Question

我写下了以下一段代码：

<?php

$username = $_POST['user'];
$password = $_POST['pass'];

$db = new PDO ('mysql:host=localhost;dbname=ozdatabase;charset=utf8', 'root', '');
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$db->setAttribute(PDO::ATTR_ERRMODE, 'ERRMODE_EXCEPTION');

$stmt = $db->prepare("SELECT id, users FROM ozusers WHERE username=? AND password=?");
$stmt->execute(array($username, $password));
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);

$id = $rows['id'];
$user = $rows['users'];

if ($id) {
    print "Logged";
}

else {
    print "not good";
}

?>

这是HTML表单：

<form id='login' action='login.php' method='post' accept-charset='UTF-8'>
<fieldset >
<LEGEND>COMMUNICATION</LEGEND>
<input type='hidden' name='submitted' id='submitted' value='1' />

<label for='username' >UserName*:</label>
<input type='text' name='user' id='username' maxlength="50" />

<label for='password' >Password*:</label>
<input type='password' name='pass' id='password' maxlength="50">

<input type='submit' name='Submit' value='Submit' />

</fieldset>
</form>

我在尝试登录页面时收到错误并写了：＆＃34;致命错误：在第15行＆＃34;

上的非对象上调用成员函数execute（）

为什么会这样？我遵循了最佳实践指南，它显示使用＆＃34; execute（）＆＃34;功能完全像...

由于

Answer 1

ERRMODE_EXCEPTION是不断引用它的引用

$db->setAttribute(PDO::ATTR_ERRMODE, ERRMODE_EXCEPTION);//Your code fails at this line

Answer 2

<?php

    //error_reporting(0);

    $username = $_POST['user'];
    $password = $_POST['pass'];

    // Connecting, selecting database
    $db = new PDO ('mysql:dbhost=localhost;dbname=ozdatabase;charset=utf8', 'root', '');
    $db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
    $db->setAttribute(PDO::ATTR_ERRMODE, ERRMODE_EXCEPTION);

    //first
    $stmt = $db->prepare("SELECT id, users FROM ozusers WHERE username = :username AND password = :password");
    $stmt->execute(array(':username'=>$username, ':password' => $password));
    $rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
    foreach($rows AS $r) {
        $id = $r['id'];
        $user = $r['users'];
    }

    if ($id) {
        print "Logged";
    }

    else {
        print "not good";
    }

?>

Answer 3

在try catch中包装数据库init以捕获任何连接失败。

try {
    $db = new PDO($dsn, $user, $password);
} catch (PDOException $e) {
    echo 'Connection failed: ' . $e->getMessage();
}

目前db-＆gt; prepare（）失败，返回false，因此不允许你在非对象上调用execute。

PHP错误 - 在非对象上调用成员函数execute（）

3 个答案: