: domingo, junio 23, 2013 4:38 PM
Para: php-db@lists.php.net; Jim Giner
Asunto: [PHP-DB] Re: Problem with query
On 6/23/2013 2:31 PM, Ethan Rosenberg, PhD wrote:
Dear List -
There is an error in my query, and I cannot find it.
This fails:
$_SESSION['Cust_Num'] = $_REQUEST['cnum
On Sun, Jun 23, 2013 at 8:31 PM, Ethan Rosenberg, PhD
erosenb...@hygeiabiomedical.com wrote:
Dear List -
There is an error in my query, and I cannot find it.
This fails:
$_SESSION['Cust_Num'] = $_REQUEST['cnum'];
$_SESSION['CustNum'] = $_REQUEST['cnum'];
echo sessionbr /; //this has
On 23 June 2013 21:37, Ethan Rosenberg, PhD
erosenb...@hygeiabiomedical.com wrote:
On 6/23/2013 2:31 PM, Ethan Rosenberg, PhD wrote:
Dear List -
There is an error in my query, and I cannot find it.
This fails:
$_SESSION['Cust_Num'] =
Turn on error reporting/logging/displaying and what errors are you getting?
And as you said ...
$result10 = mysqli_query($cxn, $sql10);
var_dump($result1); // this returns NULL
is your actual code, maybe ...
?php
$a = 'set variable a to this message';
var_dump($b);
?
gives you a
Dear Ethan
It seems you are trying to build a query.But you are not getting field
names. If you required field names then change the following line to
foreach ( $allowed_fields AS $field = $_POST['field'])
to
foreach ( $allowed_fields AS $field)
This would convert the variable field to value.
At 12:13 AM 2/10/2012, Amit Tandon wrote:
Dear Ethan
It seems you are trying to build a query.But you are not getting field
names. If you required field names then change the following line to
foreach ( $allowed_fields AS $field = $_POST['field'])
to
foreach ( $allowed_fields AS $field)
This
Dear Ethan
The line you are getting is because the $_POST[fieldname] is blank. So for
the following line
if ( ! empty( $_POST['field'] ) )
change it to
if ( ! empty( $_POST[$field] ) )
Your line : Program is searxchinbg for variable name field
New line : The Program is seacging for varable
At 12:48 AM 2/10/2012, Amit Tandon wrote:
Dear Ethan
The line you are getting is because the
$_POST[fieldname] is blank. So for the following line
 if ( ! empty( $_POST['field'] ) )
change it to
 if ( ! empty( $_POST[$field] ) )
Your line : Program is searxchinbg for variable name field
At 12:48 AM 2/10/2012, Amit Tandon wrote:
Dear Ethan
The line you are getting is because the
$_POST[fieldname] is blank. So for the following line
 if ( ! empty( $_POST['field'] ) )
change it to
 if ( ! empty( $_POST[$field] ) )
Your line : Program is searxchinbg for variable name field
Hello!
I have a problem with a mysql query!
after the query the result variable is "resource id 2"
what is wrong!
Nothing.
Read the manual, even the mysq_fetch_array($result) function.
Mage
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