更新脚本不工作且未显示错误

Question

我是一个更新数据库的新手(我跳过了一些步骤，以便更快地做一些事情)我有这段代码，但数据库不更新，我没有得到任何错误。

<?php
require_once '../includes/configuration.php';

if (isset($_POST['enter']))
{
    $_POST['employee_id_passport'] = $employee_id_passport;
    $_POST['first_name'] = $first_name;
    $_POST['surname'] = $surname;
    $_POST['mobile_number'] = $mobile_number;
    $sql_query_update = "UPDATE employee_master 
                             SET first_name = '$first_name', 
                             surname = '$surname', 
                             mobile_number = '$mobile_number' 
                             WHERE employee_id_passport = '$employee_id_passport'";
            $result = mysql_query($sql_query_update, $connection);
    echo "this";

}
else
    {

            //display form with employee details
            if (isset($_GET['idno']))
            {
                $employee_id_passport = $_GET['idno'];

                $sql_get_employee_details_query = "SELECT * FROM employee_master WHERE employee_id_passport = '$employee_id_passport'";
                $result = mysql_query($sql_get_employee_details_query, $connection);
                $row = mysql_fetch_assoc($result);
                echo $row['employee_id_passport'] . "<br /> " . $row['surname'] . "<br /> " . $row['first_name'] . "<br /> " . $row['mobile_number'] . "<br /> ";



                echo "<!DOCTYPE html PUBLIC \"-//W3C//DTD XHTML 1.0 Transitional//EN\"";
                echo "\"http:www.w3.org/TR/xhthml1/DTD/xhtml1-transitional.dtd\">";

                echo "<html xmls=\"http://www.w3.org/1999/xhtml\" xml:lang=\"en\" lang=\"en\">";

                echo "<head>";
                echo    "<title> </title>";

                echo    "<meta http-equiv=\"Content-Type\" content=\"text/html; charset=iso-8859-1\" />";
                echo "</head>";

                echo "<body>";
                echo   "<form name=\"view_employee\" method=\"POST\" action=\"" .  $_SERVER['PHP_SELF'] . " \"   />";
                echo        "Employee ID/Passport: <input type=\"text\" name=\"id_passport\" disabled=\"disabled\" value=\" " .  $row['employee_id_passport'] . "\" /> <br />";
                echo        "First Name: <input type=\"text\" name=\"first_name\" value=\" ".  $row['first_name'] .  "\" /> <br />";
                echo        "Surname: <input type=\"text\" name=\"surname\" value=\" " .  $row['surname'] . "\" /> <br />";
                echo        "Mobile Number: <input type=\"text\" name=\"mobile_number\" value=\" " .   $row['mobile_number'] . "\"/> <br />";
                echo        "<input type=\"submit\" value=\"enter\" name=\"submit\" />";
                echo    "</form>";
                echo "</body>";

                echo "<html>";
            }
    }

?>

Answer 1

看起来您已经将sql查询字符串赋给了一个变量，但实际上并没有运行它。

if (isset($_POST['enter']))
{
    $_POST['employee_id_passport'] = $employee_id_passport;
    $_POST['first_name'] = $first_name;
    $_POST['surname'] = $surname;
    $_POST['mobile_number'] = $mobile_number;
    $sql_query_update = "UPDATE employee_master SET first_name = '$first_name', surname = '$surname', mobile_number = '$mobile_number' WHERE employee_id_passport = '$employee_id_passport'";
    mysql_query( $sql_query_update ); // This line will run the update query
    echo "this";

}

添加'mysql_query( $sql_query_update );‘行应该会为您执行查询。

Answer 2

if (false === mysql_query($sql_query_update)) {
    throw new Exception(mysql_error());
}

您的脚本也容易受到SQL注入的攻击...你知道吗，我不想再说这句话了。阅读SQL注入，并使用PDO、预准备语句和参数绑定。