#43568 [Opn]: mysqli statement stmt->bind_param doesn't work when called by call_user_func

ID: 43568 Updated by: uw@php.net Reported By: gabriel at oxeva dot fr Status: Open Bug Type: MySQLi related Operating System: Linux 2.6 PHP Version: 5.3CVS-2007-12-11 (snap) New Comment: Hold on, a PHP internals expert should have a look. sort() is another example of a function which uses pass by reference. The manual says about call_user_func_array(): "Note: Referenced variables in param_arr are passed to the function by a reference, others are passed by a value. In other words, it does not depend on the function signature whether the parameter is passed by a value or by a reference. " From a discussion with Johannes: sapi/cli/php -r '$a = array(2, 1); var_dump(call_user_func_array("sort", array($a)));' PHP 5.2 -> bool(true) PHP 5.3 -> NULL Looks like its not a mysqli issue but rather a general PHP issue. Previous Comments: ------------------------------------------------------------------------ [2008-01-04 13:49:02] uw@php.net Forgot to say: IMHO this should be reported over at Zend Framework but not here. I've tested 5_2, 5_3, 6_0 and I see no differences in handling. ------------------------------------------------------------------------ [2008-01-04 13:42:14] uw@php.net IMHO this is not a bug. However, I leave it to Johannes/Andrey to review and close the issue. bind_param() is about binding variables not constant values: "stmt->bind_param() — Binds variables to a prepared statement as parameters". The signature of the function tells you that you have to pass input parameter variables by reference: bool mysqli_stmt_bind_param ( mysqli_stmt $stmt , string $types , mixed &$var1 [, mixed &$... ] ) Using references to "real variables" instead of "constant values" does work fine. Note the small difference between references to variables in PHP and memory pointer in C. In PHP your references have to point to existing variables. You can't make them point to arbitrary places in memory like in C. If a PHP function asks you to provide a reference to an existing variable, you must not try to pass a (constant) value to it. But this is exactly what you do in your call_user_func_array() example. Use the function like this: $int = 1; mysqli_stmt_bind_param($stmt, 'i', $int); $stmt->bind_param('i', $int); And, if you want to use call_user_func_array(), the syntax is: $int = 1; $format = 'i'; $params = array(0 => $format, 1 => &$int); call_user_func_array('mysqli_stmt_bind_param', $params); call_user_func_array(array($stmt, 'bind_param'), $params); // - or - $int = 1; $params = array(0 => 'i', 1 => &$int); call_user_func_array('mysqli_stmt_bind_param', $params); call_user_func_array(array($stmt, 'bind_param'), $params); // - or - $int = 1; call_user_func_array('mysqli_stmt_bind_param', array('i', &$int)); Neither of the following does work with PHP 5_3. Recall that the function is about binding variables not binding (constant) values, recall that you have to pass a reference to a variable: WRONG: mysqli_stmt_bind_param($stmt, 'i', 1); WRONG: call_user_func_array('mysqli_stmt_bind_param', array('i', 1)); ------------------------------------------------------------------------ [2007-12-11 17:11:46] gabriel at oxeva dot fr Description: ------------ The Mysqli method stmt->bind_param() doesn't work when called by call_user_func_array. Too bad Zend Framework is using this method with the Mysqli driver. In the following example, call_user_func_array always return NULL, but directly calling $stmt->bind_param() works and return the correct bool(true) value. Note: this works in PHP 5.2 Note: The same error occurs using either libmysqlclient or mysqlnd. Final note: ATM the PDO_mysql driver works. But leaving a bug like this in the mysqli driver is not an option :-) Reproduce code: --------------- $my = new mysqli('localhost', 'test', 'test', 'testdb'); $stmt = $my->prepare('INSERT INTO car (brand, brand_id, car_type) VALUES (?, ?, ?)'); $param = array('sss', 'GMC', 1, 'SUV'); $return = call_user_func_array(array($stmt, 'bind_param'), $param); var_dump($return); $stmt->execute(); echo $stmt->errno, ':', $stmt->error; Expected result: ---------------- var_dump($return) prints bool(true) No error returned in $stmt->errno / $stmt->error Actual result: -------------- var_dump($return) prints NULL Statement error is returned: 2031:No data supplied for parameters in prepared statement ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=43568&edit=1