Markus JäNtti wrote:
> I'm working myself through Julie C. Meloni's book "PHP" and now I'm
> stuck at chapter 12.
> My script gives me this error: You have an error in your SQL syntax
> near '()' at line 1
> even tho I've even tried replacing my own work with the file from the
> book's companion-files.
> I'd be very happy if someone would take the time to look through this
> and tell me what the problem might be. Hard to move forward when I
> don't understand this.
>
The $_POST array exists of keys and values. The keys are strings (in the
case of $_POST), so you'll have to access them like strings. Use
$_POST['table_name'] instead of $_POST[table_name]. The same goed also for
$_POST[field_name], $_POST[field_type] and $_POST[field_length].
HTH
Erwin
> <?
> //indicate the database you want to use
> $db_name ="testDB";
>
> //connect to database
> $connection = @mysql_connect("localhost","john","doe99") or
> die(mysql_error());
> $db = @mysql_select_db($db_name,$connection) or die(mysql_error());
>
> //start creating the SQL statement
> $sql ="CREATE TABLE $_POST[table_name] ((";
>
> //continue the SQL statement for each new field
> for ($i =0;$i < count($_POST[field_name]);$i++){
> $sql .= $_POST[field_name][$i]."".$_POST[field_type][$i];
> if ($_POST [field_length][$i] != "") {
> $sql .= "(".$_POST [field_length][$i]."),";
> } else {
> $sql .= ",";
> }
> }
>
> //clean up the end of the string
> $sql = substr($sql,0,-1);
> $sql .= ")";
>
> //execute the query
> $result = mysql_query($sql,$connection) or die(mysql_error());
>
> //get a good message for display upon success
> if ($result) {
> $msg ="<P>".$_POST[table_name]." has been created!</P>";
> }
>
> ?>
>
> <HTML>
> <HEAD>
> <TITLE>Create a Database Table:Step 3</TITLE>
> </HEAD>
> <BODY>
> <h1>Adding table to <? echo "$db_name"; ?>...</h1>
>
> <? echo "$msg"; ?>
>
> </BODY>
> </HTML>
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