如何創建一個php函數來更新提交數據庫？

Question

我創建了這個函數來在點擊提交按鈕時更新數據庫。但它似乎並不奏效。該腳本旨在通過id使用用戶的登錄細節從數據庫中繪製用戶的值，並填充表單。該功能用於幫助用戶在表單文本字段中更改用戶的詳細信息時進行更新。

請幫我調試它。

libraries.php 
function db(){ //handles database connection 

//connect to the database server or die and spit out connection error 
$conn = mysql_connect('localhost','root', '') or die("Cannot connect to the database server now". mysql_error()); 
//select database table or die and spit out database selection error 
mysql_select_db('newbishop',$conn) or die("Error in selecting database now ".mysql_errno()); 
    return $conn; 
} 


personalsettings.php 
<?php 
include_once('libraries.php'); // contains the database function 
session_checker(); 

db(); 
$categoryid = $_SESSION['id']; 
$select = "SELECT * FROM users WHERE categoryid ='$categoryid' LIMIT 1"; 
$row1 = dbprocess ($select); 
$rows = mysql_fetch_assoc($row1); 
$pname1 = $rows['pname']; 
$email1 = $rows['email']; 
$user1 = $rows['user']; 
$pass1 = $rows['pass']; 
$salt1 = $rows['salt']; 
$phone1 = $rows['phone']; 
$accesslevel = $rows['accesslevel']; 
$position = $rows['position']; 

function update(){ 
    db(); // database function 
    $pname = $_POST['pname']; 
    $categoryid = $_POST['categoryid']; 
    $email = $_POST['email']; 
    $phone = $_POST['phone']; 
    $user = $_POST['users']; 
    $pass = $_POST['pass']; 

    function createSalt(){ 
    $string = md5(uniqid(rand(), true)); 
    return substr($string, 0, 3); 
    }; 

    $salt = createSalt(); 
    $hash = hash('sha256', $salt . $pass); 

    $sql = "UPDATE users SET user=?,pass=?,salt=?,pname=?,email=?,phone=? WHERE categoryid=?"; 
    $q = $conn->prepare($sql); 
    $q->execute(array($user,$hash,$salt,$pname,$email,$phone,$categoryid)); 
} 

?> 

形式 編輯個人設置

<input name="users" type="text" id="users" class="users" autocomplete="off" value="<?php echo $user1; ?>" /> 

<input type="text" autocomplete="off" name="pass" id="pass" placeholder="Create password" class="passwd" value="<?php echo $pass; ?>"/> 
    <input type="hidden" name="salt" id="salt" value="<?php echo $salt1; ?>"/> 


    <input name="pname" type="text" id="lname" placeholder="Name of Group" class="input-block-level" value="<?php echo $pname1; ?>"/> 

    <input type="hidden" id="categoryselect" name="categoryselect"/> 
    <input name="categoryid" type="text" id="resultselect" readonly class="input-block-level" value="<?php echo $_SESSION['id']; ?>"/> 


    <input type="text" name="email" id="email" placeholder="Email Address" class="input-block-level" value="<?php echo $email1; ?>"/> 


    <input type="text" name="phone" id="phone" placeholder="Enter Phone Number" class="input-block-level" value="<?php echo $phone1; ?>"/> 


    <input type="text" name="accesslevel" id="accesslevel" class="input-block-level" value="<?php echo $accesslevel; ?>" readonly/> 



    <input type="text" name="position" id="position" class="input-block-level" value="<?php echo $position; ?>" readonly/> 


    <button type="submit" class="btn btn-small btn-primary" name="register" id="register" value="Register" onclick="update()">Submit</button> 
    </form> 

Answer 1

db(); // database function 

看來，如果你已經忘記了將函數的返回值賦值給$conn。

正確的應該是：

$conn = db(); 

Answer 2

嘗試?>之前在該行的底部添加以下代碼：

if isset($_POST['user']) { update(); } 

Answer 3

這只是一個粗略的例子，我沒有測試它。它在這裏，就是爲了讓你可以得到照片。

功能

function update_db($data, $update){ 
    $conn = new PDO('mysql:host=locahost; dbname=xxxx', 'xxx', 'xxx'); 
    $stmt = $conn->prepare("UPDATE table, SET something=? WHERE id = ? "); 
    $stmt->execute(array($data, $update)); 

    if($stmt->rowCount() > 0){ 
    return 'Updated';} 
    else { 
    return 'Something is wrong'; 
    } 
} 

調用它

if(isset($_GET['data'])) { 
$data = $_GET['data']; 
$update = 'text'; 


echo update_db($data, $update) 

} 

Answer 4

感謝所有。我已經解決了這個問題。數據庫的問題在於被調用的函數無法訪問數據庫。