Javascript certificate.php:5未捕获的引用错误：未定义TestsDASD

我的php代码

<?php
if (isset($_POST['submit'])) {
    $name = $_POST['name'];
    $address = $_POST['address'];
    $fname = $_POST['fname'];
    $mname = $_POST['mname'];
    $all_info = $name . "<br>" . $address . "<br>" . $fname . "<br>" . $fname;
    // echo $all_info;
}

// echo $name,$email,$message;

?>

<script>

const name = <?php echo $name ?>;
console.log(name);
alert(name);


</script>

<script>

const name = TEstasdasd;
console.log(name);
alert(name);


</script>

我的js代码

<?php
if (isset($_POST['submit'])) {
    $name = $_POST['name'];
    $address = $_POST['address'];
    $fname = $_POST['fname'];
    $mname = $_POST['mname'];
    $all_info = $name . "<br>" . $address . "<br>" . $fname . "<br>" . $fname;
    // echo $all_info;
}

// echo $name,$email,$message;

?>

<script>

const name = <?php echo $name ?>;
console.log(name);
alert(name);


</script>

<script>

const name = TEstasdasd;
console.log(name);
alert(name);


</script>

常量名称=；
console.log（名称）；
警报（名称）；

错误

certificate.php:5未捕获的引用错误：未定义TestsDASD

eror源代码

<?php
if (isset($_POST['submit'])) {
    $name = $_POST['name'];
    $address = $_POST['address'];
    $fname = $_POST['fname'];
    $mname = $_POST['mname'];
    $all_info = $name . "<br>" . $address . "<br>" . $fname . "<br>" . $fname;
    // echo $all_info;
}

// echo $name,$email,$message;

?>

<script>

const name = <?php echo $name ?>;
console.log(name);
alert(name);


</script>

<script>

const name = TEstasdasd;
console.log(name);
alert(name);


</script>

const name=TEstasdasd；
console.log（名称）；
警报（名称）；

我无法修复我的错误有人能帮我吗

---

const name=TEstasdasd在这一行中，您遇到了错误。因为你没有在任何地方定义它，试着调用那个常量

尝试如下

const name = 'TEstasdasd';

尝试将名称用引号括起来，即
const name='TEstasdasd'