回声问题

Question

目前我在下面显示了此代码，当用户所在国家/地区=美国但​​ID无法正常运行时，可能会显示广告。

$ip = $_SERVER['REMOTE_ADDR'];
$details = json_decode(file_get_contents("http://ipinfo.io/{$ip}/json"));
$location = $details->country;
$details->country;

    if($location == 'CA') {

    echo '<script type="text/javascript">';
    echo "<!-- window['pubs_ad_padid']"; echo '= "1005";';
    echo "window['pubs_ad_subid'] = ''; //Insert value to pass through your sub id //-->";
    echo "var se = document.createElement('script');";
    echo 'se.type = "text/javascript";';
    echo "se.src = (document.location.protocol ==="; echo ' "https:" ? "https:" : "http:") + "//d.example.com/js/pub.js";';
    echo "document.getElementsByTagName('head')[0].appendChild(se);";
    echo "</script>";
}

这里是没有包含所有回声爵士乐的普通代码。

<script type="text/javascript">
<!--
window['pubs_ad_padid'] = "1005";
window['pubs_ad_subid'] = ''; //Insert value to pass through your sub id
//-->
var se = document.createElement('script');
se.type = "text/javascript";
se.src = (document.location.protocol === "https:" ? "https:" : "http:") + "//d.example.com/js/pub.js";
document.getElementsByTagName('head')[0].appendChild(se);
</script>

Answer 1

您是否尝试过这样：

<?php
if($location == 'CA') {
?>
<script type="text/javascript">
<!--
window['pubs_ad_padid'] = "1005";
window['pubs_ad_subid'] = ''; //Insert value to pass through your sub id
//-->
var se = document.createElement('script');
se.type = "text/javascript";
se.src = (document.location.protocol === "https:" ? "https:" : "http:") + "//d.example.com/js/pub.js";
document.getElementsByTagName('head')[0].appendChild(se);
</script>
<?>}</?>

这是它的基础知识。如果符合条件，那将只会回显脚本。如果需要，它还允许您更轻松地在其中添加另一个PHP回显。而且你不必为所有这些引用弄乱。这可能是你的问题所在。