # Circumvent prototypes.
&NAME; # Makes current #_ visible to called subroutine.
Of which in my opinion, only the first one is transparent, and the other ones a bit obscure.
This is all to do with how you're invoking T and how perl is interpreting the results.
If we deparse your example we get:
BEGIN { $^W = 1; }
sub T {
use strict;
print $/, 'called T(#' . join(',', #_) . '#)';
42;
}
use strict;
print $/, 'results:', join(' ,', T(1), T(1, 1 + T()));
This is clearly not what you've got in mind, but does explain why you get the result you do.
I would suggest in your original example - rather that + you may wish to consider using | as it looks very much like MB_ICONQUESTION is intended to be a flag.
So:
use strict;
use warnings;
use Win32 qw( MB_ICONQUESTION );
print MB_ICONQUESTION;
Win32::MsgBox( "world", 4 | MB_ICONQUESTION , "hello" );
Or
use strict;
use warnings;
use Win32 qw( MB_ICONQUESTION );
print MB_ICONQUESTION;
Win32::MsgBox( "world", MB_ICONQUESTION | 4 , "hello" );
Produce the same result.
This is because of precence when invoking subroutines without brackets - you can do:
print "one", "two";
And both are treated as arguments to print. Perl assumes that arguments after a sub are to be passed to it.
+4 is enumerated as an argument, and passed to T.
sub test { print #_,"\n";};
test 1;
test +1;
If we deparse this, we see perl treats it as:
test 1;
test 1;
So ultimately - there is a bug in Win32 that you have found, that would be fixable by:
sub MB_ICONQUESTION() {0x00000020}
Win32::MsgBox "world", 4 + MB_ICONQUESTION, "hello";
Win32::MsgBox "world", MB_ICONQUESTION + 4, "hello";
Or perhaps:
use constant MB_ICONQUESTION => 0x00000020;
Or as noted - the workaround in your code - don't use + and instead use | which is going to have the same result for bit flag operations, but because of operator precedence is never going to be passed into the subroutine. (Or of course, always specify the parenthesis for your constants)

2 Sub references as arguments in perl

I have perl function I dont what does it do?
my what does min in perl?
#ARVG what does mean?
sub getArgs
{
my $argCnt=0;
my %argH;
for my $arg (#ARGV)
{
if ($arg =~ /^-/) # insert this entry and the next in the hash table
{
$argH{$ARGV[$argCnt]} = $ARGV[$argCnt+1];
}
$argCnt++;
}
return %argH;}
Code like that makes David sad...
Here's a reformatted version of the code doing the indentations correctly. That makes it so much easier to read. I can easily tell where my if and loops start and end:
sub getArgs {
my $argCnt = 0;
my %argH;
for my $arg ( #ARGV ) {
if ( $arg =~ /^-/ ) { # insert this entry and the next in the hash table
$argH{ $ARGV[$argCnt] } = $ARGV[$argCnt+1];
}
$argCnt++;
}
return %argH;
}
The #ARGV is what is passed to the program. It is an array of all the arguments passed. For example, I have a program foo.pl, and I call it like this:
foo.pl one two three four five
In this case, $ARGV is set to the list of values ("one", "two", "three", "four", "five"). The name comes from a similar variable found in the C programming language.
The author is attempting to parse these arguments. For example:
foo.pl -this that -the other
would result in:
$arg{"-this"} = "that";
$arg{"-the"} = "other";
I don't see min. Do you mean my?
This is a wee bit of a complex discussion which would normally involve package variables vs. lexically scoped variables, and how Perl stores variables. To make things easier, I'm going to give you a sort-of incorrect, but technically wrong answer: If you use the (strict) pragma, and you should, you have to declare your variables with my before they can be used. For example, here's a simple two line program that's wrong. Can you see the error?
$name = "Bob";
print "Hello $Name, how are you?\n";
Note that when I set $name to "Bob", $name is with a lowercase n. But, I used $Name (upper case N) in my print statement. As it stands, now. Perl will print out "Hello, how are you?" without a care that I've used the wrong variable name. If it's hard to spot an error like this in a two line program, imagine what it would be like in a 1000 line program.
By using strict and forcing me to declare variables with my, Perl can catch that error:
use strict;
use warnings; # Another Pragma that should always be used
my $name = "Bob";
print "Hello $Name, how are you doing\n";
Now, when I run the program, I get the following error:
Global symbol "$Name" requires explicit package name at (line # of print statement)
This means that $Name isn't defined, and Perl points to where that error is.
When you define variables like this, they are in scope with in the block where it's defined. A block could be the code contained in a set of curly braces or a while, if, or for statement. If you define a variable with my outside of these, it's defined to the end of the file.
Thus, by using my, the variables are only defined inside this subroutine. And, the $arg variable is only defined in the for loop.
One more thing:
The person who wrote this should have used the Getopt::Long module. There's a major bug in their code:
For example:
foo.pl -this that -one -two
In this case, my hash looks like this:
$args{'-this'} = "that";
$args{'-one'} = "-two";
$args{'-two'} = undef;
If I did this:
if ( defined $args{'-two'} ) {
...
}
I would not execute the if statement.
Also:
foo.pl -this=that -one -two
would also fail.
#ARGV is a special variable (refer to perldoc perlvar):
#ARGV
The array #ARGV contains the command-line arguments intended for the
script. $#ARGV is generally the number of arguments minus one, because
$ARGV[0] is the first argument, not the program's command name itself.
See $0 for the command name.
Perl documentation is also available from your command line:
perldoc -v #ARGV

