当点击删除按钮时，需要从 table 中删除图像，而是显示错误消息

Question

我有一个画廊，我希望用户可以在其中更新图像标题并根据需要删除图像。我的更新标题代码按预期工作，我假设删除代码不会太远。

<?php
    $query = "SELECT * FROM `tblImage`";

    $result = $conn -> query($query);

    while($row = $result -> fetch_assoc())
    {
?>

<form method="post" action="">
    <a href="<?php echo $row['fldFilePath']; ?>" data-lightbox="gallery" data-title="<?php echo $row['fldName']; ?>"><img src="<?php echo $row['fldFilePath']; ?>" class="ImgRound"></a>
    <label>Image Name: <?php echo $row['fldName']; ?></label>
    <input name ="img-title" type ="text" placeholder="Enter New Image Title...">
    <button type="submit" value ="<?php echo $row['fldName'] ?>" name="update_title" class="ImgRound size-btn">Update</button>
    <button type="submit" value ="<?php echo $row['fldName'] ?>" name="delete" class="ImgRound size-btn">Delete</button>
</form>
<?php
    if(isset($_POST['update_title']))
        {
                $imgTitle = $_POST['img-title'];

    $stmt = $conn->prepare("UPDATE `tblImage` SET `fldName` = ? WHERE `fldName` = ?");
    $stmt->bind_param("ss", $imgTitle, $_POST['update_title']) or die($stmt->error);
        if ($stmt->execute())
            {
                header("Refresh:0");
            }   
        else {

                die($stmt->error);
             }

         }
    }
?>

只要我在最后一次关闭之前添加以下代码 } :

if(isset($_POST['delete']))
    {
        $imgTitle = $_POST['img-title'];
        $stmt = $conn->prepare("DELETE FROM `tblImage` WHERE `tblImage`.`fldName` = ?"); 
        $stmt->bind_param("ss", $imgTitle, $_POST['delete']) or die($stmt->error);
        if ($stmt->execute())
            {
                header("Refresh:0");
            }   
        else {
                die($stmt->error);
            }
    }

出现的错误是：警告：mysqli_stmt::bind_param()：变量数与准备语句中的参数数不匹配

对我做错了什么有什么指导吗？也许是查询，或者我应该将值放在文件路径的删除按钮中作为图像名称的对应 - fldName?

如有任何帮助，我们将不胜感激。

Answer 1

$stmt = $conn->prepare("DELETE FROM `tblImage` WHERE `tblImage`.`fldImageID` = ?"); 
$stmt->bind_param("ss", $imgTitle, $_POST['delete']) or die($stmt->error);

你有一个问号（参数），所以两个变量太多了。

这应该变成

$stmt = $conn->prepare("DELETE FROM `tblImage` WHERE `tblImage`.`fldImageID` = ?"); 
$stmt->bind_param("s", $_POST['delete']) or die($stmt->error);

Answer 2

您似乎只有一个参数 ?，但您有 "ss"，请删除其中一个并重新运行查询。

当点击删除按钮时，需要从 table 中删除图像，而是显示错误消息

When Delete button hit, image needs to be deleted from table, error message showing instead

php

mysql

sql

prepared-statement