Dear all,

I new to PHP/MySql.

here i am trying to create a database.


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

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

// Create table
mysql_select_db("my_db", $con);
$sql = "CREATE TABLE Persons
(
FirstName varchar(15),
LastName varchar(15),
Age int
)";

// Execute query
mysql_query($sql,$con);

mysql_close($con);

?>

1. I couldn't identify that whats done. In PHPMyAdmin there is no database with name "my_db".

2. I could see only one message on my page the "connected......" which is from line no: 9. if the CREATE DATABASE is unsucess then the other message "Error creating database:" should be displayed.

Regards:

Muhammad Nadeem

thats because you are using the die construct on both arms of the if statement.
die halts execution of the script.

usually dies is used to report an error message, not a success condition.

Dear,

Please guide me what to do?

I tried the both die and echo but the same address.

I am new to PHP and MySql.


waiting for you.

Muhammad Nadeem

so what happened when you used to echo function?

ironically you don't use the or die construct or examine the MySQL error / status whenyou try to create the table

?>