How to convert single quoted string to double quoted string so it gets interpreted?

For example I have,
my $str = '\t';
print "My String is ".$str;
I want the output to interpret the tab character and return something like:
"My String is \t"
I am actually getting the value of the string from the database, and it returns it as a single quoted string.
String::Interpolate does exactly that
$ perl -MString::Interpolate=interpolate -E 'say "My String is [". interpolate(shift) . "]"' '\t'
My String is [ ]
'\t' and "\t" are string literals, pieces of Perl code that produces strings ("\","t" and the tab character respectively). The database doesn't return Perl code, so describing the problem in terms of single-quoted literals and double-quoted literals makes no sense. You have a string, period.
The string is formed of the characters "\" and "t". You want to convert that sequence of characters into the tab character. That's a simple substitution.
s/\\t/\t/g
I presume you don't want to deal with just \t. You can create a table of the sequences.
my %escapes = (
"t" => "\t",
"n" => "\n",
"\" => "\\",
);
my $escapes_pat = join('', map quotemeta, keys(%escapes));
$escapes_pat = qr/[$escapes_pat]/;
s/\\($escapes_pat)/$escapes{$1}/g;
You can follow the technique in perlfaq4's answer to How can I expand variables in text strings?:
If you can avoid it, don't, or if you can use a templating system, such as Text::Template or Template Toolkit, do that instead. You might even be able to get the job done with sprintf or printf:
my $string = sprintf 'Say hello to %s and %s', $foo, $bar;
However, for the one-off simple case where I don't want to pull out a full templating system, I'll use a string that has two Perl scalar variables in it. In this example, I want to expand $foo and $bar to their variable's values:
my $foo = 'Fred';
my $bar = 'Barney';
$string = 'Say hello to $foo and $bar';
One way I can do this involves the substitution operator and a double /e flag. The first /e evaluates $1 on the replacement side and turns it into $foo. The second /e starts with $foo and replaces it with its value. $foo, then, turns into 'Fred', and that's finally what's left in the string:
$string =~ s/(\$\w+)/$1/eeg; # 'Say hello to Fred and Barney'
The /e will also silently ignore violations of strict, replacing undefined variable names with the empty string. Since I'm using the /e flag (twice even!), I have all of the same security problems I have with eval in its string form. If there's something odd in $foo, perhaps something like #{[ system "rm -rf /" ]}, then I could get myself in trouble.
To get around the security problem, I could also pull the values from a hash instead of evaluating variable names. Using a single /e, I can check the hash to ensure the value exists, and if it doesn't, I can replace the missing value with a marker, in this case ??? to signal that I missed something:
my $string = 'This has $foo and $bar';
my %Replacements = (
foo => 'Fred',
);
# $string =~ s/\$(\w+)/$Replacements{$1}/g;
$string =~ s/\$(\w+)/
exists $Replacements{$1} ? $Replacements{$1} : '???'
/eg;
print $string;
Well, I just tried below workaround it worked. Please have a look
my $str1 = "1234\n\t5678";
print $str1;
#it prints
#1234
# 5678
$str1 =~ s/\t/\\t/g;
$str1 =~ s/\n/\\n/g;
print $str1;
#it prints exactly the same
#1234\n\t5678

