2015-11-19 56 views
-1

我想在PHP上做一個代碼,當時間從上午6點到下午5點59分將是早晨,並將顯示一個太陽,而下午​​6點到5:59我會顯示月亮。如果其他語句基於時間

到目前爲止,這是我的代碼:

它總是顯示月亮圖像。

<?php 
date_default_timezone_set("Asia/Manila"); 
    echo date("l, "); 
    echo date("d F Y "); 


    echo date("G:i:sa "); 

    if (date('G'>17) && ('f' >=59) && ('sa' >=59)) 
    { 
     echo $timeImage = "<img src=sun.png>"; 

    } 

    else if(date('G'<6) && ('f' <= 01) && ('sa' <= 01)) 
    { 
     echo $timeImage = "<img src=moon.png>"; 

    } 


?> 
+0

' 'F'> = 59'使沒有太大意義。在數字上下文中,它將歸結爲:'0> = 59' – Rizier123

+0

您的'if'語句沒有意義,條件完全混亂。弗朗茲的方法是更好,並工作:) –

+0

我的歉意,我是新來的PHP,我試圖做到這一點的基礎上我知道:)那麼,感謝您的幫助:)非常感謝 – NoobCoder

回答

2
<?php 
$seconds = time() - strtotime("today"); //seconds since midnight for now 
$sec_a = 6*60*60; //seconds since midnight for 6am 
$sec_b = 18*60*60; //seconds since midnight for 6pm 

if($seconds >= $sec_a and $sec_b >= $seconds) //current seconds are larger than 6am or equal but smaller than 6pm 
{ 
    echo "daytime"; 
} else { 
    echo "nighttime"; 
}