PDO选择语句

Question

希望有人可以给我一些帮助...我使用PDO从数据库中获取一些数据，但每次脚本运行时都会返回并且没有错误，但不会显示所需的数据，我知道我正在寻找的数据也在那里..任何帮助将不胜感激，这是我目前的，谢谢。

$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

准备语句时使用的占位符，然后执行查询传递参数，如：

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

Answer 2

在方括号内：

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

然后使用FETCH_ASSOC获取结果$ yourFileName as {$ yourFileName}

Answer 3

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

其余的都是一样的