我有一个画廊,我希望用户能够更新图像标题并删除图像,如果他们愿意。我的更新标题代码按预期工作,我认为删除代码不会相差无几。
<?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);
}
}
出现错误是:警告:mysqli_stmt :: bind_param():变量数与预准备语句中的参数数不匹配
对我所做错误的任何指导?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 - fldName
?
任何帮助将不胜感激。
$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),因此两个变量太多了。
这应该成为
$stmt = $conn->prepare("DELETE FROM `tblImage` WHERE `tblImage`.`fldImageID` = ?");
$stmt->bind_param("s", $_POST['delete']) or die($stmt->error);
看来你只有一个参数?
但是你有"ss"
,删除其中一个并重新运行查询。