希望有人可以给我一些帮助......我正在使用PDO从数据库中获取一些数据,但每次脚本运行时它都会返回,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..任何帮助都会受到赞赏,这就是我现在所拥有的,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>";
答案 0 :(得分:0)
$stmt = $db->prepare("SELECT * FROM details WHERE fileName = ?");
$stmt->execute(array($yourFileName));
$contents = $stmt->fetchAll();
其余的是相同的
答案 1 :(得分:-1)
在准备语句时使用占位符,然后执行传递参数的查询,如下所示:
$contents = $db->prepare("SELECT * FROM details WHERE fileName = ?");
$contents->execute(array($yourFileName));
然后使用FETCH_ASSOC获取结果:
while($row = $contents->fetch(PDO::FETCH_ASSOC)){
//your code here;
}
答案 2 :(得分:-1)
将$ yourFileName括号括起来{$ yourFileName}