Re: [PHP] list($bar['CompanyCode'], $CompanyDB) = mysql_fetch_row($sth) fails.

2004-07-29 Thread Matthew Sims
> 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

Re: [PHP] list($bar['CompanyCode'], $CompanyDB) = mysql_fetch_row($sth) fails.

2004-07-28 Thread Jason Davidson
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 variab

Re: [PHP] list($bar['CompanyCode'], $CompanyDB) = mysql_fetch_row($sth) fails.

2004-07-28 Thread Justin Patrin
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) = mysql_f

[PHP] list($bar['CompanyCode'], $CompanyDB) = mysql_fetch_row($sth) fails.

2004-07-28 Thread Daevid Vincent
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'], $CompanyDB) = mysql_fetch_row($sth); But this works: list($foo, $CompanyDB) = SQL_ROW($sth); And of course