PHP,MySql,mysqli insert_id返回0 Always
内容导读
互联网集市收集整理的这篇技术教程文章主要介绍了PHP,MySql,mysqli insert_id返回0 Always,小编现在分享给大家,供广大互联网技能从业者学习和参考。文章包含3392字,纯文字阅读大概需要5分钟。
内容图文
![PHP,MySql,mysqli insert_id返回0 Always](/upload/InfoBanner/zyjiaocheng/903/772641efff21441b919cd34df808d212.jpg)
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.
内容总结
以上是互联网集市为您收集整理的PHP,MySql,mysqli insert_id返回0 Always全部内容,希望文章能够帮你解决PHP,MySql,mysqli insert_id返回0 Always所遇到的程序开发问题。 如果觉得互联网集市技术教程内容还不错,欢迎将互联网集市网站推荐给程序员好友。
内容备注
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 gblab@vip.qq.com 举报,一经查实,本站将立刻删除。
内容手机端
扫描二维码推送至手机访问。