php - want to see SQL error in report -


i have query

insert table (id) values (5);

the table has record such id. query fails.

my mysqli class extension looks this:

<?php  class my_mysqli extends mysqli {      function __construct($config) {         $this->db = parent::__construct($config['host'], $config['user'], $config['pass'], $config['db']);     }      function exe($sql) {         if ( ! $st = $this->db->prepare($sql)) {             trigger_error($st->error); // 1 isn't triggered         }         if ( ! $st->execute()) {             trigger_error($st->error); // 1 triggered         }         // ..then parse results , close     } }

right after $mysqli->execute() log $mysqli->error , get:

*unknown prepared statement handler (0) given mysqld_stmt_execute*

but see sql error instead:

duplicate entry '5' key 'primary'

there not sense in first block actually. you're doing:

if ( ! $st = $this->db->prepare($sql)) {     trigger_error($st->error); // 1 isn't triggered }

"if there no $st object - call object's method".

next 1 better anyway - there no error method or property in mysqli_stmt class.

function exe($sql) {     if ( ! $st = $this->db->prepare($sql)) {         throw new exception($this->db->error);     }     if ( ! $st->execute()) {         throw new exception($this->db->error);     } }

exceptions better can caught , contain stack trace out of box.

by way, there no point in using prepare() without parameters. so, code have 

function exe($sql) {     if ( ! $this->db->query($sql) ) {         throw new exception($this->db->error);     } }

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