What is wrong here??

Anon

hi, i am constantly adding the same tables (names and values) to different databases. I came up with a code that will ask for the database name and then auto create the tables(these tables and there values never change)

This code doesnt seem to want to work, this is what i have:

<<<<<<<<<<<form>>>>>>>>>>>
<?

$form_block = "

<FORM METHOD=\"POST\" ACTION=\"creatdbtbl2.php\">

// create letters2 table
// 2 fields letter_id letter_name

$form_block .= "

$form_block .= "

<strong>Enter the database where information should be placed:</strong><INPUT TYPE=\"text\" name=\"db_name\">
<TD ALIGN=CENTER COLSPAN=3><INPUT TYPE=\"submit\" VALUE=\"Create Table\"></TD>

</TABLE>
</FORM>

";

?>

<HEAD><TITLE>Create a Database Table: Step 2</TITLE>
</HEAD>
<BODY>
<H1>Define fields for <? echo "$table_name"; ?></H1>
<? echo "$form_block"; ?>
</BODY>

------To do the processing my sql part of the code is:

$sql = "CREATE TABLE letters2 (";

for ($i = 0; $i < 3; $i++) {

$sql .= "$field_name[$i] $field_type[$i]";

if ($auto_increment[$i] == "Y") {
	$additional = "NOT NULL auto_increment";
} else {
	$additional = "";
}

if ($primary[$i] == "Y") {
	$additional .= ", primary key ($field_name[$i])";
} else {
	$additional .= "";
}



if ($field_length[$i] != "") {
	$sql .= " ($field_length[$i]) $additional ,";
} else {
	$sql .= " $additional ,";
}

}

$sql = substr($sql, 0, -1);

$sql .= ")";

// execute the query

$result = mysql_query($sql,$connection) or die(mysql_error());

if ($result) {

$msg = "<P>letters2 has been created!</p>

}

------- Here is the error

You have an error in your SQL syntax near 'letter_id int ,letter_name varchar (50) )' at line 1

Sorry this is so long, but if anyone can help i really appreciate it.

Thanks

Anon

Your post is way too long for me to look deep into it. Anyway, your error is typically a SQL syntax error.

My advice: echo the sql command before querying it, the error should be obvious.

If you still don't understand where it is, copy the sql command and paste it into MySqlAdmin. The error might be better explained.

Ronan

Anon

Thanks for your response Ronan, i know my code is not easy on the eyes. Anyway, it never fails - i will be working on a code and with problems for hours and then when i am frustrated enough i will post here, and as soon as i do, I figure it out by myself!!! go figure. Anyway if anyone was curious - my problem was i was looking at my array starting at [1] instead of [0]. instead of [1], [2], i changed it to [0],[1] and it works fine...