如何解决"Parameter must be an array or an object that implements Countable?"

Question

警告：count()：参数必须是一个数组或对象在C:\examplep\htdocs\try\process.php的第30行实现了Countable。

这就是我的代码所说的。它看起来很好，但当我按下编辑时，就会显示这个错误。我不明白。有人能告诉我第30行发生了什么事吗？

这是我的流程.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");
}
?>

Answer 1

事实上，您永远不需要计算任何东西。这一步是多余的。

如果你想一想，你可以先提取数据，然后在条件下使用它。更重要的是，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'];
}

你也不应该在or die上使用这种可怕的做法

Answer 2

$result->num_rows的这种方式怎么样？

if(isset($_GET['edit'])){
$id = $_GET['edit'];
$update = true;
$result = $mysqli->query("SELECT * FROM isalonusers WHERE user_id=$id") or die($mysqli->error);
if(isset($result->num_rows) && $result->num_rows > 0){
$row = $result->fetch_array(MYSQLI_ASSOC);
$username = $row['username'];
$password = $row['password'];
}
}

参见参考文献：http://php.net/manual/en/mysqli.query.php