对讲PHP变量破坏脚本

Question

我有一些变量，我试图插入内部通信参数：

<?php
$user_id = "1234";
$intercom_name = "example";
$intercom_email = "example";
$intercom_name = "example";
$company_id = "example";
$company_name= "example";
?>

    <script>
    window.intercomSettings = {
    app_id: "XYZ123",
    id: "<?php echo $user_id; ?>",
    name: "<?php echo $intercom_name; ?>",
    email: "<?php echo $intercom_email; ?>",
    company: {
        id: "<?php echo $company_id; ?>",
        name: "<?php echo $company_name; ?>",
      }
    };
    </script>

然而，以这种方式使用php变量似乎打破了脚本，有人知道为什么吗？

Answer 1

你的代码看起来不错但是你可以尝试使用json_encode（）将javascript中的初始化对象导入php

<?php
$object=array(
    "app_id"=>"XYZ123",
    "id"=> "1234",
    "name"=> "example",
    "email"=> "example@example.com",
    "company"=> array(
        "id"=> "example",
        "name"=> "example",
    )
);
?>

<script>
window.intercomSettings = <?php echo json_encode($object); ?>;
</script>