Question

下面是我的货币转换器代码，如果没有cookie，它将设置一个新的cookie和变量，检查是否已请求更改cookie，并检查现有的cookie并为其设置变量。我使用了一个3rd party变量，因为在新的cookie集上，PHP在刷新之前无法看到并因此读取新的cookie。请检查是否可以简化：D

/* No Cookie (Default Currency GBP) */
if(!isset($_COOKIE['currency'])) {
    setcookie(currency, GBP, time() + (86400 * 365), "/");
    $currency = "GBP";
    $currencyid = "1";
}

/* Set Varible Based On Exsiting Cookie and Change Varibles */
if (isset($_COOKIE['currency']) == "GBP" ){
    $currency = "GBP";
    $currencyid = "1";
}
elseif (isset($_COOKIE['currency']) == "USD" ){
    $currency = "USD";
    $currencyid = "2";
}

/* Change Cookie and Change Varibles (Because Check Below Wont Find New Varible of Cookie) */
if (isset($_GET['GBP'])) {
    setcookie(currency, GBP, time() + (86400 * 365), "/"); 
    $currency = "GBP";
    $currencyid = "1";
}

elseif (isset($_GET['USD'])) {
    setcookie(currency, USD, time() + (86400 * 365), "/");
    $currency = "USD";
    $currencyid = "2";
}

Answer 1

您已将两个条件合并为一个表达式并破坏了预期的逻辑...

if (isset($_COOKIE['currency']) == "GBP" ){

应该是：

if (isset($_COOKIE['currency']) && $_COOKIE['currency'] == "GBP" ){

或其他一些单独的版本来维护逻辑。如前所述，您的代码正在询问是否true|false == GBP。

...实际上，根据您的第一个条件块，isset()是毫无意义的。

有人可以检查一下，看看是否有可能做得更小更简单

1 个答案: