Question

我有一个非常糟糕的学习php日。有人能告诉我为什么这不起作用？我在第23行调用了非对象广告未定义变量的成员函数（）这是代码：

<?php
// connect to the MySQL server
$conn = new mysqli('localhost', 'username', 'password', 'Org_db');

// check connection
 if (mysqli_connect_errno()) {
 exit('Connect failed: '. mysqli_connect_error());
 }
 //for insert user info into users table columns
 $id = '';
 $name = 'Bob Grundy';
  $pass = md5("secure");
  $email = 'bob@acgs.org';
  $reg_date = date("m/d/Y");
 //line 23
 if (!$stmt = $mysqli->prepare("INSERT INTO `users` (`id`, `name`, `pass`, `email`, 
 `reg_date`)     VALUES ( ?,?,?,?,?)")) {
echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
 }
 /* Prepared statement, bind and execute */


if (!$stmt->bind_param("issss", $id, $name, $pass, $email, $reg_date)) {
echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error;
}

if (!$stmt->execute()) {
echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
}
?>

Answer 1

您定义了$conn = new mysqli(..但在使用!$stmt = $mysqli->prepare("..之后，$mysqli变量应更改为$conn，例如：

 if (!$stmt = $conn->prepare("INSERT INTO `users` (`id`, `name`, `pass`, `email`, 
 `reg_date`)     VALUES ( ?,?,?,?,?)")) {
echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
 }

在非对象和变量undefined上调用成员函数prepare（）

1 个答案: