[PHP-DB] Syntax Error
Can anyone help me figure out what is wrong with this syntax. Thanks.
list($qh,$num) = dbQuery(SELECT password AS passwd1,
$DATABASE_PASSWORD_FUNCTION('$password') AS passwd2 .
FROM $USER_TABLE WHERE username='$username');
$data = dbResult($qh);
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Re: [PHP-DB] Syntax Error
Hi is the syntax error reported for the php or the SQL? Better yet, supply the actual error message, so we can see for ourselves Niel -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php ** IMPORTANT NOTICE This communication is for the exclusive use of the intended recipient(s) named above. If you receive this communication in error, you should notify the sender by e-mail or by telephone (+44) 191 224 4461, delete it and destroy any copies of it. This communication may contain confidential information and material protected by copyright, design right or other intellectual property rights which are and shall remain the property of Piranha Studios Limited. Any form of distribution, copying or other unauthorised use of this communication or the information in it is strictly prohibited. Piranha Studios Limited asserts its rights in this communication and the information in it and reserves the right to take action against anyone who misuses it or the information in it. Piranha Studios Limited cannot accept any liability sustained as a result of software viruses and would recommend that you carry out your own virus checks before opening any attachment. GWAVAsigAdmID:CBE4B095E96FA633FC7643E6228B7CCE ** IMPORTANT NOTICE This communication is for the exclusive use of the intended recipient(s) named above. If you receive this communication in error, you should notify the sender by e-mail or by telephone (+44) 191 224 4461, delete it and destroy any copies of it. This communication may contain confidential information and material protected by copyright, design right or other intellectual property rights which are and shall remain the property of Piranha Studios Limited. Any form of distribution, copying or other unauthorised use of this communication or the information in it is strictly prohibited. Piranha Studios Limited asserts its rights in this communication and the information in it and reserves the right to take action against anyone who misuses it or the information in it. Piranha Studios Limited cannot accept any liability sustained as a result of software viruses and would recommend that you carry out your own virus checks before opening any attachment. GWAVAsig -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Syntax Error
JD wrote:
Can anyone help me figure out what is wrong with this syntax. Thanks.
list($qh,$num) = dbQuery(SELECT password AS passwd1,
$DATABASE_PASSWORD_FUNCTION('$password') AS passwd2 .
FROM $USER_TABLE WHERE username='$username');
$data = dbResult($qh);
No need to repost your question.
What is the $DATABASE_PASSWORD_FUNCTION variable ?
What is the $USER_TABLE variable ?
Since there are 4 variables in this query ($DATABASE_PASSWORD_FUNCTION,
$password, $USER_TABLE, $username), we have no idea what's actually
being run.
The easiest way to work out what's wrong is to print the query out and
run it manually.
$query = SELECT password AS passwd1,
$DATABASE_PASSWORD_FUNCTION('$password') AS passwd2 FROM $USER_TABLE
WHERE username='$username';
echo wordwrap($query) . br/\n;
list ($qh, $num) = dbQuery($query);
Post us the real sql query that's being run through that and we might be
able to help you.
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it and destroy any copies of it.
This communication may contain confidential information and material
protected by copyright, design right or other intellectual property
rights which are and shall remain the property of Piranha Studios
Limited. Any form of distribution, copying or other unauthorised use
of this communication or the information in it is strictly prohibited.
Piranha Studios Limited asserts its rights in this communication and
the information in it and reserves the right to take action against
anyone who misuses it or the information in it.
Piranha Studios Limited cannot accept any liability sustained as a
result of software viruses and would recommend that you carry out your
own virus checks before opening any attachment.
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IMPORTANT NOTICE
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named above. If you receive this communication in error, you should
notify the sender by e-mail or by telephone (+44) 191 224 4461, delete
it and destroy any copies of it.
This communication may contain confidential information and material
protected by copyright, design right or other intellectual property
rights which are and shall remain the property of Piranha Studios
Limited. Any form of distribution, copying or other unauthorised use
of this communication or the information in it is strictly prohibited.
Piranha Studios Limited asserts its rights in this communication and
the information in it and reserves the right to take action against
anyone who misuses it or the information in it.
Piranha Studios Limited cannot accept any liability sustained as a
result of software viruses and would recommend that you carry out your
own virus checks before opening any attachment.
GWAVAsig
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Re: [PHP-DB] Syntax Error
JD wrote:
Can anyone help me figure out what is wrong with this syntax. Thanks.
list($qh,$num) = dbQuery(SELECT password AS passwd1,
$DATABASE_PASSWORD_FUNCTION('$password') AS passwd2 .
FROM $USER_TABLE WHERE username='$username');
$data = dbResult($qh);
No need to repost your question.
What is the $DATABASE_PASSWORD_FUNCTION variable ?
What is the $USER_TABLE variable ?
Since there are 4 variables in this query ($DATABASE_PASSWORD_FUNCTION,
$password, $USER_TABLE, $username), we have no idea what's actually
being run.
The easiest way to work out what's wrong is to print the query out and
run it manually.
$query = SELECT password AS passwd1,
$DATABASE_PASSWORD_FUNCTION('$password') AS passwd2 FROM $USER_TABLE
WHERE username='$username';
echo wordwrap($query) . br/\n;
list ($qh, $num) = dbQuery($query);
Post us the real sql query that's being run through that and we might be
able to help you.
--
Postgresql php tutorials
http://www.designmagick.com/
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Re: [PHP-DB] Syntax Error
Hi is the syntax error reported for the php or the SQL? Better yet, supply the actual error message, so we can see for ourselves Niel -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: AW: [PHP-DB] syntax error - -- agghhh!
it's always eaiser to count the parens and braces backwards to make sure they're all there. i do it all the time. ... or you use a cool Editor (like UE32) which will do this job for you =8) Greetinx, Mike (Germany) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] syntax error - -- agghhh!
Hi
Anyone have any idea of what the syntax error in this statement is?
else if (($op == ds) ($action == sub) ($password == $passnog)
($password)
(!eregi(^([a-z0-9_]|\\-|\\.)+@(([a-z0-9_]|\\-)+\\.)+[a-z]{2,4}$,$email))
{
It worked fine until I added the tricky part:
(!eregi(^([a-z0-9_]|\\-|\\.)+@(([a-z0-9_]|\\-)+\\.)+[a-z]{2,4}$,$email)
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AW: [PHP-DB] syntax error - -- agghhh!
Marc S. Bragg [mailto:[EMAIL PROTECTED]] wrote:
Hi
Anyone have any idea of what the syntax error in this statement is?
else if (($op == ds) ($action == sub) ($password ==
$passnog)
($password)
(!eregi(^([a-z0-9_]|\\-|\\.)+@(([a-z0-9_]|\\-)+\\.)+[a-z]{2,4
}$,$email))
{
It worked fine until I added the tricky part:
(!eregi(^([a-z0-9_]|\\-|\\.)+@(([a-z0-9_]|\\-)+\\.)+[a-z]{2,4
}$,$email)
You are missing a ) to close the if expression
Thomas
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