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How to see the recently uploaded file or image from the remote server


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Hello

 

I have a very small qustion .

jst have a look on this link>>http://ice.host-care.com/~com/n/upload.html

and now my quistion is tht how can i able to see the recently uploaded image automatically

 

here is the code which i used for this and im not getting the out put which i want

 

<HTML>

<HEAD>

<TITLE>upload</TITLE> <META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=iso-8859-1">

</HEAD>

 

<BODY BGCOLOR="#FFFFFF" TEXT="#000000">

<H1> </H1><TITLE></TITLE><?php

if (($_FILES["file"]["type"] == "image/gif")

|| ($_FILES["file"]["type"] == "image/jpeg")

|| ($_FILES["file"]["type"] == "image/pjpeg")

&& ($_FILES["file"]["size"] < 20000))

  {

  if ($_FILES["file"]["error"] > 0)

    {

    echo "Return Code: " . $_FILES["file"]["error"] . "<br />";

    }

  else

    {

    echo "Upload: " . $_FILES["file"]["name"] . "<br />";

    echo "Type: " . $_FILES["file"]["type"] . "<br />";

    echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />";

    echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />";    if (file_exists("upload/" . $_FILES["file"]["name"]))

      {

      echo $_FILES["file"]["name"] . " already exists. ";

      }

    else

      {

      move_uploaded_file($_FILES["file"]["tmp_name"],

      "upload/" . $_FILES["file"]["name"]);

      echo "Stored in: " . "upload/" . $_FILES["file"]["name"];

      }

    }

  }

else

  {

  echo "Invalid file";

  }

?>

</BODY>

</HTML>

 

------------------------------

 

Friends did you get what i mean and please try to help me for this asap.

Naveen

 

Re:

Sir,

Im trying to get the image link instantly  on the next page which should comes after uploading

did you get what i mean??

Naveen

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