Error in syntax, I cant figure out where.

breenthemachine · #1 (**permalink**) 06-30-06, 02:09 AM

Here is the code where I am recieving my error.

do_createtable.php

PHP Code:


			
<?

//indicate the database you want to use

$db_name = "testDB";



//connect to database

$connection = @mysql_connect("localhost","breen","area") 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 ($_POST[auto_increment][$i] == "Y")  {

$additional = "NOT NULL auto_increment";

} else {

    $additional = "";

}



if ($_POST[primary][$i] == "Y") {

$additional .= ", primary key (".$_POST[field_name][$i].")";

 } else {

    $additional = "";

}







    if ($_POST [field_length][$i] != "") {

        $sql .= " (".$_POST[field_length][$i].") $additional ,";

    } else {

        $sql .= " $additional ,";

    }

}



//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>

The error I get is:

You have an error in your SQL syntax near ')' at line 1

Obviously, my code is dying. What it should be doing is adding a table to "testDB". I use a form page where I enter in the field name, the field type, the field length, and whether it is a primary key and/or auto increment. Then I click the submit button where it should create my table no problem. I get this error instead.

If anyone could point me in the right direction, that would be great.

Thanks.

UnrealEd · #2 (**permalink**) 06-30-06, 03:21 AM

try echoing the query before entering it into mysql. most of te time you'll see what's wrong

just a small note: if you are going to allow user's to create a table, i don't think that's very wise: it is better to do it yourself, for security reasons.

Show me the query, and i'll see what's wrong
UnrealEd

breenthemachine · #3 (**permalink**) 06-30-06, 12:04 PM

Quote:

Originally Posted by UnrealEd

try echoing the query before entering it into mysql. most of te time you'll see what's wrong

just a small note: if you are going to allow user's to create a table, i don't think that's very wise: it is better to do it yourself, for security reasons.

Show me the query, and i'll see what's wrong
UnrealEd

I totally understand the severity in security. Actually, this is not for users to use, its for my use. Im trying to teach myself PHP, since that is one language on my list to learn, and I have a book telling me to enter this code, and obviously, the book is wrong. I have checked over the code mulitple times to verify if i have what the book has, and I even tried using previous programming knowledge to see if I could figure out the error. I dont know if this is what you wanted but I put this line

PHP Code:


			
echo "$sql";

and commented out the execution statement. This is the result I get when its ran.

CREATE TABLE)
Adding table to testDB...

I double checked everything to try to see if the syntax is correct and I guess if there is a mistake, im just not that PHP savvy to find it. Thanks for your help.

mab · #4 (**permalink**) 06-30-06, 12:25 PM

UnrealEd: Hope you don't mind if I jump in here.

It appears that the $_POST variables are empty, probably due to a mismatch in the names used or some problem in the form code. If you post the form code it would help.

You can also add the following, right after the <? opening PHP tag, to get PHP to tell you more info about what is going on and to show you the actual POST variables -

PHP Code:


			
error_reporting(E_ALL); // show all notices, warnings, and errors

echo "<pre>";

print_r($_POST); // show the actual post variables

echo "</pre>";

breenthemachine · #5 (**permalink**) 06-30-06, 10:13 PM

Thanks. That worked. I entered in the code you told me to enter, and looked at the do_showfielddef.php and noticed this error:

PHP Code:


			
<INPUT TYPE=\"hidden \" NAME=\"table_name \" VALUE=\"$_POST[table_name]\">

The extra spaces were what was giving me troubles. Thanks for all of your help.