[Jack Drake]

I'm having trouble creating a table through php in mysql.
FYI: The database connects successfully.

This is the code that is to supposed to send the query

Code

<?php
    $con = mysqli_connect("localhost", "username", "password", "database");

    if(!$con)
    {
        echo "failed to connect to the database.";
    }
    else
    {
    echo '<script>alert("success, connected to database");</script>';
    }

    $stmt = "CREATE TABLE database.table(name TEXT, number INT)";
        if(!mysqli_query($stmt))
        {
                echo "could not create the tutors table " . mysqli_error();
        }
?>

The database does exist and the username and password do work.
What is the issue?

[Charger]

Well, what’s the error displayed?

[Squito]

Mysqli_query requires 2 parameters, you only provided one.

[Mohinder]

I think you have to do something like:

Code

$conn->query($query);

Since you created the connection in variable $conn, and that variables is unused.

[xMAAT]

You can solve that with 2 options:

• You can use object-oriented style doing this:
  
  PHP Code

  
<?php 
$con->query($stmt); 

  But remember that you have to create the connection with the object MySQLi:
  
  PHP Code

  
<?php 
$con = new mysqli("localhost","user","pass","db_name"); 

  
  


• The second option is to add in this string:
  
  PHP Code

  
<?php 
if(!mysqli_query($stmt)) 

  That:
  
  PHP Code

  
<?php 
if(!mysqli_query($con,$stmt)) 


[jascorp]

Your ask problem about creating table, but you provided connection string, and you put FYI as connection was successful without problems. So, which one is the script problem?

[itzDecon]

When using MySQLi procedurally instead of the object-orientated, you need to specify the connection as the first argument. (https://www.php.net/manual/en/mysqli.query.php)

Or you can look into switching to object orientated as xMAAT has described which uses classes/method vs functions.

[Jack Drake]

Thanks a lot, the issue was that I was missing a parameter from the query call, but I only figured it out through your guys' help.

My new problem is trying to understand the issue with inserting into a table in this database.

This is the code I have:

Code

<?php
   $con = new mysqli("localhost", "username", "password", "database");
   if($con->connect_error)
   {
       echo "failed to connect to the database.";
   }
   else
   {
   echo '<script>alert("success, connected to database");</script>';
   }
       $stmt = "insert into test (name text, num text) values ('name', 'last')";
        if($con->query($stmt) !== true)
       {
        echo $con->error;
       }
?>

The error I receive is this:

Code

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'text, num text) values ('name', 'last')' at line 1

What is the issue.

[Nachfolger]

Thanks a lot, the issue was that I was missing a parameter from the query call, but I only figured it out through your guys' help.

My new problem is trying to understand the issue with inserting into a table in this database.

This is the code I have:

Code

<?php
   $con = new mysqli("localhost", "username", "password", "database");
   if($con->connect_error)
   {
       echo "failed to connect to the database.";
   }
   else
   {
   echo '<script>alert("success, connected to database");</script>';
   }
       $stmt = "insert into test (name text, num text) values ('name', 'last')";
        if($con->query($stmt) !== true)
       {
        echo $con->error;
       }
?>

The error I receive is this:

Code

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'text, num text) values ('name', 'last')' at line 1

What is the issue.

Specify the column names only, not their character types.

Code

$stmt = "insert into test (name, num) values ('name', 'last')";

If you're inserting things dynamically, or through user input, start using prepared statements.

[Jack Drake]

Is it true that preparing the statement actually decreasing the chance of SQL Injection?

[Nachfolger]

Is it true that preparing the statement actually decreasing the chance of SQL Injection?

Do you think that's just nazi-propaganda or something? Yes, using prepared/parameter statements with proper validation will absolutely decrease the chance of SQL injection.

[Mohinder]

Do you think that's just nazi-propaganda or something? Yes, using prepared/parameter statements with proper validation will absolutely decrease the chance of SQL injection.

This made me laugh o//

[xHeaven]

@OP, please start learning PHP with more simple things. 2 weeks ago you tried to use the Selly API, yet you can't even make a proper MySQL connection, and doesn't even understand variable scopes. Try to learn 1 by 1, not all at once.