Question

我想从数据库表中检索记录并从章节表中获取id，title，我想在select标签中显示章节名称。

当用户选择任何一章时，我想获得章节的id并通过表格发布。

我尝试添加select查询但是我没有得到如何在select标签中显示这个并获取选择章节的id。

     <?php
    ini_set('display_errors', 1);
error_reporting(1); 
ini_set('error_reporting', E_ALL);

$dbh = new PDO('mysql:host;dbname','', '');

    $stmt = $dbh->prepare("SELECT * FROM chapters");
    $stmt->execute();
    $stmt->fetchAll(PDO::FETCH_ASSOC);
?>


<!DOCTYPE html>
<html>
    <head>
        <body>


    <form action="fileUpload.php" method="post" enctype="multipart/form-data">
    <p> Select image to upload:</p>
    <input name = "file" type="file" id="fileToUpload"><br><br>

    <input type="submit" value = "Upload Image">

    </form>

        </body>

    </head>

</html>

有人可以帮忙吗？谢谢..

编辑：

    <!DOCTYPE html>
<html>
    <head>
        <body>
    <?php
         <select>

          foreach ($stmt as $row)
        {
            <option value=$row['chapterName']>$row['chapterName']</option>
        }

            </select>
    ?>

Answer 1

首先，head加价无效，body之后您无法关闭<!DOCTYPE html> <html> <head> <title>This is a title </title> META TAGs </head> <body BODY CONTENT </body> </html>标记这就是html标记的样子：

<?php
    ini_set('display_errors', 1);
    error_reporting(1);
    ini_set('error_reporting', E_ALL);

    $dbh = new PDO('mysql:host=host;dbname=db', 'airman', 'pass');
    ?>
<!DOCTYPE html>
<html>
    <head> </head>
    <body>
        <form action="fileUpload.php" method="post" enctype="multipart/form-data">
            <p> Select image to upload:</p>
            <input name = "file" type="file" id="fileToUpload"><br><br>
            <input type="submit" value = "Upload Image">
            <select name="chapters">
                <?php
                    $stmt = $dbh->query("SELECT * FROM chapters")->fetchAll(FETCH_ASSOC);
                    foreach($stmt as $row):?>
                <option value="<?php echo $row['chapterid'];?>"><?php echo $row['chapterName'];?></option>
                <?php
                    endforeach;?>
            </select>
        </form>
    </body>
</html>

现在这就是你的代码的样子：

  foreach ($stmt as $row)
        {
            <option value=$row['chapterName']>$row['chapterName']</option>
        }

你在这里得到500错误的原因：

echo

它因为你混合php和html错误。它要么使用<?php ini_set('display_errors', 1); error_reporting(1); ini_set('error_reporting', E_ALL); $dbh = new PDO('mysql:host=host;dbname=db', 'airman', 'pass'); ?> <!DOCTYPE html> <html> <head> </head> <body> <form action="fileUpload.php" method="post" enctype="multipart/form-data"> <p> Select image to upload:</p> <input name = "file" type="file" id="fileToUpload"><br><br> <input type="submit" value = "Upload Image"> <select name="chapters"> <?php $stmt = $dbh->query("SELECT * FROM chapters"); while($row = $stmt->fetchall(FETCH_ASSOC)):?> <option value="<?php echo $row['chapterid'];?>"><?php echo $row['chapterName'];?></option> <?php endwhile;?> </select> </form> </body> </html>

回显html标签

或：

<?php
    ini_set('display_errors', 1);
    error_reporting(1);
    ini_set('error_reporting', E_ALL);

    $dbh = new PDO('mysql:host=host;dbname=db', 'airman', 'pass');
    ?>
<!DOCTYPE html>
<html>
    <head> </head>
    <body>
        <form action="fileUpload.php" method="post" enctype="multipart/form-data">
            <p> Select image to upload:</p>
            <input name = "file" type="file" id="fileToUpload"><br><br>
            <input type="submit" value = "Upload Image">
            <select name="chapters">
                <?php
                    $stmt = $dbh->prepare("SELECT * FROM chapters");
                    $stmt->execute();
                    $results = $stmt->fetchall(PDO::FETCH_ASSOC);

                    if(count($results > 0)){
                        foreach($results as $row):?>
                         <option value="<?php echo $row['chapterid'];?>"><?php echo $row['chapterName'];?></option>
                    <?php
                        endforeach;
                    }else{?>

                        <option value="0">No data found</option>
                    <?php}?>


            </select>
        </form>
    </body>
</html>

或

{{1}}

如何从数据库中获取记录并以html格式显示？

1 个答案: