Why does this fail when using an array element, but using a variable will
work? Why should PHP care what the variable is I'm trying to store into?
list($bar['CompanyCode'], $CompanyDB) = mysql_fetch_row($sth);
Wouldn't it be easier to simply do:
$result = mysql_fetch_row($sth);
And then
On Wed, 28 Jul 2004 20:47:37 -0700, Daevid Vincent [EMAIL PROTECTED] wrote:
Linux. PHP5.
Why does this fail when using an array element, but using a variable will
work? Why should PHP care what the variable is I'm trying to store into?
list($bar['CompanyCode'], $CompanyDB) =
check the manual for list. It mentions using only numerical array
indices for list. There is a warning on it even i beleive.
Jason
On Wed, 28 Jul 2004 20:47:37 -0700, Daevid Vincent [EMAIL PROTECTED] wrote:
Linux. PHP5.
Why does this fail when using an array element, but using a variable
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