Question

I’m a beginner programmer, and I’ve installed XAMPP with the intention of learning a bit of PHP. I have a working knowledge of SQL.

I’ve been following the PHP tutorial at w3schools. The problem I am currently having is this: I’m using the script here to create a database and a table within it. What I’m using is almost verbatim, except I’ve replaced the user with “root” and I’ve deleted the password.

After running the script through the browser, the database my_db appears in the datafile for mysql in XAMPP.

However, there is no sign of a table, and when I try to select the table, I get

Table ‘my_db.persons’ doesn’t exist

What is going on? Is there something wrong with the code I took verbatim, or is it something with permissions?

It’s weird that the database is created but not a table…

Answer 1

What has happenned is that your “CREATE TABLE” will have failed.

Get into the habit of checking for errors after EVERY mySQL query (in your code): there are some times you can ignore errors, but it’s rare. So program alonghte lines of:

Create query: $sql =
Set parameters: $aParams =   (or bind paramters)
Execute query
If errors
    If debug: Show error and query
    If live: Log error and query

I’m not giving the solution is code, as one problem with the tutorial you follow is that it uses mysql_() functions that are going to be depreciated shortly. You should use PDO (PHP Database Objects) or mysqli() functions otherwise your code will not work in a few releases time.

With PDO, you can set error handling to use exceptions, and you wrap every call in try {} catch {} and this makes the habit of catching and reporting errors very easy.

    $sql = 'CREATE TABLE....';
    $aParams = array(
        ':param_name' => $param_value,
        ':param_name2' => $param_value2
           );

    try {
        $stmnt = $db->prepare($sql);
        $stmnt->execute($aParams);
        $stmnt = null;
    } catch (Exception $e) {
        // Error log here; $e contins line of error and the actual error, you have $sql and $aParams
        LogDBError($e, $sql, $aParams);
    }