2

如何使用 LIKE 运算符找到文字 %?

#!/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;

# Output:
# 3, 'e * f'
# 1 rows
4

1 回答 1

4

看起来你不能用当前版本的DBD::CSV.

您正在使用DBD::CSV模块来访问数据。它使用SQL::Statement模块来处理表达式。我搜索了它的源代码,发现以下代码处理了LIKEsql 语句条件:

## 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} );
}

$self->{PATTERNS}->{$right} =~ s/%/.*/g;线。它将LIKE模式转换为正则表达式。它不会对任何转义符号进行任何检查。所有%符号都被盲目地转换为.*模式。这就是为什么我认为它尚未实施。

好吧,可能有人会抽出时间来解决这个问题。

于 2010-01-06T19:14:23.540 回答