Image path not storing in sql database

Hello,
I have read here on the forum how to upload an image to server and store path in your database, the image uploads correctly to the correct folder on my server but the image path does not get stored on my sql database (not local hosting). I receive the error: The file has been uploaded, and your information has been added to the directory. Column 'image' cannot be null.
My database has the following columns:
id
datum
image
sectie
My code is as follows:
<?php require_once('Connections/dbTroch.php'); ?>
<?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "")
if (PHP_VERSION < 6) {
    $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
$theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);
switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? doubleval($theValue) : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
return $theValue;
mysql_select_db($database_dbTroch, $dbTroch);
$query_rs_aanbod = "SELECT * FROM tblSlideshow ORDER BY id ASC";
$rs_aanbod = mysql_query($query_rs_aanbod, $dbTroch) or die(mysql_error());
$row_rs_aanbod = mysql_fetch_assoc($rs_aanbod);
$totalRows_rs_aanbod = mysql_num_rows($rs_aanbod);
$editFormAction = $_SERVER['PHP_SELF'];
if (isset($_SERVER['QUERY_STRING'])) {
$editFormAction .= "?" . htmlentities($_SERVER['QUERY_STRING']);
$target = "images/slides/"; //This is the directory where images will be saved//
$target = $target . basename( $_FILES['image']['name']); //change the image and name to whatever your database fields are called//
if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "add-photos-aanbod")) {
$insertSQL = sprintf("INSERT INTO tblSlideshow (image, sectie) VALUES (%s, %s)",
                       GetSQLValueString($_POST['file'], "text"),
                       GetSQLValueString($_FILES['image']['name'], "text"));
//This code writes the photo to the server//
if(move_uploaded_file($_FILES['image']['tmp_name'], $target))
//And confirms it has worked//
echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory";
else {
//Gives error if not correct//
echo "Sorry, there was a problem uploading your file.";
mysql_select_db($database_dbTroch, $dbTroch);
$Result1 = mysql_query($insertSQL, $dbTroch) or die(mysql_error());
?>
<!doctype html>
<html>
<meta charset="utf-8" />
<meta name="viewport" content="width=device-width, initial-scale=1.0" />
<title>Troch Project Solutions - Admin - Toevoegen</title>
<link rel="stylesheet" href="css/foundation.css" />
<link rel="stylesheet" href="css/layout.css" />

<script type="text/javascript" src="//use.typekit.net/vob8gxg.js"></script>
<script type="text/javascript">try{Typekit.load();}catch(e){}</script>

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script src="js/vendor/modernizr.js"></script>
</head>
<body>
    <div class="row">
      <div class="large-8 medium-8 small-8 large-centered medium-centered small-centered columns intro">
          <h2 class="subheader text-center">Admin</h2>
          <p><a>Log uit</a></p>
          <p><a href="admin.php">Terug naar Admin menu</a>
          <h4>image toevoegen naar aanbod slideshows:</h4>
          <form action="<?php echo $row_rs_aanbod['']; ?>" method="POST" name="add-photos-aanbod" id="add-photos-aanbod" enctype="multipart/form-data">
          <table>
                <tbody>
                     <tr>
                          <td><label for="image">Kies foto:</label></td>
                          <td><input name="image" type="file" id="image" value="<?php echo $row_rs_aanbod['image']; ?>" /></td>
                    <tr>
                          <td>Sectie:</td>
                          <td><select name="sectie" id="sectie" option value="sectie">
                            <?php
do {
?>
                            <option value="<?php echo $row_rs_aanbod['sectie']?>"><?php echo $row_rs_aanbod['sectie']?></option>
                            <?php
} while ($row_rs_aanbod = mysql_fetch_assoc($rs_aanbod));
$rows = mysql_num_rows($rs_aanbod);
if($rows > 0) {
      mysql_data_seek($rs_aanbod, 0);
      $row_rs_aanbod = mysql_fetch_assoc($rs_aanbod);
?>
                          </select></td>
                    </tr>
                    <tr>
                        <td><input type="Submit" name="Add" id="add" value="Toevoegen" /></td>
                    </tr>
            </tbody>
        </table>
        <input type="hidden" name="MM_insert" value="add-photos-aanbod" />
    </form>
      </div>
    </div>
<script src="js/vendor/jquery.js"></script>
<script src="/js/vendor/fastclick.js"></script>
<script src="js/foundation.min.js"></script>
<script>
            $(document).foundation();
        </script>
</body>
</html>
<?php
mysql_free_result($rs_aanbod);
?>
I cannot work out what is wrong and I would appreciate any help on this. Thanks

Your form field and array variable names do not match
<td><input name="image" type="file" id="image" value="<?php echo $row_rs_aanbod['image']; ?>" /></td>
GetSQLValueString($_POST['file'], "text"),

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Now you can share the preview from Aperture with iPhoto, but at this point you basically have the adjusted image in two places.
And in reference to the other response that suggests that iPhoto and Aperture use different databases, I'd completely expect that to be the case. What I'm wondering about is can Aperture initialize its database from iPhoto's database; i.e., create its equivalent to iPhoto's "library" file (where all the iPhoto metadata is stored) from iPhoto's AND like iPhoto reference the original files in their original location.
Yes Aperture can initialize a database of its own based on the iPhoto library and that library can be setup to use referenced masters, see above.
I don't see why I'd need to 'export' any image from iPhoto (unless of course I'd "edited" it in iPhoto - in which case the *modified* file IS I think stored IN iPhoto's library) but that's my point... I don't WANT to edit the files with iPhoto... I want to edit the files with Aperture.
You're right you won't need to export from iPhoto if you setup both with referenced masters. But like I said above to get back to iPhoto you will have to export from Aperture unless you can live with using the previews in iPhoto. But again you will basically have two identical library structures, at least at start, as time goes on they will diverge.
I hope this better clarifies what I'm trying to explain.
It does, I hope my reply also helps explain the difficulties with this setup and why it's not as cut and dry as it first appears.
I have read *many* other posts that point out what you're pointing out but they all seem to make assumptions about wanting to modify files in Aperture which were previously *modified* in iPhoto. The unmodified "original" photos (assuming both iPhoto and Aperture have the ability that iPhoto has to reference the original files from a location *external* to iPhoto) are then by definition are already shared between iPhoto and Aperture so there's no need to 'export' from one to the other, no?
As I've said whats shared is the original unmodified master image. Once either program modifies it it is no longer shared.

Image path not storing in sql database

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