php – MySqli命令不同步;你现在不能运行这个命令
我已将我的注册脚本从
mysql转换为
mysqli.我工作得很好,但它现在给了我错误
Commands out of sync; you can't run this command now 这是我用来注册用户的功能 function register_user($register_data) { global $myConnection; array_walk($register_data,'array_sanitize'); //Make the array readable and seperate the fields from data $fields = '`' . implode('`,`',array_keys($register_data)) . '`'; $data = ''' . implode('','',$register_data) . '''; //Insert the data and email an activation email to the user mysqli_query($myConnection,"INSERT INTO `members` ($fields) VALUES ($data)") or die(mysqli_error($myConnection)); email($register_data['mem_email'],'Activate your account',"Hello " . $register_data['mem_first_name'] . ",nnThank you for creating an account with H Fencing. Please use the link below to activate your account so we can confirm you identity:nnhttp://blah.blah.co.uk/activate.php?mem_email=" . $register_data['mem_email'] . "&email_code=" . $register_data['email_code'] . "nn - David & Jay "); } 电子邮件使用我的阵列中的正确数据发送正常.但是没有数据插入数据库,我得到上面提到的错误.我之前从未遇到过这个错误.
从这里: 更新 如果为查询创建变量并将变量直接粘贴到MySQL Workbench之类,则可以在执行之前检查语法. <?php function myConnection(){ $myConnection = mysqli_connect('localhost','my_user','my_password','my_db'); return $myConnection; } function register_user($register_data) { array_walk($register_data,'array_sanitize'); //Make the array readable and seperate the fields from data $fields = '`' . implode('`,array_keys($register_data)) . '`'; $data = "'" . implode("','",$register_data) . "'"; //Insert the data and email an activation email to the user $query = "INSERT INTO `members` ($fields) VALUES ($data)"; $myNewConnection = myConnection(); if($result = mysqli_query($myNewConnection,$query)){ email($register_data['mem_email'],nnThank you for creating an account with H Fencing. Please use the link below to activate your account so we can confirm you identity:nnhttp://blah.blah.co.uk/activate.php?mem_email=" . $register_data['mem_email'] . "&email_code=" . $register_data['email_code'] . "nn - David & Jay "); mysqli_free_result($result); return ("Success"); } else { echo $query; die(mysqli_error($myNewConnection)); } } ?> (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |