Question

我想从数据库中删除一行，具体取决于图像的文件名。可以说我下面有两张表：

Image Table:

ImageId  ImageFile

01       cat.png
02       dog.png
03       dog_2.png

Image_Question Table:

ImageId   SessionId   QuestionId
01        AAA              4
02        ABD              1
03        RTD              11

让我们在我的应用程序中说删除文件图像dog_2.png，然后我希望它删除图像表中状态为dog_2.png的行（这很正常）并且能够从Image_Question中删除该行表包含与Image Table中的ImageId和ImageFile名称相关联的相同ImageId的表（这不起作用）。

因此，对于上面的示例，删除后的2个表现在看起来像这样：

Image Table:

ImageId  ImageFile

01       cat.png
02       dog.png

Image_Question Table:

ImageId   SessionId   QuestionId
01        AAA              4
02        ABD              1

但它不会从Image_Question表中删除该行，如何才能删除此行？

下面是从图像表中删除行的完整代码，它包含从Image_Question表中删除行时已设置但尚未完全完成的大部分代码：

$image_file_name = $_GET["imagefilename"];
$img = "ImageFiles/$image_file_name";

echo "$image_file_name was Deleted";
unlink("ImageFiles/$image_file_name");

$imagedeletesql = "DELETE FROM Image WHERE ImageFile = ?";

if (!$delete = $mysqli->prepare($imagedeletesql)) {
    // Handle errors with prepare operation here
}

//Don't pass data directly to bind_param; store it in a variable
$delete->bind_param("s",$img);

$delete->execute();

if ($delete->errno) {
    // Handle query error here
}

$delete->close();

$imagequestiondeletesql = "DELETE FROM Image_Question WHERE ImageId = ?";

if (!$deleteimagequestion = $mysqli->prepare($imagequestiondeletesql)) {
    // Handle errors with prepare operation here
}

// Don't pass data directly to bind_param; store it in a variable
$deleteimagequestion->bind_param("s",....);

$deleteimagequestion->execute();

if ($deleteimagequestion->errno) {
    // Handle query error here
}

$deleteimagequestion->close();

Answer 1

您可以使用JOIN查询从单个查询中的多个表中删除。我认为这种陈述对你有用：

$sql = "
DELETE img, img_q
FROM Image AS img
LEFT JOIN Image_Question AS img_q
    ON img_q.ImageId = img.ImageId
WHERE img.ImageFile = ?";

不删除第二个表中的行

1 个答案: