Question

<?php

// 1. create a database connection

$dbhost = "localhost" ;
$dbuser = "widget_cms";
$dbpass = "secretpassword";
$dbname = "widget_corp";

$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);

// test if connection occured
if(mysqli_connect_errno())
{
    die("database connection failed:". mysqli_connect_error() . 
        " " . mysqli_connect_errno() . " "  );
}


?>

<?php 
    //often these are $_POST values through a web form.
    // 2. Perform the database query

    $id = 5;
    $menu_name = "delete me";
    $position = 4;
    $visible = 1;

    $query = "UPDATE subjects SET ";
    $query .= "menu_name = '{ menu_name}', ";
    $query .= "position = {$position}, ";
    $query .= "visible = {$visible}, ";
    $query .= "WHERE id = {$id} ";

    $result =  mysqli_query($connection, $query);

    if ($result)
    {
        //success
        //redirect to somepage.php
        echo "success!";
    }
    else
    {
        //failure
        //$message = "subjet update  failed.";
        die("database query failed. " . mysqli_error($connection));
    }
?>

<DOCTYPE HTML>
    <head>
        <title> databases </title>
    </head>

    <body>



    </body> 

</html>


<?php
    // 5. close the database connection
    mysqli_close($connection);
?>

当我尝试在firefox中运行此页面时，这是弹出的错误。而且我不知道如何解决它。请帮忙！

＆＃34;数据库查询失败。您的SQL语法有错误;检查与您的MariaDB服务器版本对应的手册，以便在＃＆lt; WHERE id = 5＆＃39;附近使用正确的语法。在第1行＆＃34;

TYSM。：）

Answer 1

事情需要改变： -

1

if (mysqli_connect_error()) {
  die('Database connection failed: '  . mysqli_connect_error()); // remove second  mysqli_connect_error()
}

2

 $query .= "menu_name = '{ $menu_name}', "; // $ missed

3

$query .= "visible = {$visible} "; // remove trailing `,`

注意： - 这不是结束，可以在代码中完成很多改进。

mariadb（xampp）

1 个答案: