Question

我有两张想要上传到服务器的图片。如果两个文件都在html中设置，那么第一个查询应该执行，如果选择了image1而image2不是第二个查询shoud执行，如果未选择image1并且选择了inage2，那么如果没有选择图像则应该执行第三个查询在image1和image2中，然后执行查询4。

请注意，我还有三个文本框，无论是否选择了image1或image2，都会更新。

我遇到的问题是

当我在image1 nad image2中没有选择任何文件时，第一个代码通过插入图像1和图像2的情感值来执行，这是nto的意思。

是否有更清洁，更有效的方法来减少代码重复。

if (isset($_FILES['newsthumb']) && isset($_FILES['newsmain'])) {

  $title            = mysql_real_escape_string(trim($_POST['title']));
  $body             = mysql_real_escape_string(trim($_POST['body']));
  $mainimage_title  = mysql_real_escape_string(trim($_POST['mainimage_title']));
  $thumbimage = $_FILES['newsthumb'];
  $mainimage  = $_FILES['newsmain'];

$unique = time();

$thumbname  = strtolower($thumbimage['name']); 
$mainmane   = strtolower($mainimage['name']);

 $thumbname  = preg_replace("/[^A-Z0-9._-]/i", "_", $thumbname);
 $mainmane   = preg_replace("/[^A-Z0-9._-]/i", "_", $mainmane);

 $thumbname = mysql_real_escape_string($thumbname);
 $mainmane  = mysql_real_escape_string($mainmane);

 $uploaddir = "images/newsimage/";

 $thumbsuccess = move_uploaded_file($thumbimage["tmp_name"], $uploaddir.$thumbname);
 $mainsuccess  = move_uploaded_file($mainimage["tmp_name"], $uploaddir.$mainmane);

 $newsarticle = "UPDATE news 
              SET title = '$title', 
                  body = '$body', 
                  mainimage_title = '$mainimage_title' , 
                  thumbnail = '$thumbname', 
                  mainimage = '$mainmane', 
                  editdate = NOW()
                  WHERE id = '$article_id'"; 
              mysql_query($newsarticle) or die (mysql_error());
            }

  if (!isset($_FILES['newsthumb']) && isset($_FILES['newsmain'])) {

  $title            = mysql_real_escape_string(trim($_POST['title']));
  $body             = mysql_real_escape_string(trim($_POST['body']));
  $mainimage_title  = mysql_real_escape_string(trim($_POST['mainimage_title']));
// Commented out since it is not defined anywhere in your original posted code.
// You will have to implement that in.
  $mainimage  = $_FILES['newsmain'];
  $unique = time();
  $mainmane   = strtolower($mainimage['name']);
  $mainmane   = preg_replace("/[^A-Z0-9._-]/i", "_", $mainmane);
  $mainmane  = mysql_real_escape_string($mainmane);
  $uploaddir = "images/newsimage/";
  $mainsuccess  = move_uploaded_file($mainimage["tmp_name"], $uploaddir.$mainmane);
  $newsarticle = "UPDATE news 
                  SET title = '$title', 
                      body = '$body', 
                      mainimage_title = '$mainimage_title' , 
                      mainimage = '$mainmane', 
                      editdate = NOW()
                      WHERE id = '$article_id'"; 
                  mysql_query($newsarticle) or die (mysql_error());
                }

if (isset($_FILES['newsthumb']) && (!isset($_FILES['newsmain']))) {

  $title            = mysql_real_escape_string(trim($_POST['title']));
  $body             = mysql_real_escape_string(trim($_POST['body']));
  $mainimage_title  = mysql_real_escape_string(trim($_POST['mainimage_title']));
// Commented out since it is not defined anywhere in your original posted code.
// You will have to implement that in.
  $thumbimage = $_FILES['newsthumb'];
  $unique = time();
  $thumbname  = strtolower($thumbimage['name']); 
  $thumbname  = preg_replace("/[^A-Z0-9._-]/i", "_", $thumbname);
  $thumbname = mysql_real_escape_string($thumbname);
  $uploaddir = "images/newsimage/";
  $thumbsuccess = move_uploaded_file($thumbimage["tmp_name"], $uploaddir.$thumbname);


  $newsarticle = "UPDATE news 
                  SET title = '$title', 
                      body = '$body', 
                      mainimage_title = '$mainimage_title' , 
                      thumbnail = '$thumbname', 
                      editdate = NOW()
                      WHERE id = '$article_id'"; 
                  mysql_query($newsarticle) or die (mysql_error());
                }

if (!isset($_FILES['newsthumb']) && (!isset($_FILES['newsmain']))) {

  $title            = mysql_real_escape_string(trim($_POST['title']));
  $body             = mysql_real_escape_string(trim($_POST['body']));
  $mainimage_title  = mysql_real_escape_string(trim($_POST['mainimage_title']));    
  $newsarticle = "UPDATE news 
                  SET title = '$title', 
                      body = '$body', 
                      mainimage_title = '$mainimage_title' , 
                      editdate = NOW()
                      WHERE id = '$article_id'"; 
                  mysql_query($newsarticle) or die (mysql_error());
                }

Answer 1

也许通过创建一个函数来获取参数和他的东西中的文件。没有重复会有所帮助。

用于检查选择的文件或用于：

if ($_FILES["file"]["error"] > 0)

更多解释：

function uploadASingleFile($file){ 
// Do your stuff here for the upload (moving the file, sql request etc.) 
}

然后，检查文件是否被选中（使用[“错误”]）并调用函数

if($_FILES["file1"]["error"] == 0){
uploadASingleFile($_FILES["file1"]); // Obviously, handle errors...
}

if($_FILES["file2"]["error"] == 0){
uploadASingleFile($_FILES["file2"]); // Obviously, handle errors...
}

代码不那么重复和清晰。我回答了你的问题吗？

同时在php上载两个图像

1 个答案: