PDO選擇語句

Question

希望有人可以給我一些幫助...我使用PDO從數據庫中獲取一些數據，但每次腳本運行時都會返回並且沒有錯誤，但不會顯示所需的數據，我知道我正在尋找的數據也在那裏..任何幫助將不勝感激，這是我目前的，謝謝。

$db = new PDO('sqlite:C:\\xampp\\htdocs\\Utils\\PDF_Utils\\PDF2Word\\details.sqlite'); 

    echo "<table border=1>"; 
    echo "<tr><td>FileID</td> 
       <td>File Name</td> 
       <td>Email From</td> 
       <td>CC</td> <td>Subject</td> 
       <td>File Size</td></tr>"; 


       $contents = $db->prepare("SELECT * FROM details WHERE fileName = 
         '$yourFileName'"); 
       $contents->execute(); 

        foreach($contents as $row) { 
         echo "<tr><td>" . $row['FileID'] . "</td>"; 
         echo "<td>" . $row['fileName'] . "</td>"; 
         echo "<td>" . $row['emailFrom'] . "</td>"; 
         echo "<td>" . $row['CC'] . "</td>"; 
         echo "<td>" . $row['subject'] . "</td>"; 
         echo "<td>" . $row['fileSize'] . "</td></tr>"; 
        } 
        echo "</table>"; 

Answer 1

準備語句時使用的佔位符，然後執行查詢傳遞參數，如：

while($row = $contents->fetch(PDO::FETCH_ASSOC)){ 
    //your code here; 
} 

Answer 2

在方括號內：

$contents = $db->prepare("SELECT * FROM details WHERE fileName = ?"); 
$contents->execute(array($yourFileName)); 

然後使用FETCH_ASSOC獲取結果$ yourFileName as {$ yourFileName}

Answer 3

$stmt = $db->prepare("SELECT * FROM details WHERE fileName = ?"); 
$stmt->execute(array($yourFileName)); 
$contents = $stmt->fetchAll(); 

其餘的都是一樣的