JSON字符串返回，包含PHP的Javascript和HTML

Question

我有问题要创建我的RESTFul API的JSON响应。以下javascript代码需要作为JSON响应发送：

<script language="JavaScript">var zflag_nid="3372"; var zflag_cid="2/1"; var zflag_sid="1"; var zflag_width="300"; var zflag_height="250"; var zflag_msize="1366x768:9,728x90:19,310x352:19";        </script><script language="JavaScript" src="http://xp2.xyz.com/jsc/xp2/fo.js"></script>

我的代码片段是：

$str = '<script language="JavaScript">var zflag_nid="3372"; var zflag_cid="2/1"; var zflag_sid="1"; var zflag_width="300"; var zflag_height="250"; var zflag_msize="1366x768:9,728x90:19,310x352:19";                 </script><script language="JavaScript" src="http://xp2.xyz.com/jsc/xp2/fo.js"></script>';
        //echo $str;
        echo json_encode(array("data"=>$str,"msg"=>"success","status"=>200));

我通过上述方式得到错误的JSON输出响应。 请帮助我获得正确的JSON响应。

我的回答是：

{"data":"
<script language=\"JavaScript\">var zflag_nid=\"3372\"; var zflag_cid=\"2\/1\"; var zflag_sid=\"1\"; var zflag_width=\"300\"; var zflag_height=\"250\"; var zflag_msize=\"1366x768:9,728x90:19,310x352:19\";<\/script>
    <script language=\"JavaScript\" src=\"http:\/\/xp2.xyz.com\/jsc\/xp2\/fo.js\"><\/script>","msg":"success","status":200}