PHP,MySql,mysqli insert_id返回0 Always
作者:互联网
系统/开发规范;
> Windows 7 Ultimate x64 SP1所有更新
> PHP版本5.5.15非线程安全(x64实验)
> MySQL Server 5.6(x64)
> Php mysqli
我正在执行一个存储过程,该过程将用户名和密码插入到具有AUTO_INCREMENT id INT(11)PK字段的表中.
PROCEDURE `user_account_create`(IN userName VARCHAR(32), IN userPasskey VARCHAR(254))
BEGIN
START TRANSACTION;
INSERT INTO user_account (`name`, passkey) VALUES (userName, userPasskey);
IF (ROW_COUNT() = 1) THEN COMMIT; ELSE ROLLBACK; END IF;
SELECT ROW_COUNT() AS affected_rows; -- Used for PHP mysqli's Connection and Statement affected_rows, and num_rows (Statement) fields.
END
简而言之,我已经在我自己的课堂上包装了mysqli;
namespace DataAccess\Broker {
final class MySqliDb {
private $conn;
public function __construct($dbHost, $dbUser, $dbPass, $dataBase) {
$this->conn = new \mysqli($dbHost, $dbUser, $dbPass, $dataBase);}
public function ExecuteStatement($cmdText, array $paramValue = null) {
$affected = -1;
$stmt = $this->CreateStatement($cmdText, paramValue);
$stmt->execute();
// echo 'insert_id' . $this->conn->insert_id;
$stmt->store_result();
$affected = $stmt->affected_rows;
stmt->close();
return $affected;
}
// ... other functions that utilse CreateStatement below
private function CreateStatement($cmdText, array $paramValue = null) {
$stmt = $this->conn->prepare($cmdText);
if ($paramValue !== null) {
$params = [];
foreach ($paramValue as $p => &$v) {$params[$p] = &$v;}
call_user_func_array([$stmt, 'bind_param'], $params);
}
return $stmt;
}
} // class
} // namespace
在index.php页面上测试它;
use \DataAccess\Broker\MySqliDb as mysqldb;
$db = new mysqldb('127.0.0.1', 'root', '', 'thedb');
$types = 'ss'; $user_name = 'its_me'; $pass_key = 'a-hashed-password';
echo 'Affected Rows: ' . $db->ExecuteStatement('CALL user_account_create(?,?)', [$types, $user_name, $pass_key]);
将产量,受影响的行:1.
插入成功.我也需要来自此命令的insert id,但mysqli连接和语句的insert_id都是0.来自var_dump;
var_dump用于连接:
object(mysqli)#2 (19) {
["affected_rows"] => int(1)
["client_info"] => string(79) "mysqlnd 5.0.11-dev - 20120503 - $Id: xxx$" ["client_version"] => int(50011)
["connect_errno"] => int(0) ["connect_error"] => NULL
["errno"] => int(0) ["error"] => string(0) "" ["error_list"] => array(0) { }
["field_count"] => int(1)
["host_info"] => string(20) "127.0.0.1 via TCP/IP" ["info"] => NULL
["insert_id"] => int(0)
["server_info"] => string(6) "5.6.20" ["server_version"] => int(50620)
["stat"] => NULL
["sqlstate"] => string(5) "HY000"
["protocol_version"] => int(10)
["thread_id"] => int(6)
["warning_count"] => int(0)}
声明的var_dump:
object(mysqli_stmt)#3 (10) {
["affected_rows"] => int(1)
["insert_id"] => int(0)
["num_rows"] => int(1)
["param_count"] => int(6)
["field_count"] => int(1)
["errno"] => int(0) ["error"] => string(0) "" ["error_list"] => array(0) { }
["sqlstate"] => string(5) "00000"
["id"] => int(1)}
有趣的是,有一个字段“id”,即从我表中的字段id中撤回所需的id.任何人都可以看到为什么insert_id返回0.
感谢致敬,
NJC
解决方法:
尝试使用变量编写存储过程,而不是两次调用row_count():
PROCEDURE `user_account_create`(IN userName VARCHAR(32), IN userPasskey VARCHAR(254))
BEGIN
START TRANSACTION;
INSERT INTO user_account (`name`, passkey) VALUES (userName, userPasskey);
IF ((@rc := ROW_COUNT) = 1) THEN COMMIT; ELSE ROLLBACK; END IF;
SELECT @rc AS affected_rows; -- Used for PHP mysqli's Connection and Statement affected_rows, and num_rows (Statement) fields.
END;
我认为第二个调用是指if语句.
标签:php,mysql,mysqli,insert-id 来源: https://codeday.me/bug/20190708/1403939.html