在php中上传文件时未定义的变量

Question

我想在我的数据库中存储2个图像.当我上传2张图片时,两者都成功存储,但在尝试上传单张图片时,则为未上传的图片提供未定义的错误.哪里我错了？

我的代码是:

         <label for="certificate">Upload Scaned Document:</label>
         <input type="file" id="uploadImage" name="image" />
         <label for="certificate">Upload Scaned QR Code</label>
         <input type="file" name="QRimage" id="File2" />

和PHP代码是

if((!empty($_FILES["image"])) && ($_FILES['image']['error'] == 0))   {
    $imageName = mysql_real_escape_string($_FILES["image"]["name"]);
    $imageData = mysql_real_escape_string(file_get_contents($_FILES["image"]["tmp_name"]));
    $imageType = mysql_real_escape_string($_FILES["image"]["type"]);

}
if((!empty($_FILES["QRimage"])) && ($_FILES['QRimage']['error'] == 0)) {
    $QRimageName = mysql_real_escape_string($_FILES["QRimage"]["name"]);
    $QRimageData = mysql_real_escape_string(file_get_contents($_FILES["QRimage"]["tmp_name"])); 
}

Answer 1

尝试这个

$imageName = "";
$imageData = "";
$QRimageName = "";
$QRimageData = "";
if(!empty($_FILES["image"]["name"])){ 
 $imageName = mysql_real_escape_string($_FILES["image"]["name"]);
 $imageData = mysql_real_escape_string(file_get_contents($_FILES["image"]["tmp_name"]));
 }
 if(!empty($_FILES["QRimage"]["name"])){
 $QRimageName = mysql_real_escape_string($_FILES["QRimage"]["name"]);
  $QRimageData = mysql_real_escape_string(file_get_contents($_FILES["QRimage"]["tmp_name"]));
     }