actually, i'm a newbie in php...i'd like to ask you guys.....how to display back all the data from the database for these coding....here's the coding....

($_POST ['action'] == 'pointofcontact'){
	
	$projectid = $_POST['projectid']; 
	
	
	$poc = '';

		//echo "Pastikan pilihan anda betul!!<br><br>";
		for ($i=1; $i <= 3; $i++){
		
			$poc = 'poc' . $i;
			
			$poc2 = $_POST[$poc];
			
			if (!empty($poc2)){
			$query = "insert into projectcontact set resourcesid = '$poc2', projectid = '$projectid'";
	
			$result = mysql_query($query);
			$projectid = $_POST['projectid'];
			}
			//echo $poc2."<br>";
			
		}
	
	if ($result){
		
		header ('Location: createproject.php?action=projectmanager&projectid='. $projectid);
		
		
		
	}


and number 2 is this :-

else if ($_POST ['action'] == 'uploaddownload'){

$projectid = $_POST['projectid']; 
$docname = $_POST['docname'];

$remarks =  $_POST ['remarks'];



$query = "insert into user_upload set docname= '$docname' ,  remarks = '$remarks' '";

$result = mysql_query ($query);


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 />"; // $_FILES["file"]["error"] - the error code resulting from the file upload
    }
  else
    {
    echo "Upload: " . $_FILES["file"]["name"] . "<br />"; // $_FILES["file"]["name"] - the name of the uploaded file
    echo "Type: " . $_FILES["file"]["type"] . "<br />"; // $_FILES["file"]["type"] - the type of the uploaded file
    echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />";//$_FILES["file"]["size"] - the size in bytes of the uploaded file
    echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />";// $_FILES["file"]["tmp_name"] - the name of the temporary copy of the file stored on the server
    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";
  }

}

i wantto display these two element....thanks in advance.....

Hi Asyieen,
I'm afraid the reason why nobody has answered so far is that it's not very clear what you actually want.

You posted a script that inserts records into projectcontact and user_upload.
And you say you want to display "these two elements". Word "element" is usually used for referencing a tag in a HTML document, while with databases we speak about tables, their records / rows and their attributes.

So what do you need? To list all records in a database table? Have a look at documentation for mysql_fetch_assoc() there's an example that does just that.

This article has been dead for over six months. Start a new discussion instead.