On Wed, 10 Sep 2003, Micah Stevens wrote:
> describe TBL;
> explain TBL;
> show columns from TBL;
Yes, they do give the correct information.
However, my question was is there a nice simple SQL command that will
return exactly what mysql_list_fields (now depreciated) so that the
$result can be used with the mysql_field_* functions.
All of the above commands give the exact same data in the same format, all
of which are incompatible with the mysql_field_* functions.
$fld_results = mysql_query("show fields from ".$db.".".$tbl) or
die(mysql_error().'PMA_mysql_list_fields(' . $db . ', ' . $tbl . ')');
$fld_results_cnt = ($fld_results) ? mysql_num_fields($fld_results) : 0;
for ($j = 0; $j < $fld_results_cnt; $j++) {
echo mysql_field_name($fld_results, $j) . "<br/>";
}
Returns:
Field
Type
Null
Key
Default
Extra
instead of what I was expecting.
> On Wed September 10 2003 12:08 pm, Peter Beckman wrote:
> > So PHP is saying mysql_list_tables is depreciated. I replaced it with
> > mysql_query("SHOW TABLES FROM DB") and everything is fine.
> >
> > Then mysql_list_fields didn't work right. So I submitted a bug
> > (http://bugs.php.net/bug.php?id=25460) about it, and they said IT TOO was
> > depreciated, which is fine. But I haven't figured out how to replace
> > mysql_list_fields with a nice simple SQL command, since a bunch of code
> > relies on mysql_list_fields to return a nice $result (ie Resource #37) to
> > use with mysql_field_flags, mysql_field_name, etc.
> >
> > Anyone got any ideas on how to do that without replacing mysql_field_flags
> > et al? Or do I have to rewrite the code entirely to use "SHOW FIELDS FROM
> > DB.TABLE" and rewrite my own _field_ functions?
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