Php 如何将两个图像取消链接在一起？

在这个函数下面，我取消了一个图像的链接。现在我想在这个函数中取消一个图像的链接，比如“imagetwo”

如何在此函数中为取消链接imagetwo编写代码

public function nameDeleteById($data){

    $delete_id = $data['delete_id'];

    $stmt = $this->pdo->prepare("SELECT * FROM name WHERE id=:delete_id");
    $stmt->bindparam(":delete_id", $delete_id);
    $stmt->execute();

    if ($stmt) {
        while ($delimg=$stmt->fetch(PDO::FETCH_ASSOC)) {
            $dellink=$delimg['image'];
            unlink($dellink);
        }
    }
    $stmt=$this->pdo->prepare("DELETE FROM name WHERE id=:delete_id");
    $stmt->bindparam(":delete_id", $delete_id);
    $stmt->execute();
    if ($stmt) {
        $msg = 'Name Deleted Secessfully';
        return $msg;
    } else {
        $msg = 'Name Not Deleted Secessfully';
        return $msg;
    }
}

---

从您的问题中不清楚，

imagetwo

到底是什么。我假设它是

name

表（？）中的另一列。如果是这样，您可以像这样增强您的功能：

public function nameDeleteById($data){

    $delete_id = $data['delete_id'];

    $stmt = $this->pdo->prepare("SELECT * FROM name WHERE id=:delete_id");
    $stmt->bindparam(":delete_id", $delete_id);
    $stmt->execute();

    if ($stmt) {
        while ($delimg=$stmt->fetch(PDO::FETCH_ASSOC)) {
            unlink($delimg['image']);
            unlink($delimg['imagetwo']);
        }
    }
    $stmt=$this->pdo->prepare("DELETE FROM name WHERE id=:delete_id");
    $stmt->bindparam(":delete_id", $delete_id);
    $stmt->execute();
    if ($stmt) {
        $msg = 'Name Deleted Successfully';
        return $msg;
    } else {
        $msg = 'Name Not Deleted Successfully';
        return $msg;
    }
}