PHP json\u编码为javascript json解析错误[重复]

我在PHP中有一个变量,我需要它在JavaScript代码中的值。如何将变量从PHP转换为JavaScript?

我有如下代码:

<?php
$val = $myService->getValue(); // Makes an API and database call

在同一页上,我有JavaScript代码,它需要 $val 要作为参数传递的变量:

<script>
    myPlugin.start($val); // I tried this, but it didn't work
    <?php myPlugin.start($val); ?> // This didn't work either
    myPlugin.start(<?=$val?>); // This works sometimes, but sometimes it fails
</script>