为什么这个php脚本是错误的,我';我收到很多错误信息
为了便于学习,我编写了这个PHP脚本,以便将数据保存到mysql数据库中 我的最后一个任务是通过我正在开发的Xcode应用程序使用Alamofire发出post请求 但作为试用,我正在使用“邮递员应用程序”测试post请求 下面是我的php脚本:为什么这个php脚本是错误的,我';我收到很多错误信息,php,mysql,xcode,post,alamofire,Php,Mysql,Xcode,Post,Alamofire,为了便于学习,我编写了这个PHP脚本,以便将数据保存到mysql数据库中 我的最后一个任务是通过我正在开发的Xcode应用程序使用Alamofire发出post请求 但作为试用,我正在使用“邮递员应用程序”测试post请求 下面是我的php脚本: <?php $host = "127.0.0.1"; $user = "root"; $password = ""; $database = "login_db";
<?php
$host = "127.0.0.1";
$user = "root";
$password = "";
$database = "login_db";
$staff = $_POST['staff_ID'];
$password = $_POST['password'];
$email = $_POST['email'];
$cpt = $_POST['isCPT'];
$name = $_POST['Name'];
$connessione = new mysqli($host,$user,$password,$database);
if ($connessione === false) {
die("errore connessione db " .$connessione->connection_error);
}
$query = "INSERT INTO users (staff_ID , password, email, isCPT, Name ) VALUES(?, ?, ?, ?, ?)";
$stmt = $connessione->prepare($query); //Prepare
$stmt->bind_param("sssss", $staff,$password,$email,$cpt,$name); //Bind
$stmt->execute();//Execute
if($stmt->affected_rows === 0) exit('Nothing was inserted.'); //Check to see if it worked
$stmt->close();
$connessione -> close();
?>
错误:
<br />
<b>Warning</b>: Undefined array key "staff_ID" in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>8</b><br />
<br />
<b>Warning</b>: Undefined array key "password" in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>9</b><br />
<br />
<b>Warning</b>: Undefined array key "email" in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>10</b><br />
<br />
<b>Warning</b>: Undefined array key "isCPT" in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>11</b><br />
<br />
<b>Warning</b>: Undefined array key "Name" in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>12</b><br />
<br />
<b>Warning</b>: Undefined variable $mysqli in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>35</b><br />
<br />
<b>Fatal error</b>: Uncaught Error: Call to a member function prepare() on null in E:\XAMPP\htdocs\test\nxSignUp.php:35
Stack trace:
#0 {main}
thrown in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>35</b><br />
警告:第8行E:\XAMPP\htdocs\test\nxSignUp.php中未定义的数组键“staff\u ID”
警告:第9行E:\XAMPP\htdocs\test\nxSignUp.php中未定义的数组键“password”
警告:第10行E:\XAMPP\htdocs\test\nxSignUp.php中未定义的数组键“email”
警告:第11行E:\XAMPP\htdocs\test\nxSignUp.php中未定义的数组键“isCPT”
警告:第12行E:\XAMPP\htdocs\test\nxSignUp.php中未定义的数组键“Name”
警告:第35行E:\XAMPP\htdocs\test\nxSignUp.php中未定义变量$mysqli
致命错误:未捕获错误:在E:\XAMPP\htdocs\test\nxSignUp.php:35中调用null上的成员函数prepare()
堆栈跟踪:
#0{main}
在第35行的E:\XAMPP\htdocs\test\nxSignUp.php中抛出
您是通过URL发送这些参数,而不是通过请求正文发送。您没有$mysqli您有$connessionEthaks,我更正了$connessione错误,但仍有许多错误显示:警告:未定义的数组键“staff\u ID”在第8行的E:\XAMPP\htdocs\test\nxSignUp.php中,看起来post请求没有传递任何数据。如果您刚刚开始学习php,那么您应该学习PDO而不是mysqli。PDO更简单,更适合初学者。从这里开始&
<br />
<b>Warning</b>: Undefined array key "staff_ID" in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>8</b><br />
<br />
<b>Warning</b>: Undefined array key "password" in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>9</b><br />
<br />
<b>Warning</b>: Undefined array key "email" in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>10</b><br />
<br />
<b>Warning</b>: Undefined array key "isCPT" in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>11</b><br />
<br />
<b>Warning</b>: Undefined array key "Name" in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>12</b><br />
<br />
<b>Warning</b>: Undefined variable $mysqli in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>35</b><br />
<br />
<b>Fatal error</b>: Uncaught Error: Call to a member function prepare() on null in E:\XAMPP\htdocs\test\nxSignUp.php:35
Stack trace:
#0 {main}
thrown in <b>E:\XAMPP\htdocs\test\nxSignUp.php</b> on line <b>35</b><br />