[英]How to properly update a MySQL table in PHP
我正在尝试使用PHP更新表中的一行,尽管这不是我第一次,但是update函数无法正常工作。
这是PHP代码:
<?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";
}
?>
顺便说一句,如果您尝试插入,它将起作用:
$q=mysql_query("insert into `phonegap_login` (`reg_date`, `firstname`, `lastname`, `email`, `password`) values ('test', 'test', 'test', 'test', 'test')");
我如何解决它? 插入有效,但更新行无效。 我正在使用reg_id = 50的行进行测试。
您正在混合mysqli_query和mysql_query。 仅使用mysqli_query。 mysql_query已弃用
$con = mysqli_connect($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,面向对象:
$mysqli = new mysqli($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;");
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