2

我在带有文本输入的表单上有一个图像上传按钮,但设计用于检查、调整大小和上传文件的脚本仅用于一次上传。

有没有办法可以使用带有以下代码的 foreach() 将每个文件提供给该脚本?

if (($_FILES['file']['type'] != 'image/jpeg') && ($_FILES['file']['type'] != 'image/jpg') && ($_FILES['file']['type'] != 'image/pjpeg')) {
            echo "<script>alert('Images Must be in jpg format and under 2 Mb');\n";
            echo sprintf("window.location='add.php'");
            echo "</script>";
            include('no-ad-footer.php');
                } else {
                chdir('admin/photos'); 
                require_once('upload.php');
                chdir('../../');

        if ($up->ValidateUpload()) { 
                $node = new sqlNode();
                $node->table = "photos";
                $node->push("int","TypeID",$_POST['TypeID']);
                $node->push("int","ListingID",$ListingID);          
                $node->push("text","Location",$new_name);           
                $node->push("int","POrder",$_POST['POrder']);
                $node->push("defined","DateofUpload","NOW()");

                if(($result = $mysql->insert($node)) === false)
                    die('Unable to push photo details into table photos line 38');          
            } else {
                echo "<font color='red'>Unable to upload image</font>";
                }
        }
4

1 回答 1

0

没有尝试过,但这应该可以:

$ftypelist = array('image/jpeg', 'image/jpg', 'image/pjpeg');

foreach($_FILES as $file){
    if(in_array($file[type], $ftypelist){
        echo "<script>alert('Images Must be in jpg format and under 2 Mb');\n";
        echo sprintf("window.location='add.php'");
        echo "</script>";
        include('no-ad-footer.php');
    } else {
        chdir('admin/photos'); 
        require_once('upload.php');
        chdir('../../');

        if ($up->ValidateUpload()) { 
            $node = new sqlNode();
            $node->table = "photos";
            $node->push("int","TypeID",$_POST['TypeID']);
            $node->push("int","ListingID",$ListingID);          
            $node->push("text","Location",$new_name);           
            $node->push("int","POrder",$_POST['POrder']);
            $node->push("defined","DateofUpload","NOW()");

            if(($result = $mysql->insert($node)) === false)
                die('Unable to push photo details into table photos line 38');          
        } else {
            echo "<font color='red'>Unable to upload image</font>";
        }
    }
}
于 2013-04-02T13:52:29.710 回答