kunyomi 5 Newbie Poster

Hi guys, I'm trying to UPDATE the image that has been already in the database with another image using a form in PHP.

I got the uploading working all fine already. Just the edit part.

With this code, I get no error, but the MySQL table is not written with the new image name and the image does not get uploaded into the 'uploads' folder.

include ("dbConfig.php");
require ("check.php");

if($_GET["cmd"]=="edit" || $_POST["cmd"]=="edit")
   if (!isset($_POST["submit"]))
      $id2 = $_GET["id2"];
      $sql = "SELECT * FROM contacts WHERE id2=$id2";
      $result = mysql_query($sql);        
      $myrow = mysql_fetch_array($result);
      <form action="<?php $v=explode('?',$_SERVER['PHP_SELF']); echo $v[0]; ?>" method="post">
      <input type=hidden name="id2" value="<?php echo $myrow["id2"]; ?>">
      Name: <INPUT TYPE="text" NAME="name" VALUE="<?php echo $myrow["name"]; ?>" SIZE=30><br>
      Email: <INPUT TYPE="text" NAME="email" VALUE="<?php echo $myrow["email"]; ?>" SIZE=30><br>
      Who: <INPUT TYPE="text" NAME="age" VALUE="<?php echo $myrow["age"]; ?>" SIZE=30><br>
      Birthday: <INPUT TYPE="text" NAME="birthday" VALUE="<?php echo $myrow["birthday"]; ?>" SIZE=30><br>
      Address: <TEXTAREA NAME="address" ROWS=10 COLS=30><?php echo $myrow["address"]; ?></TEXTAREA><br>
      Number: <INPUT TYPE="text" NAME="number" VALUE="<?php echo $myrow["number"]; ?>" SIZE=30><br>
      Contact Image: <INPUT NAME="uploadedfile" VALUE="<?php echo $myrow["uploadedfile"]; ?>" TYPE="file" /><br>

      <input type="hidden" name="cmd" value="edit">
      <input type="submit" name="submit" value="submit">
$target = "C:/Program Files/xampp/htdocs/cas/uploads/";
$target = $target . basename( $_FILES['uploadedfile']['name']);

   if (isset($_POST['submit'])) {
   	  $id2 = mysql_real_escape_string(stripslashes($_POST["id2"]));
      $name = mysql_real_escape_string(stripslashes($_POST["name"]));
	  $email = mysql_real_escape_string(stripslashes($_POST["email"]));
	  $age = mysql_real_escape_string(stripslashes($_POST["age"]));
	  $birthday = mysql_real_escape_string(stripslashes($_POST["birthday"]));
	  $address = mysql_real_escape_string(stripslashes($_POST["address"]));
	  $number = mysql_real_escape_string(stripslashes($_POST["number"]));
	  $sql = "UPDATE contacts SET name='$name', email='$email', age='$age', birthday='$birthday', address='$address', number='$number', uploadedfile='$uploadedfile' WHERE id2=$id2";
      $result = mysql_query($sql) or die(mysql_error());
      echo "<div align=\"center\"><b>Thank you! Information updated.</b><br><br>";
	  echo "<a href=\"http://localhost/cas/contacts.php\">Contacts</a>"; echo " | ";
	  echo "<a href=\"http://localhost/cas/members.php\">Member Home</a>";echo " | ";
	  echo "<a href=\"http://localhost/cas/logout.php\">Logout</a></div>";echo "  ";
	  //check image upload status
	  if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target))
echo "Thank you, " . $_SESSION["valid_user"];echo ". Not you? <a href=\"http://localhost/cas/logout.php\">Click here.</a>";
echo "<br>";
//Tells you if its all ok
echo "<br>";

echo "The file <b>". basename( $_FILES['uploadedfile']['name']). "</b> has been uploaded.";
echo "</div>";

else {

//Gives and error if its not
echo "<div align=\"center\">";
echo "Sorry, there was a problem uploading your file.";
echo "</div>";
Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.