Question

这是我的编辑页面更新脚本。我有个问题。在每次更新时，脚本都会更新映像名称并插入MySQL。我希望该系统在文件名为NULL（isset）时不会插入或更新文件名。这是我的代码：

    <?php
require 'aed-config.php';
require 'class.upload.php';


function gen_random_string($length=16)
{
    $chars ="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz1234567890";//length:36
    $final_rand='';
    for($i=0;$i<$length; $i++)
    {
        $final_rand .= $chars[ rand(0,strlen($chars)-1)];

    }
    return $final_rand;
}
$pic_name = gen_random_string();

$image = new Upload( $_FILES[ 'image' ] );
    if ( $image->uploaded ) {

        $image->file_new_name_body = $pic_name;
        $image->image_convert = 'jpg';
        $image->image_resize = true;
        $image->image_ratio_crop = true;
        $image->image_x = 460;
        $image->image_y = 300;
        $image->Process( '../img' );

        $image->allowed = array ( 'image/*' );
    }


$pic = $pic_name.'.jpg';
$title = $_POST['title'];
$content = $_POST['content'];
$id = $_POST['memids'];

$sql = "UPDATE about_us 
        SET title=?, content=?, pic=?
        WHERE id=?";
$q = $db->prepare($sql);
$q->execute(array($title,$content,$pic,$id));
header("location: about.php");

?>

Answer 1

你可以这样做，比如使用简单的if条件

if($pic_name != null AND isset($_FILES['image'])){
    $q->execute(array($title,$content,$pic,$id));
}

Answer 2

如果你不需要，首先不要生成文件名并创建一个Upload对象。

if ($_FILES['image'])
{
    // Generate filename and create your Upload object as well as the $pic variable
}

然后是查询部分..

$executeArray = array($title, $content);
$sql = "UPDATE about_us
        SET title = ?, content = ?, ";

if (isset($pic)
{
    $executeArray[] = $pic;
    $sql .= "pic = ?, ";
}

$sql .= "WHERE id = ?";
$executeArray[] = $id;

$q->execute($executeArray);

PHP编辑，如果文件没有改变

2 个答案: