Perl Database SQL query variable passing

[sonu2die4]

Hi,
I have the following perl script working for me.I am accesing the database from my perl script using use Net::Telnet(); package.
I am not using DBI package.as I stated earlier the following program is printing the output in a nice form.However I want to pass
a variable in the

 $t->cmd("SELECT * FROM TABLE_NAME WHERE INSTANCE_NAME LIKE '%hostname%' ;");

Here I want hostname to be replaced by a variable name like

$t->cmd("SELECT * FROM TABLE_NAME WHERE INSTANCE_NAME LIKE '%VARIABLE_NAME%' ;");.

I want just pass a variable in the above sql query so that I can change the VARIABLE_NAME.P lease help me out with it.
Thank You
Vivek

$IPAddress = " ";
$Login = " ";
$password = " ";
$Node = "hostname";
 
use Net::Telnet();
$t= new Net::Telnet (Timeout => 3000 , Prompt => '/[%#\$>?:] $/' );
$t->open("$IPAddress");
print "\nConected!";
$t->waitfor('/login: $/i');
$t->print($Login);
print "\nEntered the Username\n";
$t->waitfor('/assword: $/i');
$t->print($Password);
print "\nEntered the Password!\n";
@output=$t->cmd("export ORACLE_SID=$Node\n");
print @output;
print "\n Done with logging into the database\n";
@output= $t->cmd("bash\n");
print FILE @output;
print @output;
@output = $t->cmd("sqlplus username/pass\n");
print @output;
@output = $t->cmd("conn cusername/pass\n");
print @output;
@output = $t->cmd("SELECT * FROM TABLE_NAME WHERE INSTANCE_NAME LIKE '%hostname%' AND PARAM_KEY_NAME LIKE '%host';");
print FILE @output;
@output =  $t->cmd("exit\n");
print @output;
@output =  $t->cmd("exit\n");
print @output;

[numberwhun]

@sonu2die4

Why are you using % to specify a variable name? In Perl, a $ is for a scalar variable and a % is for a hash. If you have a $variable that you want to use, then use it in place of %hostname% as that is not correct. This is not dos scripting or any other like it.

So, you can write your statement like so:

@output = $t->cmd("SELECT * FROM TABLE_NAME WHERE INSTANCE_NAME LIKE '$hostname' AND PARAM_KEY_NAME LIKE '$host';");
 

Also, please take note of the following:

1. Code tags are required around ANY code you post in the forums.
2. Any personal or private information (ie: usernames, passwords, IP addresses, url's, email addresses) are strictly forbidden in the general technical forums. This is for your safety and security and we expect you to please abide by this.

Regards,

Jeff

[eWish]

The % is a wildcard for MySQL. It can be used like he is using it.

--Kevin

[numberwhun]

@eWish
Ok, but if he is trying to pass a variable from a Perl script, he should pass the variable itself, not using the syntax he is using. I have seen plenty of code where I work (and written some) to use a variable in place of a parameter in the SQL line of the code, but have never used the % in the Perl code.

[eWish]

Here is some code I wrote a while ago to illustrate how to use a placeholder in the LIKE clause, which could be practiced here as well.

my $var = 'Car';
my $like_var = '%' . $var . '%';
 
my $test = $dbh->prepare(qq|SELECT name FROM categories WHERE name LIKE ?|);
   $test->execute($like_var);
 
   while (my @rows = $test->fetchrow_array()){
       foreach (@rows) {
          print $_, '<br>' ;
 
       }
   }
$test->finish();
 
my $regex_test = qq|^Car|;
 
my $another_test = $dbh->prepare(qq|SELECT name FROM categories WHERE name REGEXP ?|);
   $another_test->execute($regex_test);
 
      while (my @rows = $another_test->fetchrow_array()){
       foreach (@rows) {
          print $_, '<br>';
       }
   }
$another_test->finish(); 

This way you don't lose the % wildcard and you make safer and don't have to worry about escaping the data. Helps reduce the sql injections.

--Kevin