Question

我有以下Perl脚本

$filters = "SELECT lead_id, status, user, title, first_name, last_name, city, phone_number FROM vicidial_list , vicidial_lists WHERE vicidial_list.list_id = vicidial_lists.list_id";
$stmtA   = "SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_SCHEMA = 'asterisk' AND TABLE_NAME = 'vicidial_list'";
$sthA    = $dbhA->prepare( $stmtA ) or die "preparing: ", $dbhA->errstr;

$sthA->execute or die "executing: $stmtA ", $dbhA->errstr;

$counter = 0;
$columns = ();

while ( my @row = $sthA->fetchrow_array ) {

    # GET the name columns by pos
    $column = "@row[0] \n";

    # Check if exists the name columns
    if ( index( $filters, @row[$counter] ) > -1 ) {

        $find    = @row[$counter];
        $replace = "vicidial_list." . @row[$counter];

        # Replace the name column
        $filters =~ s/$find/$replace/g;

    }
}

print $filters . "\n\n";

#Close connection
$sthA->finish();

在有评论# Check if exists the name columns的部分中，我想知道一些字符串之前和之后的内容。

例如，如果我选择MY NAME IS CRISTOPHER，则NAME会返回MY之前的IS，然后是select * from( select *, row_number() over (partition by id order by id,issuingstate desc) as rn from #temp where (case when PrimaryLocation ='Y' then 1 when PrimaryLocation ='N' and locationstate = issuingstate then 1 when PrimaryLocation ='N' and issuingstate is NULL then 1 else 0 end) = 1 ) a where rn = 1。

Answer 1

如果你的问题是关于在字符串中找到一个模式，在perl中，那么一个答案可能是正则表达式：

my $string='MY NAME IS';
my $pattern='NAME';
if ($string =~ /$pattern/) {
    print "Before: $`\n";
    print "After: $'\n";
}

另请参阅特殊RE variables的完整列表。

在给定单词的任一侧查找字符串中的单词

1 个答案: