Bug #10412 Updated: field name in mysql_lsit_fields

ID: 10412 Updated by: torben Reported By: dkokenge@netscape.net Old-Status: Open Status: Closed Bug Type: Scripting Engine problem PHP Version: 4.0.4pl1 Assigned To: Comments: Please read the following page in the manual, especially the last paragraph: http://www.php.net/manual/en/language.variables.variable.php The problem is that $$fname[] is ambiguous and you need to tell PHP what it means. The page above explains in more detail. Previous Comments: --------------------------------------------------------------------------- [2001-04-19 18:59:30] dkokenge@netscape.net When using "mysql_list_fields" for a file you can extract the field names of the schema. For example: $fname. Then turn it into the variable for the schema name as $$fname. $$fname = 'something'; // **** works $$fname[0] = 'something'; // **** don't work --------------------------------------------------- Example --------------------------------------------------- $schema = mysql_list_fields("wine","wpo_dtl"); $nf = mysql_num_fields($schema); // get num of fields $xx = mysql_fetch_array($schema); while ($i < $nf) { $fname = mysql_fieldname($schema,$i); //***** this works ***** $$fname = 'something'; // ******* this don't $x = 0; $$fname[$x] = 'something'; } What am I doing wrong??? --------------------------------------------------------------------------- ATTENTION! Do NOT reply to this email! To reply, use the web interface found at http://bugs.php.net/?id=10412&edit=2