javascript - 未捕获的语法错误 : Unexpected identifier (alert connected)

Question

这个语句有什么错误

<?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>';

    }

}
?>

enter image description here

问题在于，当我尝试将重复值注册到数据库中的唯一列时，警报中的错误消息不会显示，并且即使在 sql 中也不会显示任何类型的错误，它只是说 Uncaught SyntaxError:意外的标识符，那么我的代码有什么问题导致错误警报不显示？

Answer 1

so what is wrong with my code

这个:

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

(可能还有其他事情，但这似乎是您当前看到的错误发生的地方。)

第一行 JavaScript 代码实际上并不是 JavaScript 代码。这只是文字。因此，对于 JavaScript 而言，Error:(及其后面的所有内容)是一个意外的标识符。

我怀疑您打算将该文本放在脚本 block 之外:

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