綁定帕拉姆失敗，空變

Question

$query = "INSERT INTO api_fails (file, code, url, date) VALUES (?, ?, ?, ".time().")"; 

$stmt = $mysqli->stmt_init(); 
if(!$stmt->prepare($query)) { 
    echo("Failed to prepare statement.<br />\n"); 
} else { 
    $file = $_GET['f']; 
    $code = $_GET['c']; 
    $url = $_GET['u']; 
    $stmt->bind_param("sis", $file, $code, $url); 
    if($stmt->execute()) { 
     echo("Inserted.<br />\n"); 
    }else { 
     echo("Didn't insert.<br />\n"); 
    } 
} 

我發現如果有下列變量都沒有設置，插入失敗...

$file = $_GET['f']; 
    $code = $_GET['c']; 
    $url = $_GET['u']; 

如何設置我的插入，這樣，如果變量ISN沒有設置它只是沒有插入到該領域？

Answer 1

如果沒有設置他們只是分配空值的變量：

$file = (isset($_GET['f'])) ? $_GET['f'] : '' ; 
$code = (isset($_GET['c'])) ? $_GET['c'] : '' ; 
$url = (isset($_GET['u'])) ? $_GET['u'] : '' ;