如何正確更新PHP中的MySQL表

Question

我試圖用PHP更新表中的一行，雖然我這不是我第一次，但更新功能不起作用。

這是PHP代碼：

<?php 
header("Access-Control-Allow-Origin: *"); 

$server = "localhost"; 
$username = "username"; 
$password = "dbpass"; 
$database = "dbname"; 

$con = mysql_connect($server, $username, $password) or die ("Could not connect: " . mysql_error()); 

mysql_select_db($database, $con); 

$q=mysqli_query("UPDATE `phonegap_login` SET `firstname` = 'Alhia' WHERE `reg_id` =50;"); 


if($q) 
{ 
echo "success"; 
} 
else 
{ 
echo "failed"; 
} 
?> 

順便說一下，如果你試圖INSERT操作，它會工作：

$q=mysql_query("insert into `phonegap_login` (`reg_date`, `firstname`, `lastname`, `email`, `password`) values ('test', 'test', 'test', 'test', 'test')"); 

如何解決呢？插入正在工作，更新行不是。 我正在使用reg_id = 50進行測試。

Answer 1

你在混合mysqli_query和mysql_query。只能使用mysqli_query。的mysql_query已被棄用

$con = mysqli_connect($server, $username, $password) or die ("Could not connect: " . mysqli_error()); 

mysqli_select_db($database, $con); 

$q=mysqli_query("UPDATE `phonegap_login` SET `firstname` = 'Alhia' WHERE `reg_id` =50;"); 

或者，面向對象：

$mysqli = new mysqli($server, $username, $password, $database); 

if ($mysqli->connect_errno) { 
    echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error; 
} 

$q = $mysqli->query("UPDATE `phonegap_login` SET `firstname` = 'Alhia' WHERE `reg_id` =50;"); 

或者，PDO：

$connectionString = 'mysql:host=' . $server . ';dbname='.$database; 
try { 
    $db = new PDO($connectionString, $user, $pass); 
} catch (PDOException $e) { 
    echo "Failed to connect to Database: (" . $e->getCode() . ") " . $e->getMessage(); 
} 
$q = $db->query("UPDATE `phonegap_login` SET `firstname` = 'Alhia' WHERE `reg_id` =50;");