php - 发布表单并使用 mysql 更新多行

Question

我搜索了整个网站，但还没有找到答案。公关也许我无法正确应用它。我有一个表格可以抓取所有附有特定 GALLERY_id 的照片。后端用户然后可以更改照片的标题并更改标签。提交表单后，查询应更新所有行。这是我到目前为止没有做任何事情的内容:

表格

if(isset($_GET['id']))
{

    $id=$_GET['id'];
    $result = $db->prepare("SELECT * FROM photos WHERE gallery_id = :gallery_id ");      
    $result->bindParam(':gallery_id', $id);
    $result->execute();

    echo '<form action="" method="POST">';
    echo "<ul id='photos'>";

    for ($i = 0; $row = $result->fetch(); $i++)
    {

        $id = $row['id'];
        $title = $row['title'];
        $tags = $row['tags'];
        $src = $row['src'];

        echo "<li><a class='lightbox' href='images/$src'><img src='images/$src' id='$id' alt='$title' /></a><br />";
        echo "<input type='text' name='photo_title' value='$title' /><br />";
        echo "<input type='text' name='photo_tags' value='$tags' />";
        echo "<input type='hidden' name='photo_id' value='$id' />";
        echo "</li>";
    }

    echo "</ul>";

}

?>
<div style="clear:both"></div>
<input type="submit" name="changeTitle" value="Save"/>
</form>

更新查询

if (isset($_POST['changeTitle']))
{
    foreach ($_POST as $p)
    {
        $id=$p['photo_id'];
        $title=$p['photo_title'];
        $tags=$p['photo_tags'];


        $sql = "UPDATE photos SET title=?, tags=? WHERE id=?";
        $query = $db->prepare($sql);
        $query->execute(array($title, $tags, $id));

    }
}

Answer 1

由于您有多个具有相同名称的 html 字段，您必须将它们作为数组提交:

echo "<input type='text' name='photo_title[]' value='$title' /><br />";
echo "<input type='text' name='photo_tags[]' value='$tags' />";
echo "<input type='hidden' name='photo_id[]' value='$id' />";

提交后，遍历任何数组变量，如

foreach ($_POST['photo_id'] as $key => $photo_id) {
    $id = $photo_id;
    $title = $_POST['photo_title'][$key];
    $tags = $_POST['photo_tags'][$key];

    $sql = "UPDATE photos SET title=?, tags=? WHERE id=?";
    $query = $db->prepare($sql);
    $query->execute(array($title, $tags, $id));
}