2016-12-21 86 views
0

這是我的代碼:期待聲明如果

<?php 
    if (empty($_COOKIE["count"]) && empty($_COOKIE["date"])) { 
     setcookie("count",1); 
     setcookie("date",date("d.m.y H:i")); 
    } else { 
     var $c=$_COOKIE["count"]; 
    } 
?> 

我得到錯誤:前面沒有var $ C期待聲明。我如何解決我的問題?

+0

http://stackoverflow.com/questions/1206105/what-does-php-keyword-var-do會幫助你 – Blag

回答

2

var關鍵字已棄用,因此用於,該代碼段不適用。

Note: The PHP 4 method of declaring a variable with the var keyword is still supported for compatibility reasons (as a synonym for the public keyword). In PHP 5 before 5.1.3, its usage would generate an E_STRICT warning.

只需正常分配變量:

$c = $_COOKIE['count']; 

否則,它被期待成爲一個類裏面。

0

您需要在if語句之前聲明變量$ c。

<?php 
     $c = 0; 
     if (empty($_COOKIE["count"]) && empty($_COOKIE["date"])) { 
      setcookie("count",1); 
      setcookie("date",date("d.m.y H:i")); 
     } else { 
      $c=$_COOKIE["count"]; 
     } 
    ?>