hy im using this script but unf cant find the bug related to update the values, its keep telling succesuful update but dsnt update in db...Click Here

You have to debug ajax calls which is easier with tools like Firebug. But anyway, I think you should check the UPDATE query from the updateMemberAjx() function. You can try to display the query instead of the success message by assigning the query to a variable in each case statement like:

case 'first_name':
    $GLOBALS['MySQL']->res("UPDATE `pd_profiles` SET `first_name`='{$sVal}' WHERE `id`='{$iId}'");
    // assign the query to a temp variable
    $temp = "UPDATE `pd_profiles` SET `first_name`='{$sVal}' WHERE `id`='{$iId}'"

and changing line 098 from:

echo 'Successfully saved';


echo $temp;

This way the query will be diplayed upon an update. You can copy it and test in phpmyadmin whether it contains any errors.

Fiddler is also a great tool for debugging ajax calls.

commented: fiddler is the best for me also :) +5

broj1, it show me blank popup, ive tryied con fiddler but i'm nothing see that wouldn be...

You have to add echo statement to every case instance in the function, something like:

function updateMemberAjx() {
    $sVal = $GLOBALS['MySQL']->escape($_POST['value']);

    $iId = (int)$_POST['id'];
    if ($iId && $sVal !== FALSE) {
        switch ($_POST['columnName']) {
            case 'first_name':
                $GLOBALS['MySQL']->res("UPDATE `pd_profiles` SET `first_name`='{$sVal}' WHERE `id`='{$iId}'");
                $temp = "UPDATE `pd_profiles` SET `first_name`='{$sVal}' WHERE `id`='{$iId}'";
            case 'last_name':
                $GLOBALS['MySQL']->res("UPDATE `pd_profiles` SET `last_name`='{$sVal}' WHERE `id`='{$iId}'");
                $temp = "UPDATE `pd_profiles` SET `last_name`='{$sVal}' WHERE `id`='{$iId}'";
            case 'email':
                $GLOBALS['MySQL']->res("UPDATE `pd_profiles` SET `email`='{$sVal}' WHERE `id`='{$iId}'");
                $temp = "UPDATE `pd_profiles` SET `email`='{$sVal}' WHERE `id`='{$iId}'";
            case 'status':
                $GLOBALS['MySQL']->res("UPDATE `pd_profiles` SET `status`='{$sVal}' WHERE `id`='{$iId}'");
                $temp = "UPDATE `pd_profiles` SET `status`='{$sVal}' WHERE `id`='{$iId}'";
            case 'role':
                $GLOBALS['MySQL']->res("UPDATE `pd_profiles` SET `role`='{$sVal}' WHERE `id`='{$iId}'");
                $temp = "UPDATE `pd_profiles` SET `role`='{$sVal}' WHERE `id`='{$iId}'";
            case 'date_reg':
                $GLOBALS['MySQL']->res("UPDATE `pd_profiles` SET `date_reg`='{$sVal}' WHERE `id`='{$iId}'");
                $temp = "UPDATE `pd_profiles` SET `date_reg`='{$sVal}' WHERE `id`='{$iId}'";
        // echo 'Successfully saved';
        echo 'DEBUG: ' . $temp;

Please note that this is very simple and a bit clumsy way of debugging. Using right developer tools is the way to go.

Its write DEBUG and blank space...

solved the problem was that the name of case wsnt equale that title of the collums, but i have another question in db i have a row that is a path of a image
how ca i modify this to show the image not the path(like in db)

  $aItem = array(
       $aInfo['photo'],   $aInfo['prodoto'], $aInfo['diametro'], $aInfo['piante_pianali'], $aInfo['piante_cc'], $aInfo['l_a'], $aInfo['l_c'],  $aInfo['l_ing'], $aInfo['flag'], 'DT_RowId' => $aInfo['id']

its the $aInfo photo row....ive tryied with img src but nothing...

If the path is in the $aInfo['photo'] row then the code would be:

echo '<img src="' . $aInfo['photo'] ">'

But you have to make sure the path is correct. Maybe you have to add something to make the path absolute (e.g. http://yoursite.com/images) or be sure that relative path is correct.

Maybe you could post sample database rows.

    foreach ($aMembers as $iID => $aInfo) {
        $aItem = array(

      $aInfo['photo'],   $aInfo['prodoto'], $aInfo['diametro'], $aInfo['piante_pianali'], $aInfo['piante_cc'], $aInfo['l_a'], $aInfo['l_c'],  $aInfo['l_ing'], $aInfo['flag'], 'DT_RowId' => $aInfo['id']
        $output['aaData'][] = $aItem;
    echo json_encode($output);

i need to integrated in this array....$aInfo[photo] it is the row of photo

Sory, but I do not understand the problem. Could you describe it form start. What you have and what you want to achieve. And post relevant code as well as some relevant sample data.

Hy thank your very much, i've solved the problem adding <img src to the photo row, now i'm trying to debug this because its not working its keep telling me not send, i dont have a clue why didnt send, ut the .csv its export
link here Click Here