Question

我正在开发一个项目，使用PHP - Mysqli，我已经使用HTML表单成功地将值插入到我的表中，现在我想更新值。在我的更新页面中，所有值都是echo out，一旦我单击“提交”按钮，它将保持不变并位于同一页面中。

它说，致命错误：在a上调用成员函数bind_param（）行中没有对象...

我的代码中有任何建议，地点和类型的错误： DB_USER＆GT;表：用户，字段:( user_id，fname，laname，uname，pword，email，mobile，address，rdate）。 user_update.php

<?php ini_set('display_errors', 1);
require 'dbcox/conf.php';
require 'function/security.php';
?>
<?php 
$user_id = $_GET['user_id'];
if(isset($_GET['user_id'])){
    $update = $db->prepare("UPDATE users SET (fname=?, lname=?, uname=?, pword=?, email=?, mobile=?, address=?, WHERE user_id=? ");
    $update->bind_param('sssssss', $fname, $lname, $uname, $pword, $email, $mobile, $address);
    $fname = $_POST['fname'];
    $lname = $_POST['lname'];
    $uname = $_POST['uname'];
    $pword = $_POST['pword'];
    $email = $_POST['email'];
    $mobile = $_POST['mobile'];
    $address = $_POST['address'];
    if($update->execute()){
        echo 'Successfully updated';
    }else{
        'Failed to update!!';
    }
}

Answer 1

您没有向bind_param提供user_id

<?php ini_set('display_errors', 1);
require 'dbcox/conf.php';
require 'function/security.php';
?>
<?php 
$user_id = $_GET['user_id'];
if(isset($_GET['user_id'])){
$update = $db->prepare("UPDATE users SET fname=?, lname=?, uname=?, pword=?, email=?, mobile=?, address=? WHERE user_id=? ");
$update->bind_param('ssssssss', $fname, $lname, $uname, $pword, $email, $mobile, $address, $user_id);
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$uname = $_POST['uname'];
$pword = $_POST['pword'];
$email = $_POST['email'];
$mobile = $_POST['mobile'];
$address = $_POST['address'];
if($update->execute()){
echo 'Successfully updated';
}else{
'Failed to update!!';
}
}

Answer 2

您的SQL语法有错误，如果echo $db->error;返回$db->prepare() false，您会发现错误。您在分配列表前有一个额外的(，在最后一个分配后有一个额外的,。它应该是：

$update = $db->prepare("UPDATE users SET fname=?, lname=?, uname=?, pword=?, email=?, mobile=?, address=? WHERE user_id=? ");

你也忘了绑定user_id param。

$update->bind_param('ssssssss', $fname, $lname, $uname, $pword, $email, $mobile, $address, $user_id);

Answer 3

<?php ini_set('display_errors', 1);
require 'dbcox/conf.php';
require 'function/security.php';
?>
<?php 
$user_id = $_GET['user_id'];
if(isset($_GET['user_id'])){
    $update = $db->prepare("UPDATE users SET (fname=?, lname=?, uname=?, pword=?, email=?, mobile=?, address=? WHERE user_id=? ");
    $update->bind_param('ssssssss', $fname, $lname, $uname, $pword, $email, $mobile, $address, $user_id);
    $fname = $_POST['fname'];
    $lname = $_POST['lname'];
    $uname = $_POST['uname'];
    $pword = $_POST['pword'];
    $email = $_POST['email'];
    $mobile = $_POST['mobile'];
    $address = $_POST['address'];
    $update->execute()
    if($update->errno()){
     echo "FAILURE!!! " . $update->error;
    }
    else 
        echo "Updated {$update->affected_rows} rows";

    $update->close();
}

我的表数据不是必须更新的

3 个答案: