无法创建用于创建具有不同名称的数据库的脚本。 PHP / MySQL的

Question

guys. I would like to make this script, or other with the same effect:

<?php
$con = mysql_connect("localhost","peter","abc123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

if (mysql_query("CREATE DATABASE my_db",$con))
  {
  echo "Database created";
  }
else
  {
  echo "Error creating database: " . mysql_error();
  }

mysql_close($con);
?>

To create databases with different names, for example: First time will create database named "news1", the next time, the script will just do sth. like that: news1+1 = news2, and will create database called news2, and so on...

I hope you got my point and I'll really appreciate it if you help me to make this... Regards, Denis Saidov.

Answer 1

To do this I'm guessing that you don't just want to loop 100 times?

You will need to save a variable to a text file then call it again.

Writing $text to a file:

$var_str = var_export($text, true);
$var = "<?php

\$$text = $var_str;

?>";
file_put_contents('filename.php', $var);

Retrieving it again:

include 'filename.php';
echo $text;

All together:

<?php
$con = mysql_connect("localhost","peter","abc123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
$db_number = get_var();
$my_db = 'news' . $db_number;
if (mysql_query("CREATE DATABASE ".$my_db,$con))
  {
  echo "Database created";
  }
else
  {
  echo "Error creating database: " . mysql_error();
  }
$new_db_number = $db_number + 1;
save_var($new_db_number);
mysql_close($con);

function save_var($var) {
$var_str = var_export($text, true);
$var = "<?php

\$$text = $var_str;

?>";
file_put_contents('filename.php', $var);
}
function get_var()
include 'filename.php';
return $text;
}
?>

PS I think that should work only typing not tested