On 23 June 2013 21:37, Ethan Rosenberg, PhD <[email protected]
> 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'] = $_REQUEST['cnum'];
>> $_SESSION['CustNum'] = $_REQUEST['cnum'];
>>
>> echo "session<br />"; //this has the proper values
>> print_r($_SESSION);
>>
>> $sql10 = "select Balance, Payments, Charges, Date from Charges where
>> Cust_Num = $_SESSION[Cust_Num] order by Date";
>> echo $sql10; //echos the correct query
>> $result10 = mysqli_query($cxn, $sql10);
>> var_dump($result1); // this returns NULL
>>
>
> Against my better judgement, here I go again.
>
> Is this the "actual" code you executed, or is it once again a typeover?
>
> Your 1st error is in these two lines:
>
>> $result10 = mysqli_query($cxn, $sql10);
>>
>> var_dump($result1); // this returns NULL
>>
>
> Yes your dump returns null. And always will.
>
>
> Any further errors might be related to your non-standard syntax for the
> session variable. Per the manual, associative arrays using string indices
> should always use ' ' around them. They work (as mentioned in the manual)
> but are wrong.
> =======
> Jim -
>
>
>
> Is this the "actual" code you executed, or is it once again a typeover?
>
> The actual code
>
>
> Any further errors might be related to your non-standard syntax for the
> session variable. Per the manual, associative arrays using string indices
> should always use ' ' around them. They work (as mentioned in the manual)
> but are wrong.
>
> Newbie is confused.
>
> Please explain.
Try ...
$sql10 = "select Balance, Payments, Charges, Date from Charges where Cust_Num
= {$_SESSION['Cust_Num']} order by Date";
--
Richard Quadling
Twitter : @RQuadling
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