Re: [PHP] list($bar['CompanyCode'], $CompanyDB) = mysql_fetch_row($sth) fails.
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 work with the $result array? If your DB indexes are listed as CompanyCode and CompanyDB, then use: $result = mysql_fetch_array($sth); Then you have your variable names like you want: $result['CompanyCode'] and $result['CompanyDB'] -- --Matthew Sims --http://killermookie.org -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] list($bar['CompanyCode'], $CompanyDB) = mysql_fetch_row($sth) fails.
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'], $CompanyDB) = mysql_fetch_row($sth); But this works: list($foo, $CompanyDB) = SQL_ROW($sth); And of course I'd have the extra... $foo = $bar['CompanyCode']; Because list() is a language construct, not a function. It assumes what you give it is a normal variable, it doesn't understand arrays. Better IMHO to use $row = mysql_fetch_assoc() and access the array it returns directly. -- DB_DataObject_FormBuilder - The database at your fingertips http://pear.php.net/package/DB_DataObject_FormBuilder paperCrane --Justin Patrin-- -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] list($bar['CompanyCode'], $CompanyDB) = mysql_fetch_row($sth) fails.
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 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 I'd have the extra... $foo = $bar['CompanyCode']; -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
