[PHP] Politics

[Amy]

prints mosfet customercontrol universally lastinfirstout unixlike techie 
perversities agreement

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[PHP] Politics

[Amy]

mouso prefixes egging gestures loozr selected incomprehensibly assumed sexism

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[PHP] Politics

[Amy]

representations emphasizing leksr matching thirds painfully wakesleep 
ekswiezeezeewie accompanied

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[PHP] Politics

[Amy]

confronted gillion presently flakey expanded courses newsreader publicaccess 
storages

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[PHP] what syntax error are they talking about?

[Amy Rogers]

I need help understand the error following message I got. The related 
script is below.  I also have a separate question about dots that 
appear in the script. I am using Julie Meloni's PHP Essentials 2nd 
Edition to learn the language, and this is taken from page 119.

(I was trying to create a new table with 4 fields to store userids and 
passwords. I used the same script successfully to create a different 
table and am confused why it is not working this time.)

Error message:
You have an error in your SQL syntax. Check the manual that 
corresponds to your MySQL server version for the right syntax to use 
near '(20),username text (10),password varchar (10))' at line 1

A separate question:
What is the meaning of the . on each end of the 
.$_POST[field_length][$i]. ?

Here's the relevant part of the script:
?php
$sql = create table $_POST[table_name] (;
//to create the fields you specified in Step 2
//create loop to populate the remainer of the sql statement
for ($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 .= ,;
	}
}
$sql = substr($sql, 0, -1);
//close the SQL statement
$sql .= );
$conn = mysql_connect(localhost, userid, password) or 
die(mysql_error());
$db = mysql_select_db (aroDB, $conn) or die(mysql_error());
$sql_result = mysql_query($sql, $conn) or die(mysql_error());
if ($sql_result) {
	echo p$_POST[table_name] has been created./p;
	}
?

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[PHP] Re: what syntax error are they talking about? - FOLLOWUP

[Amy Rogers]

Thanks to Jay B, the puzzle is no more.  Turns out assigning a field 
length to the text field type caused the problem. I left out the field 
length for this data type and the script now works.

He also offered a helpful explanation about this data type:
http://dev.mysql.com/doc/mysql/en/BLOB.html
In most respects, you can regard a TEXT column as a VARCHAR column that
can be as big as you like. 
The maximum size of a BLOB or TEXT object is determined by its type,
but the largest value you actually can transmit between the client and
server is determined by the amount of available memory and the size of
the communications buffers. 
Thanks, Jay.
Amy
On Sep 30, 2004, at 12:01 PM, Amy Rogers wrote:
I need help understand the error following message I got. The related 
script is below.  I also have a separate question about dots that 
appear in the script. I am using Julie Meloni's PHP Essentials 2nd 
Edition to learn the language, and this is taken from page 119.

(I was trying to create a new table with 4 fields to store userids and 
passwords. I used the same script successfully to create a different 
table and am confused why it is not working this time.)

Error message:
You have an error in your SQL syntax. Check the manual that 
corresponds to your MySQL server version for the right syntax to use 
near '(20),username text (10),password varchar (10))' at line 1

A separate question:
What is the meaning of the . on each end of the 
.$_POST[field_length][$i]. ?

Here's the relevant part of the script:
?php
$sql = create table $_POST[table_name] (;
//to create the fields you specified in Step 2
//create loop to populate the remainer of the sql statement
for ($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 .= ,;
	}
}
$sql = substr($sql, 0, -1);
//close the SQL statement
$sql .= );
$conn = mysql_connect(localhost, userid, password) or 
die(mysql_error());
$db = mysql_select_db (aroDB, $conn) or die(mysql_error());
$sql_result = mysql_query($sql, $conn) or die(mysql_error());
if ($sql_result) {
	echo p$_POST[table_name] has been created./p;
	}
?

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