我目前能够上传多个图像,然后将文件路径插入数据库中的一行。理想情况下,我需要找到一种方法,将这些文件路径作为带有自己ID的单独条目上传。原因是我插入的图像路径要绑定到插入到单独表中的任务。
function upload_file_new_task(){
global $db;
if(isset($_POST['create'])) {
$path = "../uploads/";
for ($i=0; $i<count($_FILES['files']['name']); $i++) {
$ext = explode('.', basename( $_FILES['files']['name'][$i]));
$path = $path . md5(uniqid()) . "." . $ext[count($ext)-1];
move_uploaded_file($_FILES['files']['tmp_name'][$i], $path);
}
$sql = "INSERT INTO upload_data (`image`) VALUES ('$path');";
$res = mysqli_query($db,$sql);
echo "<p>Post Created $date</p>";
}
}
因此,图像被上传到/uploads文件夹中,然后路径作为一行加载到数据库中,id在ID
字段中,路径被加载到image
字段中。
类似的东西
function upload_file_new_task()
{
global $db;
if(isset($_POST['create']))
{
$path = "../uploads/";
for ($i=0; $i<count($_FILES['files']['name']); $i++)
{
$ext = pathinfo($_FILES['files']['name'][$i], PATHINFO_EXTENSION);
$path1 = $path . md5(uniqid()) . "." . $ext;
move_uploaded_file($_FILES['files']['tmp_name'][$i], $path1);
$sql = "INSERT INTO upload_data (`image`) VALUES ('$path1');";
$res = mysqli_query($db,$sql);
}
echo "<p>Post Created $date</p>";
}
}