I am getting a failed query error message. Could someone take a look and let
me know.
//F U N C T I O N S
//=
function AddSignupRequest($Signup_FName, $Signup_LName, $Signup_Address1,
$Signup_Address2, $Signup_City,
$Signup_State, $Signup_Zip, $Signup_Email, $Signup_Phone,
I did some more testing and I found that I forgot the first field in the
columns section of the Statement. So now it looks like this:
I also copy and pasted the statement into PHPMYADMIN and it worked fine. It
just doesn't work here
function AddSignupRequest($Signup_FName, $Signup_LName,
[snip]
/* Performing SQL query */
$query = INSERT INTO SignupRequests ('FirstName', 'LastName',
'Address1', 'Address2',
'City', 'State', 'Zip', 'Email', 'Phone', 'ContactMethod', 'Date',
'IP', 'Status', 'Comments')
VALUES ('', 'Chris', 'Crane', '655 Talcottville Road', 'Apt.
Initially there was an error with too many values verses columns. But I
think it was fixed. I am double checking now.
Jay Blanchard [EMAIL PROTECTED] wrote in message
003201c24d07$b8560470$8102a8c0@000347D72515">news:003201c24d07$b8560470$8102a8c0@000347D72515...
[snip]
/* Performing SQL
There are 15 columns and 15 pieces of data. In my second post I fixed this
error and pasted it into PHPMYADMIN and it worked fine, but not here...
Chris Crane [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Initially there was an error with too many values
Why don't you do
$result = mysql_query($query) or die(mysql_error());
or
$result = mysql_query($query);
echo mysql_error();
That way, instead of Query Failed you'll get something meaningful...
probably something that will solve the problem.
Justin French
on 26/08/02 11:55 PM, Chris Crane
Thank you. I will try that.
Justin French [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Why don't you do
$result = mysql_query($query) or die(mysql_error());
or
$result = mysql_query($query);
echo mysql_error();
That way, instead of Query Failed you'll
I got it working. It did not like the single quotes around the column names.
Chris Crane [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Thank you. I will try that.
Justin French [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Code:
?
$Query = UPDATE feRegUsers SET Constructor='2001-09-17',
Enertec='2001-09-17', Seatec='2001-09-17' WHERE ID LIKE '288';
mysql_quory($Query);
print mysql_error();
?
That results
You have an error in your SQL syntax near 'Constructor='2001-09-17',
Enertec='2001-09-17', Seatec='2001-09-17''
Hi,
I have a MySQL statement, namely:
SELECT COUNT(*) AS Count FROM images i LEFT JOIN categories c ON
i.cat_id=c.id LEFT JOIN subcategories s ON i.subcat_id=s.id
I want to change the LEFT JOINs to INNER JOINs like so:
SELECT COUNT(*) AS Count FROM images i INNER JOIN categories c ON
No, that is exactly as the PHPLib Debug function is returning... :|
-Original Message-
From: Samantha Savvakis [mailto:[EMAIL PROTECTED]]
Sent: Monday, 22 January 2001 3:21 PM
To: Matt Stone
Subject: RE: [PHP] MySQL Query Error
I'm not sure why are getting that error.
Is the statement
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