上传文件时没有文件名

Question

我在网上找到了一个从表单上传文件（图片）的代码。它工作得很好，但我希望它将文件名写入表格，就像表格中的其他数据一样。我正在使用此代码

           ?php
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file1"]["name"]);
$extension = end($temp);

if ((($_FILES["file1"]["type"] == "image/gif")
|| ($_FILES["file1"]["type"] == "image/jpeg")
|| ($_FILES["file1"]["type"] == "image/jpg")
|| ($_FILES["file1"]["type"] == "image/pjpeg")
|| ($_FILES["file1"]["type"] == "image/x-png")
|| ($_FILES["file1"]["type"] == "image/png"))
&& in_array($extension, $allowedExts)) {
  if ($_FILES["file"]["error"] > 0) {
    echo "Return Code: " . $_FILES["file1"]["error"] . "<br>";
  } else {
    echo "Upload: " . $_FILES["file1"]["name"] . "<br>";
    echo "Type: " . $_FILES["file1"]["type"] . "<br>";
    echo "Size: " . ($_FILES["file1"]["size"] / 1024) . " kB<br>";
    echo "Temp file: " . $_FILES["file1"]["tmp_name"] . "<br>";
    if (file_exists("upload/" . $_FILES["file1"]["name"])) {
      echo $_FILES["file1"]["name"] . " already exists. ";
    } else {
      copy($_FILES["file1"]["tmp_name"],
      "upload/" . $_FILES["file1"]["name"]);
      echo "Stored in: " . "upload/" . $_FILES["file1"]["tmp_name"]; 
    }
  }
} else {
  echo "Invalid file";
}
?>