Question

我的方法是创建三个输入文件，将其发布到数据库表中的三个不同列。

继承我的参数：

$additional_comments=$_POST['additional_comments']; 
$image_one = $_FILES['image_one']['type'];
$image_two = $_FILES['image_two']['type'];
$image_three = $_FILES['image_three']['type'];

    if($image=="image/jpeg" || $image=="image/jpg" || $image=="image/gif" || $image=="image/png") 
        {           
            $gambar_satu        = $foldername . basename($_FILES['image_one']['name']);  
            $gambar_dua         = $foldername . basename($_FILES['image_two']['name']);  
            $gambar_tiga        = $foldername . basename($_FILES['image_three']['name']);  

                if  (   move_uploaded_file 
                            ($_FILES['image']['tmp_name'], $gambar_satu) 
                            ($_FILES['image']['tmp_name'], $gambar_dua) 
                            ($_FILES['image']['tmp_name'], $gambar_tiga) 
                    ) 

                    {
                        $stmt = $mysqli->prepare
                            ("INSERT INTO table...

在if (move_uploaded_file...我试图插入三个参数，它们是$gambar_satu，$gambar_dua，$gambar_tiga。

我的问题是这样做的正确方法是怎样的？提前致谢。

Answer 1

直接从php网站引用

<?php
$uploads_dir = '/uploads';
foreach ($_FILES["pictures"]["error"] as $key => $error) {
if ($error == UPLOAD_ERR_OK) {
    $tmp_name = $_FILES["pictures"]["tmp_name"][$key];
    $name = $_FILES["pictures"]["name"][$key];
    move_uploaded_file($tmp_name, "$uploads_dir/$name");
 }
}
?>

希望它有所帮助！

move_uploaded_file用于不同的输入文件

1 个答案: