duanlu2935 2014-02-17 19:14
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使用PHP移动上传的文件[关闭]

Here is my file system.

---public_html
-------upload
-----------images(under upload)
-------uploadimage(under public_html)
-------other folder

all my html php js files are in the upload folder, i can upload image to the images folder. Is there a way i can upload the image to the uploadimage folder ( which is outside of upload folder)

here is my php code for the moving

if (file_exists("images/" . $_FILES["file"]["name"]))
  {
  echo $_FILES["file"]["name"] . " already exists. ";
  }
else
  {
  move_uploaded_file($_FILES["file"]["tmp_name"],
  "images/" . $_FILES["file"]["name"]);
  echo "Stored in: " . "images/" . $_FILES["file"]["name"];
  }

when i change images to uploadimage the server gives warning.

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1条回答 默认 最新

  • douyan4900 2014-02-17 19:28
    关注

    You can choose any folder you want, just as long as all of them have proper permissions set to them.

    Based on example #1 on PHP.net

    You will want to use: ../ depending on the file's execution location.

    I.e.: $uploads_dir = '../images/uploadimage';

    <?php
    $uploads_dir = '../uploadimage';
    //$uploads_dir = '../images/uploadimage';
    foreach ($_FILES["file"]["error"] as $key => $error) {
        if ($error == UPLOAD_ERR_OK) {
            $tmp_name = $_FILES["file"]["tmp_name"][$key];
            $name = $_FILES["file"]["name"][$key];
            move_uploaded_file($tmp_name, "$uploads_dir/$name");
        }
    }
    ?>
    

    Edit, after seeing OP's code

    (Code was not initially posted in original question)

    and as you edited, being under public_html you would do:

    $uploads_dir = '../uploadimage';
    

    So, in your case and using your posted code, you can use the following:

    if (file_exists("../uploadimage/" . $_FILES["file"]["name"]))
      {
      echo $_FILES["file"]["name"] . " already exists. ";
      }
    else
      {
      move_uploaded_file($_FILES["file"]["tmp_name"],
      "../uploadimage/" . $_FILES["file"]["name"]);
      echo "Stored in: " . "../uploadimage/" . $_FILES["file"]["name"];
      }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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