PHP Basic Function Help

[Inuyasha1782]

Okay, I'm still in the process of learning the basics of PHP. Currently I am interacting with MySql to 'spice' up my site.

At the moment, I am trying to create a function for simplifying the creation of tables via a function.

This would require the table name and it's values. The problem is the values can be 2 or 10, and are usually not the same. So my idea was to handle this with an array, but I need to know a few things first:

-Can you use an array as a paramter for a function
-How do you use that array as a paramter
-How do you get the upper bound of an array

If anyone could help explain the above it would be nice

<?php
include 'config.php';   //Sets up the database information
include 'connect.php'; //Connects the the database

function create_table($tablename, $values){
$query = 'CREATE TABLE $tablename, $values';
$result = mysql_query($query);
}

include 'close.php'; //Closes the connection
?>

Note: My idea was to create the values string by creating a loop that ran through all the values of the array and added them to another string like ", $Value1, $Value2" etc.

[CornedBee]

1) Yes, like any other parameter.
2) Just like any other array.
3) Use count().

[Inuyasha1782]

Sooo something like this:

<?php
include 'config.php';   //Sets up the database information
include 'connect.php'; //Connects the the database

function create_table($tablename, $values[]){
while($x <= count($values)){
$VString = $VString . ", " . $values[$x];
++$x;
}
$query = 'CREATE TABLE $tablename($VString)';
$result = mysql_query($query);
}

include 'close.php'; //Closes the connection
?>

[CornedBee]

Remove the [] from the argument list. An array is an argument just like any other. (In fact, you don't even have a way to enforce the passing of an array, safe through testing with is_array inside the function body. That's the thing about weakly typed languages.)
Also, look at the join() function, it simplifies merging arrays into strings. As it is now, you get an SQL syntax error for starting the field list with a ','.
Or not: in fact, you'll get an error now because variable substitution is only done in strings delimited by double quotes - the query you have now will contain the literal strings $tablename and $VString.

[Inuyasha1782]

So how come this won't create a table then:

function create_table($tablename, $values){
  while($x <= count($values)){
    if($VString <> ''){
      $VString = $VString . ", " . $values[$x];
    }else{
      $VString = $values[$x];
    }
    ++$x;
  }
  $query = 'CREATE TABLE ' . $tablename . '(' . $VString . ')';
  $result = mysql_query($query);
  print 'Created table $tablename';
} 

That is contained in "sqlfunctions.php". I call it with:

<?php
include 'Php/sqlfunctions.php'; //Where the function is
include 'Php/connect.php';  //Connects to the DB

$name = "newlogin"; //The table name
$Values[0] = "id int(6) NOT NULL auto_increment";
$Values[1] = "user varchar(15) NOT NULL";
$Values[2] = "password varchar(15) NOT NULL";
$Values[3] = "PRIMARY KEY (id)";
$Values[4] = "UNIQUE id (id)";
$Values[5] = "KEY id_2 (id)";
create_table($name, $Values[]); //The call to the function

mysql_close($conn);
?>

It doesn't create a new table. Yes I am connected properly, else I would be getting a connection error.

[CornedBee]

Any other errors? Do you have create table permissions?
And have you output the completed SQL statement?

Btw, this;

  while($x <= count($values)){
    if($VString <> ''){
      $VString = $VString . ", " . $values[$x];
    }else{
      $VString = $values[$x];
    }
    ++$x;
  }

can be simplified to this:

$VString = join(', ', $values);

[Inuyasha1782]

No other errors, I get the final print stament and it's done.