用php提取IP作为JAVASCRIPT变量用提示"缺少';'"解决方案

用php提取IP作为JAVASCRIPT变量用提示"缺少';'"
<script type="text/javascript" language="javascript">  
<?php
function GetIP(){
 if(!empty($_SERVER["HTTP_CLIENT_IP"])){
  $cip = $_SERVER["HTTP_CLIENT_IP"];
 }
 elseif(!empty($_SERVER["HTTP_X_FORWARDED_FOR"])){
  $cip = $_SERVER["HTTP_X_FORWARDED_FOR"];
 }
 elseif(!empty($_SERVER["REMOTE_ADDR"])){
  $cip = $_SERVER["REMOTE_ADDR"];
 }
 else{
  $cip = "无法获取！";
 }
 return $cip;
}
?>  
var frip=<?php echo GetIP()?>;
</script>
如果不用
var frip=<?php echo GetIP()?>;
是正常的
否则提示
function GetIP(){
 if(!empty($_SERVER["HTTP_CLIENT_IP"])){
缺少“；”

[解决办法]
var frip="<?php echo GetIP()?>";