Why does defined sdf return true in this Perl example?

I tried this example in Perl. Can someone explain why is it true?
if (defined sdf) { print "true"; }
It prints true.
sdf could be any name.
In addition, if there is sdf function defined and it returns 0, then it does not print anything.
print (sdf); does not print sdf string but
if (sdf eq "sdf")
{
print "true";
}
prints true.
The related question remains if sdf is a string. What is it not printed by print?
sdf is a bareword.
perl -Mstrict -e "print qq{defined\n} if defined sdf"
Bareword "sdf" not allowed while "strict subs" in use at -e line 1.
Execution of -e aborted due to compilation errors.
For more fun, try
perl -Mstrict -e "print sdf => qq{\n}"
See Strictly speaking about use strict:
The subs aspect of use strict disables the interpretation of ``bare words'' as text strings. By default, a Perl identifier (a sequence of letters, digits, and underscores, not starting with a digit unless it is completely numeric) that is not otherwise a built-in keyword or previously seen subroutine definition is treated as a quoted text string:
#daynames = (sun, mon, tue, wed, thu, fri, sat);
However, this is considered to be a dangerous practice, because obscure bugs may result:
#monthnames = (jan, feb, mar, apr, may, jun,
jul, aug, sep, oct, nov, dec);
Can you spot the bug? Yes, the 10th entry is not the string 'oct', but rather an invocation of the built-in oct() function, returning the numeric equivalent of the default $_ treated as an octal number.
Corrected: (thanks #ysth)
E:\Home> perl -we "print sdf"
Unquoted string "sdf" may clash with future reserved word at -e line 1.
Name "main::sdf" used only once: possible typo at -e line 1.
print() on unopened filehandle sdf at -e line 1.
If a bareword is supplied to print in the indirect object slot, it is taken as a filehandle to print to. Since no other arguments are supplied, print defaults to printing $_ to filehandle sdf. Since sdf has not been opened, it fails. If you run this without warnings, you do not see any output. Note also:
E:\Home> perl -MO=Deparse -e "print sdf"
print sdf $_;
as confirmation of this observation. Note also:
E:\Home> perl -e "print asdfg, sadjkfsh"
No comma allowed after filehandle at -e line 1.
E:\Home> perl -e "print asdfg => sadjkfsh"
asdfgsadjkfsh
The latter prints both strings because => automatically quotes strings on the LHS if they consist solely of 'word' characters, removing the filehandle interpretation of the first argument.
All of these examples show that using barewords leads to many surprises. You should use strict to avoid such cases.
This is a "bareword". If it is allowed, it has the value of "sdf", and is therefore not undefined.
The example isn't special:
telemachus ~ $ perl -e 'if (defined sdf) { print "True\n" };'
True
telemachus ~ $ perl -e 'if (defined abc) { print "True\n" };'
True
telemachus ~ $ perl -e 'if (defined ccc) { print "True\n" };'
True
telemachus ~ $ perl -e 'if (defined 8) { print "True\n" };'
True
None of those is equivalent to undef which is what defined checks for.
You might want to check out this article on truth in Perl: What is Truth?
defined returns true if the expression has a value other than the undefined value.
the defined function returns true unless the value passed in the argument is undefined. This is useful from distinguishing a variable containing 0 or "" from a variable that just winked into existence.