警告：count()：参数必须是数组或对象
在 C:\xampp\htdocs\try\process.php 第 30 行实现 Countable。
这就是我的代码所说的。看起来很好，但是当我按编辑时，会显示此错误。我不明白。有人能指出第 30 行发生了什么吗？
这是我的 process.php
<?php
require("1password.php");
$id = 0;
$update = false;
$username='';
$password='';
if (!session_id()) { session_start(); }
$mysqli = new mysqli("localhost","root","","id7508046_isalon") or die(mysqli_error($mysqli));
if(isset($_POST['save'])){
$username = $_POST['username'];
$password = $_POST['password'];
$passwordHash = password_hash($password, PASSWORD_DEFAULT);
$mysqli->query("INSERT INTO isalonusers (username, password) values ('$username', '$passwordHash')") or die($mysqli->error);
$_SESSION['message'] = "New account saved!";
$_SESSION['msg_type'] = "success";
header("location: userlist.php");
}
if(isset($_GET['delete'])){
$id = $_GET['delete'];
$mysqli->query("DELETE FROM isalonusers WHERE user_id=$id") or die($mysqli->error());
$_SESSION['message'] = "User Account Deleted!";
$_SESSION['msg_type'] = "danger";
header("location: userlist.php");
}
if(isset($_GET['edit'])){
$id = $_GET['edit'];
$update = true;
$result = $mysqli->query("SELECT * FROM isalonusers WHERE user_id=$id") or die($mysqli->error());
if(count($result)==1){
$row = $result->fetch_array();
$username = $row['username'];
$password = $row['password'];
}
}
if(isset($_POST['update'])){
$id = $_POST['id'];
$username = $_POST['username'];
$password = $_POST['password'];
$passwordHash = password_hash($password, PASSWORD_DEFAULT);
$mysqli->query("UPDATE isalonusers SET username ='$username', password='$passwordHash' WHERE user_id=$id") or die($mysqli->error());
$_SESSION['message'] = "User Account has been updated!";
$_SESSION['msg_type'] = "warning";
header("location: userlist.php");
}
?>
事实上，你永远不需要计算任何东西。这一步只是多余的。
如果您稍微考虑一下，您可以简单地先获取数据，然后在条件中使用它。更重要的是，您必须使用参数化查询。所以代码应该是
$stmt = $mysqli->prepare("SELECT * FROM isalonusers WHERE user_id=?");
$stmt->bind_param("i", $id);
$stmt->execute();
$result = $stmt->get_result();
// here goes your problem with "count"
$row = $result->fetch_array(MYSQLI_ASSOC)
if($row) {
$username = $row['username'];
$password = $row['password'];
}
你也不应该使用它可怕的练习or die
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