How do I find a literal % with the LIKE-operator?
#!/usr/bin/perl
use warnings;
use strict;
use DBI;
my $table = 'formula';
my $dbh = DBI->connect ( "DBI:CSV:", undef, undef, { RaiseError => 1 } );
my $AoA = [ [ qw( id formula ) ],
[ 1, 'a + b' ],
[ 2, 'c - d' ],
[ 3, 'e * f' ],
[ 4, 'g / h' ],
[ 5, 'i % j' ], ];
$dbh->do( qq{ CREATE TEMP TABLE $table AS IMPORT ( ? ) }, {}, $AoA );
my $sth = $dbh->prepare ( qq{ SELECT * FROM $table WHERE formula LIKE '%[%]%' } );
$sth->execute;
$sth->dump_results;开发者_运维知识库
# Output:
# 3, 'e * f'
# 1 rows
Looks like you can't do this with current version of DBD::CSV
.
You are using DBD::CSV
module to access data. It uses SQL::Statement
module to handle expresions. I've searched its source code and found that following code handles LIKE
sql statement condition:
## from SQL::Statement::Operation::Regexp::right method
unless ( defined( $self->{PATTERNS}->{$right} ) )
{
$self->{PATTERNS}->{$right} = $right;
## looks like it doen't check any escape symbols
$self->{PATTERNS}->{$right} =~ s/%/.*/g;
$self->{PATTERNS}->{$right} = $self->regexp( $self->{PATTERNS}->{$right} );
}
Look at $self->{PATTERNS}->{$right} =~ s/%/.*/g;
line. It converts LIKE
pattern to regexp. And it doesn't do any check of any escape symbols. All %
symbols are blindly translated to .*
pattern. That's why I think it's not implemented yet.
Well, may be someone'll find time to fix this issue.
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