Question

本声明中的错误是什么

<?php

session_start();

$host = "localhost";
$uname = "root";
$pass = "";
$database = "mcndb";
$cons = mysqli_connect($host, $uname, $pass, $database);
mysqli_select_db($cons, $database);


if (!$cons) {
    die("Connection failed: " . mysqli_connect_error());
}

if (isset($_POST['username'])) {
    date_default_timezone_set("Asia/Bangkok");
    $today = date("Y/m/d  h:i:s:A");
    $username = mysqli_real_escape_string($cons, $_POST['username']);
    $password = mysqli_real_escape_string($cons, $_POST['password']);
    $firstname = mysqli_real_escape_string($cons, $_POST['firstname']);
    $middlename = mysqli_real_escape_string($cons, $_POST['middlename']);
    $lastname = mysqli_real_escape_string($cons, $_POST['lastname']);
    //  $age=mysqli_real_escape_string($cons,$_POST['age']);
    $gender = mysqli_real_escape_string($cons, $_POST['gender']);
    $email = mysqli_real_escape_string($cons, $_POST['usremail']);
    $companyname = mysqli_real_escape_string($cons, $_POST['companyname']);
    $position = mysqli_real_escape_string($cons, $_POST['position']);

    $contactnumber = mysqli_real_escape_string($cons, $_POST['contactnumber']);
    $addresss = mysqli_real_escape_string($cons, $_POST['address']);

    $sql = "INSERT INTO tbltry (username,password,email,firstname,middlename,lastname,gender, company_name,position,contact_number,address_of_company,dateofregister) 
            VALUES ('$username', '$password', '$email','$firstname', '$middlename','$lastname','$gender', '$companyname','$position','$contactnumber','$addresss','$today')";


    if ($cons->query($sql) == TRUE) {
        echo '<script>';
        echo 'alert("Successfully created an account")';
        echo '</script>';
    } else {
        echo '<script>';
        echo "Error: " . $sql . "<br>" . $cons->error;
        echo 'alert("Account already exist")';
        echo '</script>';

    }

}
?>

google.maps.Rectangle

这个问题是，当我尝试在数据库中向唯一列注册一个dulicate值时，警报中的错误消息不显示，即使在sql中它也没有显示任何类型的错误它只是说Uncaught SyntaxError：意外的标识符，所以我的代码错误警告不显示错误？

Answer 1

所以我的代码有什么问题

此：

<script>
  Error: some text<br>some more text
  alert("Account already exist")
</script>

（可能还有其他一些事情，但这似乎是你目前看到的错误发生的地方。）

第一行JavaScript代码实际上不是JavaScript代码。这只是文字。所以Error:（以及它之后的所有内容）就JavaScript而言是一个意外的标识符。

我怀疑你打算把那个文本放在脚本块之外：

echo "Error: " . $sql . "<br>" . $cons->error;
echo '<script>';
echo 'alert("Account already exist")';
echo '</script>';

未捕获的SyntaxError：意外的标识符（已连接警报）

1 个答案: