对讲PHP变量打破脚本

Question

我有一些变量我试图插入对讲参数：以这种方式

<?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> 

但是使用PHP变量似乎打破剧本，没有人知道为什么吗？

Answer 1

您的代码看起来不错，但或者你也可以用json_encode（）JavaScript中初始化对象尝试到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>